Consider the three connected objects shown in Figure P5.88. Assume first that the inclined plane is friction-less and that the system is in equilibrium. In terms of m , g , and θ , find (a) the mass M and (b) the tensions T, and T2. Now assume that the value of Af is double the value found in part (a). Find (c) the acceleration of each object and (d) the tensions T 1 and T 2 . Next, assume that the coefficient of static friction between m and 2m and the inclined plane is m, and that the system is in equilibrium. Find (e) the maximum value of M and (0 the minimum value of M . (g) Compare the values of T 2 when M has its minimum and maximum values.

Question

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Answer 1

Textbook Question

Chapter 5, Problem 5.88AP

Consider the three connected objects shown in Figure P5.88. Assume first that the inclined plane is friction-less and that the system is in equilibrium. In terms of m, g, and θ, find (a) the mass M and (b) the tensions T, and T2. Now assume that the value of Af is double the value found in part (a). Find (c) the acceleration of each object and (d) the tensions T₁ and T₂. Next, assume that the coefficient of static friction between m and 2m and the inclined plane is m, and that the system is in equilibrium. Find (e) the maximum value of M and (0 the minimum value of M. (g) Compare the values of T₂ when M has its minimum and maximum values.

Chapter 5, Problem 5.88AP, Consider the three connected objects shown in Figure P5.88. Assume first that the inclined plane is

(a)

Expert Solution

To determine

The expression for the mass of $M$ of the object.

Answer to Problem 5.88AP

The expression for the mass $M$ is $3msinθ$ .

Explanation of Solution

The free body diagram of the three connected objects is shown in Figure below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 5, Problem 5.88AP , additional homework tip 1

Figure (1)

From Figure (1), the equilibrium forces acts on the object of mass $2m$ in inclined plane is,

$∑Fnet=T1−2mgsinθ$

Here,

$m$ is the mass of object.

$g$ is the acceleration due to gravity.

$θ$ is the angle of inclined plane.

$T1$ is the tension in the rope of mass $2m$ .

From the Newton’s second law of motion, the net force on the object of mass $2m$ is,

$∑Fnet=(2m)a=2ma$

Here,

$a$ is the acceleration of the object.

Substitute $T1−2mgsinθ$ for $∑Fnet$ in the above equation.

$2ma=T1−2mgsinθ$ (1)

From Figure (1), the equilibrium forces act on the object of mass $m$ in inclined plane is,

$∑Fnet=T2−T1−mgsinθ$

Here,

$T2$ is the tension in the rope of mass $m$ .

From the Newton’s second law of motion, the net force on the object of mass $m$ is,

$∑Fnet=ma$

Substitute $T2−T1−mgsinθ$ for $∑Fnet$ in the above equation.

$ma=T2−T1−mgsinθ$ (2)

Add the equation (1) with equation (2).

$2ma+ma=T1−2mgsinθ+(T2−T1−mgsinθ)3ma=T2−3mgsinθT2=3m(a+gsinθ)$

From Figure (1), the equilibrium forces act on the object of mass $M$ is,

$∑Fnet=Mg−T2$

Here,

$M$ is the mass of hanging object.

Substitute $3m(a+gsinθ)$ for $T2$ in the above equation.

$∑Fnet=Mg−3m(a+gsinθ)$

From the Newton’s second law of motion, the net force on the object of mass $M$ is,

$∑Fnet=Ma$

Substitute $Mg−3m(a+gsinθ)$ for $∑Fnet$ in the above equation.

$Mg−3m(a+gsinθ)=MaM(g−a)=3m(a+gsinθ)M=3m(a+gsinθ)g−a$

The system is in equilibrium so value of the acceleration is zero.

Substitute $0$ for $a$ in the above equation.

$M=3m(0+gsinθ)g−0=3mgsinθg=3msinθ$

Conclusion:

Therefore, the expression for the mass $M$ is $3msinθ$ .

(b)

Expert Solution

To determine

The expressions for tensions $T1$ and $T2$ .

Answer to Problem 5.88AP

The expression for the tension $T1$ is $2mgsinθ$ and the expression for the tension $T2$ is $3mgsinθ$ .

Explanation of Solution

From part (a), the expression for the mass of $M$ is,

$M=3msinθ$

From part (a), the equilibrium forces act on the object is,

$∑Fnet=Mg−T2$

The system is in equilibrium, the value of acceleration is zero so the net force acts on the system is also zero.

Substitute $0$ for $∑Fnet$ in the above equation.

$Mg−T2=0T2=Mg$

Substitute $3msinθ$ for $M$ in the above equation.

$T2=(3msinθ)g=3mgsinθ$

Thus, the expression for the tension $T2$ is $3mgsinθ$ .

From part (a), the equation (2) is,

$ma=T2−T1−mgsinθ$

Substitute $0$ for $a$ in the above equation.

$m(0)=T2−T1−mgsinθT2−T1=mgsinθ$

Substitute $3mgsinθ$ for $T2$ in the above equation.

$3mgsinθ−T1=mgsinθT1=2mgsinθ$

Thus, the expression for the tension $T1$ is $2mgsinθ$ .

Conclusion:

Therefore, the expression for the tension $T1$ is $2mgsinθ$ and the expression for the tension $T2$ is $3mgsinθ$ .

(c)

Expert Solution

To determine

The acceleration of each object.

Answer to Problem 5.88AP

The acceleration of each object is $gsinθ1+2sinθ$ .

Explanation of Solution

Given info: The value of mass $M$ is double.

From part (a), the expression for the mass when it is double represents as,

$M=2(3msinθ)=6msinθ$

From part (a), the equation (1) is,

$2ma=T1−2mgsinθ$

Rearrange the above equation.

$T1=2ma+2mgsinθ$

From part (a), the equation (2) is,

$ma=T2−T1−mgsinθ$

Rearrange the above equation.

$T2−T1=ma+mgsinθ$

Substitute $2ma+2mgsinθ$ for $T1$ in the above equation.

$T2−(2ma+2mgsinθ)=ma+mgsinθT2=3ma+3mgsinθ$ (3)

From Figure (1), the equilibrium forces act on the object of mass $M$ is,

$Ma=Mg−T2T2=M(g−a)$

Substitute $6msinθ$ for $M$ in the above equation.

$T2=6msinθ(g−a)$ (4)

Subtract the equation (3) from equation (4).

$T2−T2=6msinθ(g−a)−(3ma+3mgsinθ)6masinθ+3ma=6mgsinθ−3mgsinθa(1+2sinθ)=gsinθa=gsinθ1+2sinθ$

Conclusion:

Therefore, the acceleration of each object is $gsinθ1+2sinθ$ .

(d)

Expert Solution

To determine

The expressions for tensions $T1$ and $T2$ .

Answer to Problem 5.88AP

The expression for the tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ and the expression for the tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

Explanation of Solution

From part (c), the expression for the acceleration is,

$a=gsinθ1+2sinθ$

From part (c), the equation for tension $T1$ is,

$T1=2ma+2mgsinθ$

Substitute $gsinθ1+2sinθ$ for $a$ in the above equation.

$T1=2m(gsinθ1+2sinθ)+2mgsinθ=2mgsinθ(11+2sinθ+1)=2mgsinθ(2+2sinθ1+2sinθ)=4mgsinθ(1+sinθ1+2sinθ)$

Thus, the expression for tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ .

From part (c), the equation (3) is,

$T2=3ma+3mgsinθ$

Substitute $gsinθ1+2sinθ$ for $a$ in the above equation.

$T2=3m(gsinθ1+2sinθ)+3mgsinθ=3mgsinθ(11+2sinθ+1)=3mgsinθ(1+1+2sinθ1+2sinθ)=6mgsinθ(1+sinθ1+2sinθ)$

Thus, the expression for tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

Conclusion:

Therefore, the expression for the tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ and the expression for the tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

(e)

Expert Solution

To determine

The maximum value of $M$ .

Answer to Problem 5.88AP

The maximum value of $M$ is $3m(sinθ+mscosθ)$ .

Explanation of Solution

Given info: The coefficient of static friction between mass $m$ , $2m$ and the inclined plane is $ms$ .

The static friction forces on the masses $m$ and $2m$ is shown in Figure below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 5, Problem 5.88AP , additional homework tip 2

Figure (2)

From the Figure (2), the normal force on the object of mass $2m$ is,

$n=2mgcosθ$

The expression for the static friction force on the object of mass $2m$ and mass $m$ is,

$fs=msn$ (3)

Here,

$ms$ is the coefficient of static friction.

$n$ is the normal force.

Substitute $2mgcosθ$ for $n$ in the above equation.

$fs=ms(2mgcosθ)=2mmsgcosθ$

From the Figure (1), the equilibrium forces acts on the object of mass $2m$ is,

$2ma=T1−2mgsinθ−fs$

Substitute $2mmsgcosθ$ for $fs$ in the above equation.

$2ma=T1−2mgsinθ−2mmsgcosθ$ (4)

Rearrange the above equation for $T1$ .

$T1=2ma+2mgsinθ+2mmsgcosθT1=2m(a+gsinθ+msgcosθ)$

Substitute $0$ for $a$ in the above equation.

$T1=2m(0+gsinθ+msgcosθ)=2mg(sinθ+mscosθ)$

From the Figure (2), the normal force on the object of mass $m$ is,

$n=mgcosθ$

Substitute $mgcosθ$ for $n$ in equation (3).

$fs=msmgcosθ$

From the Figure (1), the equilibrium forces acts on the object of mass $m$ is,

$ma=T2−T1−mgsinθ−fs$

Substitute $msmgcosθ$ for $fs$ in the above equation.

$ma=T2−T1−mgsinθ−msmgcosθ$

Add equation (4) with the above equation.

$2ma+ma=T1−2mgsinθ−2mmsgcosθ+T2−T1−mgsinθ−msmgcosθ3ma=−3mgsinθ−3mmsgcosθ+T2T2=3ma+3mgsinθ+3mmsgcosθT2=3m(a+gsinθ+msgcosθ)$

Substitute $0$ for $a$ in the above equation.

$T2=3m(0+gsinθ+msgcosθ)=3mg(sinθ+mscosθ)$

The equilibrium forces on the block of mass $M$ when mass is maximum represents as,

$T2=Mmaxg$

Substitute $3mg(sinθ+mscosθ)$ for $T2$ in the above equation.

$3mg(sinθ+mscosθ)=MmaxgMmax=3m(sinθ+mscosθ)$

Conclusion:

Therefore, the maximum value of $M$ is $3m(sinθ+mscosθ)$ .

(f)

Expert Solution

To determine

The minimum value of $M$ .

Answer to Problem 5.88AP

The minimum value of $M$ is $3m(sinθ−mscosθ)$ .

Explanation of Solution

From part (e), the expression of tension $T2$ for the minimum mass is,

$T2=3mg(sinθ−mscosθ)$

The equilibrium forces acts on block for minimum mass of $M$ is,

$T2=Mming$

Substitute $3mg(sinθ−mscosθ)$ for $T2$ in the above equation.

$3mg(sinθ−mscosθ)=MmingMmin=3m(sinθ−mscosθ)$

Conclusion:

Therefore, the minimum value of $M$ is $3m(sinθ−mscosθ)$ .

(g)

Expert Solution

To determine

The difference between the tension $T2$ for maximum and minimum mass.

Answer to Problem 5.88AP

The difference between the tension for maximum and minimum mass is $6msmgcosθ$ .

Explanation of Solution

From part (e), the expression for maximum mass is,

$Mmax=3m(sinθ+mscosθ)$

From part (f), the expression for the minimum mass is,

$Mmin=3m(sinθ−mscosθ)$

From part (e), the expression for the tension for maximum mass is,

$T2,max=Mmaxg$

From part (f), the expression for the tension for maximum mass is,

$T2,min=Mming$

Compare both the above equation.

$T2,max−T2,min=Mmaxg−Mming$

Substitute $3m(sinθ+mscosθ)$ for $Mmax$ and $3m(sinθ−mscosθ)$ for $Mmin$ in the above equation.

$T2,max−T2,min=(3m(sinθ+mscosθ))g−(3m(sinθ−mscosθ))g=6msmgcosθ$

Conclusion:

Therefore, the difference between the tension for maximum and minimum mass is $6msmgcosθ$ .

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Answer 2

Textbook Question

Chapter 5, Problem 5.88AP

Consider the three connected objects shown in Figure P5.88. Assume first that the inclined plane is friction-less and that the system is in equilibrium. In terms of m, g, and θ, find (a) the mass M and (b) the tensions T, and T2. Now assume that the value of Af is double the value found in part (a). Find (c) the acceleration of each object and (d) the tensions T₁ and T₂. Next, assume that the coefficient of static friction between m and 2m and the inclined plane is m, and that the system is in equilibrium. Find (e) the maximum value of M and (0 the minimum value of M. (g) Compare the values of T₂ when M has its minimum and maximum values.

Chapter 5, Problem 5.88AP, Consider the three connected objects shown in Figure P5.88. Assume first that the inclined plane is

(a)

Expert Solution

To determine

The expression for the mass of $M$ of the object.

Answer to Problem 5.88AP

The expression for the mass $M$ is $3msinθ$ .

Explanation of Solution

The free body diagram of the three connected objects is shown in Figure below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 5, Problem 5.88AP , additional homework tip 1

Figure (1)

From Figure (1), the equilibrium forces acts on the object of mass $2m$ in inclined plane is,

$∑Fnet=T1−2mgsinθ$

Here,

$m$ is the mass of object.

$g$ is the acceleration due to gravity.

$θ$ is the angle of inclined plane.

$T1$ is the tension in the rope of mass $2m$ .

From the Newton’s second law of motion, the net force on the object of mass $2m$ is,

$∑Fnet=(2m)a=2ma$

Here,

$a$ is the acceleration of the object.

Substitute $T1−2mgsinθ$ for $∑Fnet$ in the above equation.

$2ma=T1−2mgsinθ$ (1)

From Figure (1), the equilibrium forces act on the object of mass $m$ in inclined plane is,

$∑Fnet=T2−T1−mgsinθ$

Here,

$T2$ is the tension in the rope of mass $m$ .

From the Newton’s second law of motion, the net force on the object of mass $m$ is,

$∑Fnet=ma$

Substitute $T2−T1−mgsinθ$ for $∑Fnet$ in the above equation.

$ma=T2−T1−mgsinθ$ (2)

Add the equation (1) with equation (2).

$2ma+ma=T1−2mgsinθ+(T2−T1−mgsinθ)3ma=T2−3mgsinθT2=3m(a+gsinθ)$

From Figure (1), the equilibrium forces act on the object of mass $M$ is,

$∑Fnet=Mg−T2$

Here,

$M$ is the mass of hanging object.

Substitute $3m(a+gsinθ)$ for $T2$ in the above equation.

$∑Fnet=Mg−3m(a+gsinθ)$

From the Newton’s second law of motion, the net force on the object of mass $M$ is,

$∑Fnet=Ma$

Substitute $Mg−3m(a+gsinθ)$ for $∑Fnet$ in the above equation.

$Mg−3m(a+gsinθ)=MaM(g−a)=3m(a+gsinθ)M=3m(a+gsinθ)g−a$

The system is in equilibrium so value of the acceleration is zero.

Substitute $0$ for $a$ in the above equation.

$M=3m(0+gsinθ)g−0=3mgsinθg=3msinθ$

Conclusion:

Therefore, the expression for the mass $M$ is $3msinθ$ .

(b)

Expert Solution

To determine

The expressions for tensions $T1$ and $T2$ .

Answer to Problem 5.88AP

The expression for the tension $T1$ is $2mgsinθ$ and the expression for the tension $T2$ is $3mgsinθ$ .

Explanation of Solution

From part (a), the expression for the mass of $M$ is,

$M=3msinθ$

From part (a), the equilibrium forces act on the object is,

$∑Fnet=Mg−T2$

The system is in equilibrium, the value of acceleration is zero so the net force acts on the system is also zero.

Substitute $0$ for $∑Fnet$ in the above equation.

$Mg−T2=0T2=Mg$

Substitute $3msinθ$ for $M$ in the above equation.

$T2=(3msinθ)g=3mgsinθ$

Thus, the expression for the tension $T2$ is $3mgsinθ$ .

From part (a), the equation (2) is,

$ma=T2−T1−mgsinθ$

Substitute $0$ for $a$ in the above equation.

$m(0)=T2−T1−mgsinθT2−T1=mgsinθ$

Substitute $3mgsinθ$ for $T2$ in the above equation.

$3mgsinθ−T1=mgsinθT1=2mgsinθ$

Thus, the expression for the tension $T1$ is $2mgsinθ$ .

Conclusion:

Therefore, the expression for the tension $T1$ is $2mgsinθ$ and the expression for the tension $T2$ is $3mgsinθ$ .

(c)

Expert Solution

To determine

The acceleration of each object.

Answer to Problem 5.88AP

The acceleration of each object is $gsinθ1+2sinθ$ .

Explanation of Solution

Given info: The value of mass $M$ is double.

From part (a), the expression for the mass when it is double represents as,

$M=2(3msinθ)=6msinθ$

From part (a), the equation (1) is,

$2ma=T1−2mgsinθ$

Rearrange the above equation.

$T1=2ma+2mgsinθ$

From part (a), the equation (2) is,

$ma=T2−T1−mgsinθ$

Rearrange the above equation.

$T2−T1=ma+mgsinθ$

Substitute $2ma+2mgsinθ$ for $T1$ in the above equation.

$T2−(2ma+2mgsinθ)=ma+mgsinθT2=3ma+3mgsinθ$ (3)

From Figure (1), the equilibrium forces act on the object of mass $M$ is,

$Ma=Mg−T2T2=M(g−a)$

Substitute $6msinθ$ for $M$ in the above equation.

$T2=6msinθ(g−a)$ (4)

Subtract the equation (3) from equation (4).

$T2−T2=6msinθ(g−a)−(3ma+3mgsinθ)6masinθ+3ma=6mgsinθ−3mgsinθa(1+2sinθ)=gsinθa=gsinθ1+2sinθ$

Conclusion:

Therefore, the acceleration of each object is $gsinθ1+2sinθ$ .

(d)

Expert Solution

To determine

The expressions for tensions $T1$ and $T2$ .

Answer to Problem 5.88AP

The expression for the tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ and the expression for the tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

Explanation of Solution

From part (c), the expression for the acceleration is,

$a=gsinθ1+2sinθ$

From part (c), the equation for tension $T1$ is,

$T1=2ma+2mgsinθ$

Substitute $gsinθ1+2sinθ$ for $a$ in the above equation.

$T1=2m(gsinθ1+2sinθ)+2mgsinθ=2mgsinθ(11+2sinθ+1)=2mgsinθ(2+2sinθ1+2sinθ)=4mgsinθ(1+sinθ1+2sinθ)$

Thus, the expression for tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ .

From part (c), the equation (3) is,

$T2=3ma+3mgsinθ$

Substitute $gsinθ1+2sinθ$ for $a$ in the above equation.

$T2=3m(gsinθ1+2sinθ)+3mgsinθ=3mgsinθ(11+2sinθ+1)=3mgsinθ(1+1+2sinθ1+2sinθ)=6mgsinθ(1+sinθ1+2sinθ)$

Thus, the expression for tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

Conclusion:

Therefore, the expression for the tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ and the expression for the tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

(e)

Expert Solution

To determine

The maximum value of $M$ .

Answer to Problem 5.88AP

The maximum value of $M$ is $3m(sinθ+mscosθ)$ .

Explanation of Solution

Given info: The coefficient of static friction between mass $m$ , $2m$ and the inclined plane is $ms$ .

The static friction forces on the masses $m$ and $2m$ is shown in Figure below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 5, Problem 5.88AP , additional homework tip 2

Figure (2)

From the Figure (2), the normal force on the object of mass $2m$ is,

$n=2mgcosθ$

The expression for the static friction force on the object of mass $2m$ and mass $m$ is,

$fs=msn$ (3)

Here,

$ms$ is the coefficient of static friction.

$n$ is the normal force.

Substitute $2mgcosθ$ for $n$ in the above equation.

$fs=ms(2mgcosθ)=2mmsgcosθ$

From the Figure (1), the equilibrium forces acts on the object of mass $2m$ is,

$2ma=T1−2mgsinθ−fs$

Substitute $2mmsgcosθ$ for $fs$ in the above equation.

$2ma=T1−2mgsinθ−2mmsgcosθ$ (4)

Rearrange the above equation for $T1$ .

$T1=2ma+2mgsinθ+2mmsgcosθT1=2m(a+gsinθ+msgcosθ)$

Substitute $0$ for $a$ in the above equation.

$T1=2m(0+gsinθ+msgcosθ)=2mg(sinθ+mscosθ)$

From the Figure (2), the normal force on the object of mass $m$ is,

$n=mgcosθ$

Substitute $mgcosθ$ for $n$ in equation (3).

$fs=msmgcosθ$

From the Figure (1), the equilibrium forces acts on the object of mass $m$ is,

$ma=T2−T1−mgsinθ−fs$

Substitute $msmgcosθ$ for $fs$ in the above equation.

$ma=T2−T1−mgsinθ−msmgcosθ$

Add equation (4) with the above equation.

$2ma+ma=T1−2mgsinθ−2mmsgcosθ+T2−T1−mgsinθ−msmgcosθ3ma=−3mgsinθ−3mmsgcosθ+T2T2=3ma+3mgsinθ+3mmsgcosθT2=3m(a+gsinθ+msgcosθ)$

Substitute $0$ for $a$ in the above equation.

$T2=3m(0+gsinθ+msgcosθ)=3mg(sinθ+mscosθ)$

The equilibrium forces on the block of mass $M$ when mass is maximum represents as,

$T2=Mmaxg$

Substitute $3mg(sinθ+mscosθ)$ for $T2$ in the above equation.

$3mg(sinθ+mscosθ)=MmaxgMmax=3m(sinθ+mscosθ)$

Conclusion:

Therefore, the maximum value of $M$ is $3m(sinθ+mscosθ)$ .

(f)

Expert Solution

To determine

The minimum value of $M$ .

Answer to Problem 5.88AP

The minimum value of $M$ is $3m(sinθ−mscosθ)$ .

Explanation of Solution

From part (e), the expression of tension $T2$ for the minimum mass is,

$T2=3mg(sinθ−mscosθ)$

The equilibrium forces acts on block for minimum mass of $M$ is,

$T2=Mming$

Substitute $3mg(sinθ−mscosθ)$ for $T2$ in the above equation.

$3mg(sinθ−mscosθ)=MmingMmin=3m(sinθ−mscosθ)$

Conclusion:

Therefore, the minimum value of $M$ is $3m(sinθ−mscosθ)$ .

(g)

Expert Solution

To determine

The difference between the tension $T2$ for maximum and minimum mass.

Answer to Problem 5.88AP

The difference between the tension for maximum and minimum mass is $6msmgcosθ$ .

Explanation of Solution

From part (e), the expression for maximum mass is,

$Mmax=3m(sinθ+mscosθ)$

From part (f), the expression for the minimum mass is,

$Mmin=3m(sinθ−mscosθ)$

From part (e), the expression for the tension for maximum mass is,

$T2,max=Mmaxg$

From part (f), the expression for the tension for maximum mass is,

$T2,min=Mming$

Compare both the above equation.

$T2,max−T2,min=Mmaxg−Mming$

Substitute $3m(sinθ+mscosθ)$ for $Mmax$ and $3m(sinθ−mscosθ)$ for $Mmin$ in the above equation.

$T2,max−T2,min=(3m(sinθ+mscosθ))g−(3m(sinθ−mscosθ))g=6msmgcosθ$

Conclusion:

Therefore, the difference between the tension for maximum and minimum mass is $6msmgcosθ$ .

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Answer 3

Textbook Question

Chapter 5, Problem 5.88AP

Consider the three connected objects shown in Figure P5.88. Assume first that the inclined plane is friction-less and that the system is in equilibrium. In terms of m, g, and θ, find (a) the mass M and (b) the tensions T, and T2. Now assume that the value of Af is double the value found in part (a). Find (c) the acceleration of each object and (d) the tensions T₁ and T₂. Next, assume that the coefficient of static friction between m and 2m and the inclined plane is m, and that the system is in equilibrium. Find (e) the maximum value of M and (0 the minimum value of M. (g) Compare the values of T₂ when M has its minimum and maximum values.

Chapter 5, Problem 5.88AP, Consider the three connected objects shown in Figure P5.88. Assume first that the inclined plane is

(a)

Expert Solution

To determine

The expression for the mass of $M$ of the object.

Answer to Problem 5.88AP

The expression for the mass $M$ is $3msinθ$ .

Explanation of Solution

The free body diagram of the three connected objects is shown in Figure below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 5, Problem 5.88AP , additional homework tip 1

Figure (1)

From Figure (1), the equilibrium forces acts on the object of mass $2m$ in inclined plane is,

$∑Fnet=T1−2mgsinθ$

Here,

$m$ is the mass of object.

$g$ is the acceleration due to gravity.

$θ$ is the angle of inclined plane.

$T1$ is the tension in the rope of mass $2m$ .

From the Newton’s second law of motion, the net force on the object of mass $2m$ is,

$∑Fnet=(2m)a=2ma$

Here,

$a$ is the acceleration of the object.

Substitute $T1−2mgsinθ$ for $∑Fnet$ in the above equation.

$2ma=T1−2mgsinθ$ (1)

From Figure (1), the equilibrium forces act on the object of mass $m$ in inclined plane is,

$∑Fnet=T2−T1−mgsinθ$

Here,

$T2$ is the tension in the rope of mass $m$ .

From the Newton’s second law of motion, the net force on the object of mass $m$ is,

$∑Fnet=ma$

Substitute $T2−T1−mgsinθ$ for $∑Fnet$ in the above equation.

$ma=T2−T1−mgsinθ$ (2)

Add the equation (1) with equation (2).

$2ma+ma=T1−2mgsinθ+(T2−T1−mgsinθ)3ma=T2−3mgsinθT2=3m(a+gsinθ)$

From Figure (1), the equilibrium forces act on the object of mass $M$ is,

$∑Fnet=Mg−T2$

Here,

$M$ is the mass of hanging object.

Substitute $3m(a+gsinθ)$ for $T2$ in the above equation.

$∑Fnet=Mg−3m(a+gsinθ)$

From the Newton’s second law of motion, the net force on the object of mass $M$ is,

$∑Fnet=Ma$

Substitute $Mg−3m(a+gsinθ)$ for $∑Fnet$ in the above equation.

$Mg−3m(a+gsinθ)=MaM(g−a)=3m(a+gsinθ)M=3m(a+gsinθ)g−a$

The system is in equilibrium so value of the acceleration is zero.

Substitute $0$ for $a$ in the above equation.

$M=3m(0+gsinθ)g−0=3mgsinθg=3msinθ$

Conclusion:

Therefore, the expression for the mass $M$ is $3msinθ$ .

(b)

Expert Solution

To determine

The expressions for tensions $T1$ and $T2$ .

Answer to Problem 5.88AP

The expression for the tension $T1$ is $2mgsinθ$ and the expression for the tension $T2$ is $3mgsinθ$ .

Explanation of Solution

From part (a), the expression for the mass of $M$ is,

$M=3msinθ$

From part (a), the equilibrium forces act on the object is,

$∑Fnet=Mg−T2$

The system is in equilibrium, the value of acceleration is zero so the net force acts on the system is also zero.

Substitute $0$ for $∑Fnet$ in the above equation.

$Mg−T2=0T2=Mg$

Substitute $3msinθ$ for $M$ in the above equation.

$T2=(3msinθ)g=3mgsinθ$

Thus, the expression for the tension $T2$ is $3mgsinθ$ .

From part (a), the equation (2) is,

$ma=T2−T1−mgsinθ$

Substitute $0$ for $a$ in the above equation.

$m(0)=T2−T1−mgsinθT2−T1=mgsinθ$

Substitute $3mgsinθ$ for $T2$ in the above equation.

$3mgsinθ−T1=mgsinθT1=2mgsinθ$

Thus, the expression for the tension $T1$ is $2mgsinθ$ .

Conclusion:

Therefore, the expression for the tension $T1$ is $2mgsinθ$ and the expression for the tension $T2$ is $3mgsinθ$ .

(c)

Expert Solution

To determine

The acceleration of each object.

Answer to Problem 5.88AP

The acceleration of each object is $gsinθ1+2sinθ$ .

Explanation of Solution

Given info: The value of mass $M$ is double.

From part (a), the expression for the mass when it is double represents as,

$M=2(3msinθ)=6msinθ$

From part (a), the equation (1) is,

$2ma=T1−2mgsinθ$

Rearrange the above equation.

$T1=2ma+2mgsinθ$

From part (a), the equation (2) is,

$ma=T2−T1−mgsinθ$

Rearrange the above equation.

$T2−T1=ma+mgsinθ$

Substitute $2ma+2mgsinθ$ for $T1$ in the above equation.

$T2−(2ma+2mgsinθ)=ma+mgsinθT2=3ma+3mgsinθ$ (3)

From Figure (1), the equilibrium forces act on the object of mass $M$ is,

$Ma=Mg−T2T2=M(g−a)$

Substitute $6msinθ$ for $M$ in the above equation.

$T2=6msinθ(g−a)$ (4)

Subtract the equation (3) from equation (4).

$T2−T2=6msinθ(g−a)−(3ma+3mgsinθ)6masinθ+3ma=6mgsinθ−3mgsinθa(1+2sinθ)=gsinθa=gsinθ1+2sinθ$

Conclusion:

Therefore, the acceleration of each object is $gsinθ1+2sinθ$ .

(d)

Expert Solution

To determine

The expressions for tensions $T1$ and $T2$ .

Answer to Problem 5.88AP

The expression for the tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ and the expression for the tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

Explanation of Solution

From part (c), the expression for the acceleration is,

$a=gsinθ1+2sinθ$

From part (c), the equation for tension $T1$ is,

$T1=2ma+2mgsinθ$

Substitute $gsinθ1+2sinθ$ for $a$ in the above equation.

$T1=2m(gsinθ1+2sinθ)+2mgsinθ=2mgsinθ(11+2sinθ+1)=2mgsinθ(2+2sinθ1+2sinθ)=4mgsinθ(1+sinθ1+2sinθ)$

Thus, the expression for tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ .

From part (c), the equation (3) is,

$T2=3ma+3mgsinθ$

Substitute $gsinθ1+2sinθ$ for $a$ in the above equation.

$T2=3m(gsinθ1+2sinθ)+3mgsinθ=3mgsinθ(11+2sinθ+1)=3mgsinθ(1+1+2sinθ1+2sinθ)=6mgsinθ(1+sinθ1+2sinθ)$

Thus, the expression for tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

Conclusion:

Therefore, the expression for the tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ and the expression for the tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

(e)

Expert Solution

To determine

The maximum value of $M$ .

Answer to Problem 5.88AP

The maximum value of $M$ is $3m(sinθ+mscosθ)$ .

Explanation of Solution

Given info: The coefficient of static friction between mass $m$ , $2m$ and the inclined plane is $ms$ .

The static friction forces on the masses $m$ and $2m$ is shown in Figure below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 5, Problem 5.88AP , additional homework tip 2

Figure (2)

From the Figure (2), the normal force on the object of mass $2m$ is,

$n=2mgcosθ$

The expression for the static friction force on the object of mass $2m$ and mass $m$ is,

$fs=msn$ (3)

Here,

$ms$ is the coefficient of static friction.

$n$ is the normal force.

Substitute $2mgcosθ$ for $n$ in the above equation.

$fs=ms(2mgcosθ)=2mmsgcosθ$

From the Figure (1), the equilibrium forces acts on the object of mass $2m$ is,

$2ma=T1−2mgsinθ−fs$

Substitute $2mmsgcosθ$ for $fs$ in the above equation.

$2ma=T1−2mgsinθ−2mmsgcosθ$ (4)

Rearrange the above equation for $T1$ .

$T1=2ma+2mgsinθ+2mmsgcosθT1=2m(a+gsinθ+msgcosθ)$

Substitute $0$ for $a$ in the above equation.

$T1=2m(0+gsinθ+msgcosθ)=2mg(sinθ+mscosθ)$

From the Figure (2), the normal force on the object of mass $m$ is,

$n=mgcosθ$

Substitute $mgcosθ$ for $n$ in equation (3).

$fs=msmgcosθ$

From the Figure (1), the equilibrium forces acts on the object of mass $m$ is,

$ma=T2−T1−mgsinθ−fs$

Substitute $msmgcosθ$ for $fs$ in the above equation.

$ma=T2−T1−mgsinθ−msmgcosθ$

Add equation (4) with the above equation.

$2ma+ma=T1−2mgsinθ−2mmsgcosθ+T2−T1−mgsinθ−msmgcosθ3ma=−3mgsinθ−3mmsgcosθ+T2T2=3ma+3mgsinθ+3mmsgcosθT2=3m(a+gsinθ+msgcosθ)$

Substitute $0$ for $a$ in the above equation.

$T2=3m(0+gsinθ+msgcosθ)=3mg(sinθ+mscosθ)$

The equilibrium forces on the block of mass $M$ when mass is maximum represents as,

$T2=Mmaxg$

Substitute $3mg(sinθ+mscosθ)$ for $T2$ in the above equation.

$3mg(sinθ+mscosθ)=MmaxgMmax=3m(sinθ+mscosθ)$

Conclusion:

Therefore, the maximum value of $M$ is $3m(sinθ+mscosθ)$ .

(f)

Expert Solution

To determine

The minimum value of $M$ .

Answer to Problem 5.88AP

The minimum value of $M$ is $3m(sinθ−mscosθ)$ .

Explanation of Solution

From part (e), the expression of tension $T2$ for the minimum mass is,

$T2=3mg(sinθ−mscosθ)$

The equilibrium forces acts on block for minimum mass of $M$ is,

$T2=Mming$

Substitute $3mg(sinθ−mscosθ)$ for $T2$ in the above equation.

$3mg(sinθ−mscosθ)=MmingMmin=3m(sinθ−mscosθ)$

Conclusion:

Therefore, the minimum value of $M$ is $3m(sinθ−mscosθ)$ .

(g)

Expert Solution

To determine

The difference between the tension $T2$ for maximum and minimum mass.

Answer to Problem 5.88AP

The difference between the tension for maximum and minimum mass is $6msmgcosθ$ .

Explanation of Solution

From part (e), the expression for maximum mass is,

$Mmax=3m(sinθ+mscosθ)$

From part (f), the expression for the minimum mass is,

$Mmin=3m(sinθ−mscosθ)$

From part (e), the expression for the tension for maximum mass is,

$T2,max=Mmaxg$

From part (f), the expression for the tension for maximum mass is,

$T2,min=Mming$

Compare both the above equation.

$T2,max−T2,min=Mmaxg−Mming$

Substitute $3m(sinθ+mscosθ)$ for $Mmax$ and $3m(sinθ−mscosθ)$ for $Mmin$ in the above equation.

$T2,max−T2,min=(3m(sinθ+mscosθ))g−(3m(sinθ−mscosθ))g=6msmgcosθ$

Conclusion:

Therefore, the difference between the tension for maximum and minimum mass is $6msmgcosθ$ .

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Answer 4

Textbook Question

Chapter 5, Problem 5.88AP

Consider the three connected objects shown in Figure P5.88. Assume first that the inclined plane is friction-less and that the system is in equilibrium. In terms of m, g, and θ, find (a) the mass M and (b) the tensions T, and T2. Now assume that the value of Af is double the value found in part (a). Find (c) the acceleration of each object and (d) the tensions T₁ and T₂. Next, assume that the coefficient of static friction between m and 2m and the inclined plane is m, and that the system is in equilibrium. Find (e) the maximum value of M and (0 the minimum value of M. (g) Compare the values of T₂ when M has its minimum and maximum values.

Chapter 5, Problem 5.88AP, Consider the three connected objects shown in Figure P5.88. Assume first that the inclined plane is

(a)

Expert Solution

To determine

The expression for the mass of $M$ of the object.

Answer to Problem 5.88AP

The expression for the mass $M$ is $3msinθ$ .

Explanation of Solution

The free body diagram of the three connected objects is shown in Figure below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 5, Problem 5.88AP , additional homework tip 1

Figure (1)

From Figure (1), the equilibrium forces acts on the object of mass $2m$ in inclined plane is,

$∑Fnet=T1−2mgsinθ$

Here,

$m$ is the mass of object.

$g$ is the acceleration due to gravity.

$θ$ is the angle of inclined plane.

$T1$ is the tension in the rope of mass $2m$ .

From the Newton’s second law of motion, the net force on the object of mass $2m$ is,

$∑Fnet=(2m)a=2ma$

Here,

$a$ is the acceleration of the object.

Substitute $T1−2mgsinθ$ for $∑Fnet$ in the above equation.

$2ma=T1−2mgsinθ$ (1)

From Figure (1), the equilibrium forces act on the object of mass $m$ in inclined plane is,

$∑Fnet=T2−T1−mgsinθ$

Here,

$T2$ is the tension in the rope of mass $m$ .

From the Newton’s second law of motion, the net force on the object of mass $m$ is,

$∑Fnet=ma$

Substitute $T2−T1−mgsinθ$ for $∑Fnet$ in the above equation.

$ma=T2−T1−mgsinθ$ (2)

Add the equation (1) with equation (2).

$2ma+ma=T1−2mgsinθ+(T2−T1−mgsinθ)3ma=T2−3mgsinθT2=3m(a+gsinθ)$

From Figure (1), the equilibrium forces act on the object of mass $M$ is,

$∑Fnet=Mg−T2$

Here,

$M$ is the mass of hanging object.

Substitute $3m(a+gsinθ)$ for $T2$ in the above equation.

$∑Fnet=Mg−3m(a+gsinθ)$

From the Newton’s second law of motion, the net force on the object of mass $M$ is,

$∑Fnet=Ma$

Substitute $Mg−3m(a+gsinθ)$ for $∑Fnet$ in the above equation.

$Mg−3m(a+gsinθ)=MaM(g−a)=3m(a+gsinθ)M=3m(a+gsinθ)g−a$

The system is in equilibrium so value of the acceleration is zero.

Substitute $0$ for $a$ in the above equation.

$M=3m(0+gsinθ)g−0=3mgsinθg=3msinθ$

Conclusion:

Therefore, the expression for the mass $M$ is $3msinθ$ .

(b)

Expert Solution

To determine

The expressions for tensions $T1$ and $T2$ .

Answer to Problem 5.88AP

The expression for the tension $T1$ is $2mgsinθ$ and the expression for the tension $T2$ is $3mgsinθ$ .

Explanation of Solution

From part (a), the expression for the mass of $M$ is,

$M=3msinθ$

From part (a), the equilibrium forces act on the object is,

$∑Fnet=Mg−T2$

The system is in equilibrium, the value of acceleration is zero so the net force acts on the system is also zero.

Substitute $0$ for $∑Fnet$ in the above equation.

$Mg−T2=0T2=Mg$

Substitute $3msinθ$ for $M$ in the above equation.

$T2=(3msinθ)g=3mgsinθ$

Thus, the expression for the tension $T2$ is $3mgsinθ$ .

From part (a), the equation (2) is,

$ma=T2−T1−mgsinθ$

Substitute $0$ for $a$ in the above equation.

$m(0)=T2−T1−mgsinθT2−T1=mgsinθ$

Substitute $3mgsinθ$ for $T2$ in the above equation.

$3mgsinθ−T1=mgsinθT1=2mgsinθ$

Thus, the expression for the tension $T1$ is $2mgsinθ$ .

Conclusion:

Therefore, the expression for the tension $T1$ is $2mgsinθ$ and the expression for the tension $T2$ is $3mgsinθ$ .

(c)

Expert Solution

To determine

The acceleration of each object.

Answer to Problem 5.88AP

The acceleration of each object is $gsinθ1+2sinθ$ .

Explanation of Solution

Given info: The value of mass $M$ is double.

From part (a), the expression for the mass when it is double represents as,

$M=2(3msinθ)=6msinθ$

From part (a), the equation (1) is,

$2ma=T1−2mgsinθ$

Rearrange the above equation.

$T1=2ma+2mgsinθ$

From part (a), the equation (2) is,

$ma=T2−T1−mgsinθ$

Rearrange the above equation.

$T2−T1=ma+mgsinθ$

Substitute $2ma+2mgsinθ$ for $T1$ in the above equation.

$T2−(2ma+2mgsinθ)=ma+mgsinθT2=3ma+3mgsinθ$ (3)

From Figure (1), the equilibrium forces act on the object of mass $M$ is,

$Ma=Mg−T2T2=M(g−a)$

Substitute $6msinθ$ for $M$ in the above equation.

$T2=6msinθ(g−a)$ (4)

Subtract the equation (3) from equation (4).

$T2−T2=6msinθ(g−a)−(3ma+3mgsinθ)6masinθ+3ma=6mgsinθ−3mgsinθa(1+2sinθ)=gsinθa=gsinθ1+2sinθ$

Conclusion:

Therefore, the acceleration of each object is $gsinθ1+2sinθ$ .

(d)

Expert Solution

To determine

The expressions for tensions $T1$ and $T2$ .

Answer to Problem 5.88AP

The expression for the tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ and the expression for the tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

Explanation of Solution

From part (c), the expression for the acceleration is,

$a=gsinθ1+2sinθ$

From part (c), the equation for tension $T1$ is,

$T1=2ma+2mgsinθ$

Substitute $gsinθ1+2sinθ$ for $a$ in the above equation.

$T1=2m(gsinθ1+2sinθ)+2mgsinθ=2mgsinθ(11+2sinθ+1)=2mgsinθ(2+2sinθ1+2sinθ)=4mgsinθ(1+sinθ1+2sinθ)$

Thus, the expression for tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ .

From part (c), the equation (3) is,

$T2=3ma+3mgsinθ$

Substitute $gsinθ1+2sinθ$ for $a$ in the above equation.

$T2=3m(gsinθ1+2sinθ)+3mgsinθ=3mgsinθ(11+2sinθ+1)=3mgsinθ(1+1+2sinθ1+2sinθ)=6mgsinθ(1+sinθ1+2sinθ)$

Thus, the expression for tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

Conclusion:

Therefore, the expression for the tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ and the expression for the tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

(e)

Expert Solution

To determine

The maximum value of $M$ .

Answer to Problem 5.88AP

The maximum value of $M$ is $3m(sinθ+mscosθ)$ .

Explanation of Solution

Given info: The coefficient of static friction between mass $m$ , $2m$ and the inclined plane is $ms$ .

The static friction forces on the masses $m$ and $2m$ is shown in Figure below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 5, Problem 5.88AP , additional homework tip 2

Figure (2)

From the Figure (2), the normal force on the object of mass $2m$ is,

$n=2mgcosθ$

The expression for the static friction force on the object of mass $2m$ and mass $m$ is,

$fs=msn$ (3)

Here,

$ms$ is the coefficient of static friction.

$n$ is the normal force.

Substitute $2mgcosθ$ for $n$ in the above equation.

$fs=ms(2mgcosθ)=2mmsgcosθ$

From the Figure (1), the equilibrium forces acts on the object of mass $2m$ is,

$2ma=T1−2mgsinθ−fs$

Substitute $2mmsgcosθ$ for $fs$ in the above equation.

$2ma=T1−2mgsinθ−2mmsgcosθ$ (4)

Rearrange the above equation for $T1$ .

$T1=2ma+2mgsinθ+2mmsgcosθT1=2m(a+gsinθ+msgcosθ)$

Substitute $0$ for $a$ in the above equation.

$T1=2m(0+gsinθ+msgcosθ)=2mg(sinθ+mscosθ)$

From the Figure (2), the normal force on the object of mass $m$ is,

$n=mgcosθ$

Substitute $mgcosθ$ for $n$ in equation (3).

$fs=msmgcosθ$

From the Figure (1), the equilibrium forces acts on the object of mass $m$ is,

$ma=T2−T1−mgsinθ−fs$

Substitute $msmgcosθ$ for $fs$ in the above equation.

$ma=T2−T1−mgsinθ−msmgcosθ$

Add equation (4) with the above equation.

$2ma+ma=T1−2mgsinθ−2mmsgcosθ+T2−T1−mgsinθ−msmgcosθ3ma=−3mgsinθ−3mmsgcosθ+T2T2=3ma+3mgsinθ+3mmsgcosθT2=3m(a+gsinθ+msgcosθ)$

Substitute $0$ for $a$ in the above equation.

$T2=3m(0+gsinθ+msgcosθ)=3mg(sinθ+mscosθ)$

The equilibrium forces on the block of mass $M$ when mass is maximum represents as,

$T2=Mmaxg$

Substitute $3mg(sinθ+mscosθ)$ for $T2$ in the above equation.

$3mg(sinθ+mscosθ)=MmaxgMmax=3m(sinθ+mscosθ)$

Conclusion:

Therefore, the maximum value of $M$ is $3m(sinθ+mscosθ)$ .

(f)

Expert Solution

To determine

The minimum value of $M$ .

Answer to Problem 5.88AP

The minimum value of $M$ is $3m(sinθ−mscosθ)$ .

Explanation of Solution

From part (e), the expression of tension $T2$ for the minimum mass is,

$T2=3mg(sinθ−mscosθ)$

The equilibrium forces acts on block for minimum mass of $M$ is,

$T2=Mming$

Substitute $3mg(sinθ−mscosθ)$ for $T2$ in the above equation.

$3mg(sinθ−mscosθ)=MmingMmin=3m(sinθ−mscosθ)$

Conclusion:

Therefore, the minimum value of $M$ is $3m(sinθ−mscosθ)$ .

(g)

Expert Solution

To determine

The difference between the tension $T2$ for maximum and minimum mass.

Answer to Problem 5.88AP

The difference between the tension for maximum and minimum mass is $6msmgcosθ$ .

Explanation of Solution

From part (e), the expression for maximum mass is,

$Mmax=3m(sinθ+mscosθ)$

From part (f), the expression for the minimum mass is,

$Mmin=3m(sinθ−mscosθ)$

From part (e), the expression for the tension for maximum mass is,

$T2,max=Mmaxg$

From part (f), the expression for the tension for maximum mass is,

$T2,min=Mming$

Compare both the above equation.

$T2,max−T2,min=Mmaxg−Mming$

Substitute $3m(sinθ+mscosθ)$ for $Mmax$ and $3m(sinθ−mscosθ)$ for $Mmin$ in the above equation.

$T2,max−T2,min=(3m(sinθ+mscosθ))g−(3m(sinθ−mscosθ))g=6msmgcosθ$

Conclusion:

Therefore, the difference between the tension for maximum and minimum mass is $6msmgcosθ$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The expression for the mass of $M$ of the object.

Answer to Problem 5.88AP

The expression for the mass $M$ is $3msinθ$ .

Explanation of Solution

The free body diagram of the three connected objects is shown in Figure below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 5, Problem 5.88AP , additional homework tip 1

Figure (1)

From Figure (1), the equilibrium forces acts on the object of mass $2m$ in inclined plane is,

$∑Fnet=T1−2mgsinθ$

Here,

$m$ is the mass of object.

$g$ is the acceleration due to gravity.

$θ$ is the angle of inclined plane.

$T1$ is the tension in the rope of mass $2m$ .

From the Newton’s second law of motion, the net force on the object of mass $2m$ is,

$∑Fnet=(2m)a=2ma$

Here,

$a$ is the acceleration of the object.

Substitute $T1−2mgsinθ$ for $∑Fnet$ in the above equation.

$2ma=T1−2mgsinθ$ (1)

From Figure (1), the equilibrium forces act on the object of mass $m$ in inclined plane is,

$∑Fnet=T2−T1−mgsinθ$

Here,

$T2$ is the tension in the rope of mass $m$ .

From the Newton’s second law of motion, the net force on the object of mass $m$ is,

$∑Fnet=ma$

Substitute $T2−T1−mgsinθ$ for $∑Fnet$ in the above equation.

$ma=T2−T1−mgsinθ$ (2)

Add the equation (1) with equation (2).

$2ma+ma=T1−2mgsinθ+(T2−T1−mgsinθ)3ma=T2−3mgsinθT2=3m(a+gsinθ)$

From Figure (1), the equilibrium forces act on the object of mass $M$ is,

$∑Fnet=Mg−T2$

Here,

$M$ is the mass of hanging object.

Substitute $3m(a+gsinθ)$ for $T2$ in the above equation.

$∑Fnet=Mg−3m(a+gsinθ)$

From the Newton’s second law of motion, the net force on the object of mass $M$ is,

$∑Fnet=Ma$

Substitute $Mg−3m(a+gsinθ)$ for $∑Fnet$ in the above equation.

$Mg−3m(a+gsinθ)=MaM(g−a)=3m(a+gsinθ)M=3m(a+gsinθ)g−a$

The system is in equilibrium so value of the acceleration is zero.

Substitute $0$ for $a$ in the above equation.

$M=3m(0+gsinθ)g−0=3mgsinθg=3msinθ$

Conclusion:

Therefore, the expression for the mass $M$ is $3msinθ$ .

(b)

Expert Solution

To determine

The expressions for tensions $T1$ and $T2$ .

Answer to Problem 5.88AP

The expression for the tension $T1$ is $2mgsinθ$ and the expression for the tension $T2$ is $3mgsinθ$ .

Explanation of Solution

From part (a), the expression for the mass of $M$ is,

$M=3msinθ$

From part (a), the equilibrium forces act on the object is,

$∑Fnet=Mg−T2$

The system is in equilibrium, the value of acceleration is zero so the net force acts on the system is also zero.

Substitute $0$ for $∑Fnet$ in the above equation.

$Mg−T2=0T2=Mg$

Substitute $3msinθ$ for $M$ in the above equation.

$T2=(3msinθ)g=3mgsinθ$

Thus, the expression for the tension $T2$ is $3mgsinθ$ .

From part (a), the equation (2) is,

$ma=T2−T1−mgsinθ$

Substitute $0$ for $a$ in the above equation.

$m(0)=T2−T1−mgsinθT2−T1=mgsinθ$

Substitute $3mgsinθ$ for $T2$ in the above equation.

$3mgsinθ−T1=mgsinθT1=2mgsinθ$

Thus, the expression for the tension $T1$ is $2mgsinθ$ .

Conclusion:

Therefore, the expression for the tension $T1$ is $2mgsinθ$ and the expression for the tension $T2$ is $3mgsinθ$ .

(c)

Expert Solution

To determine

The acceleration of each object.

Answer to Problem 5.88AP

The acceleration of each object is $gsinθ1+2sinθ$ .

Explanation of Solution

Given info: The value of mass $M$ is double.

From part (a), the expression for the mass when it is double represents as,

$M=2(3msinθ)=6msinθ$

From part (a), the equation (1) is,

$2ma=T1−2mgsinθ$

Rearrange the above equation.

$T1=2ma+2mgsinθ$

From part (a), the equation (2) is,

$ma=T2−T1−mgsinθ$

Rearrange the above equation.

$T2−T1=ma+mgsinθ$

Substitute $2ma+2mgsinθ$ for $T1$ in the above equation.

$T2−(2ma+2mgsinθ)=ma+mgsinθT2=3ma+3mgsinθ$ (3)

From Figure (1), the equilibrium forces act on the object of mass $M$ is,

$Ma=Mg−T2T2=M(g−a)$

Substitute $6msinθ$ for $M$ in the above equation.

$T2=6msinθ(g−a)$ (4)

Subtract the equation (3) from equation (4).

$T2−T2=6msinθ(g−a)−(3ma+3mgsinθ)6masinθ+3ma=6mgsinθ−3mgsinθa(1+2sinθ)=gsinθa=gsinθ1+2sinθ$

Conclusion:

Therefore, the acceleration of each object is $gsinθ1+2sinθ$ .

(d)

Expert Solution

To determine

The expressions for tensions $T1$ and $T2$ .

Answer to Problem 5.88AP

The expression for the tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ and the expression for the tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

Explanation of Solution

From part (c), the expression for the acceleration is,

$a=gsinθ1+2sinθ$

From part (c), the equation for tension $T1$ is,

$T1=2ma+2mgsinθ$

Substitute $gsinθ1+2sinθ$ for $a$ in the above equation.

$T1=2m(gsinθ1+2sinθ)+2mgsinθ=2mgsinθ(11+2sinθ+1)=2mgsinθ(2+2sinθ1+2sinθ)=4mgsinθ(1+sinθ1+2sinθ)$

Thus, the expression for tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ .

From part (c), the equation (3) is,

$T2=3ma+3mgsinθ$

Substitute $gsinθ1+2sinθ$ for $a$ in the above equation.

$T2=3m(gsinθ1+2sinθ)+3mgsinθ=3mgsinθ(11+2sinθ+1)=3mgsinθ(1+1+2sinθ1+2sinθ)=6mgsinθ(1+sinθ1+2sinθ)$

Thus, the expression for tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

Conclusion:

Therefore, the expression for the tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ and the expression for the tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

(e)

Expert Solution

To determine

The maximum value of $M$ .

Answer to Problem 5.88AP

The maximum value of $M$ is $3m(sinθ+mscosθ)$ .

Explanation of Solution

Given info: The coefficient of static friction between mass $m$ , $2m$ and the inclined plane is $ms$ .

The static friction forces on the masses $m$ and $2m$ is shown in Figure below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 5, Problem 5.88AP , additional homework tip 2

Figure (2)

From the Figure (2), the normal force on the object of mass $2m$ is,

$n=2mgcosθ$

The expression for the static friction force on the object of mass $2m$ and mass $m$ is,

$fs=msn$ (3)

Here,

$ms$ is the coefficient of static friction.

$n$ is the normal force.

Substitute $2mgcosθ$ for $n$ in the above equation.

$fs=ms(2mgcosθ)=2mmsgcosθ$

From the Figure (1), the equilibrium forces acts on the object of mass $2m$ is,

$2ma=T1−2mgsinθ−fs$

Substitute $2mmsgcosθ$ for $fs$ in the above equation.

$2ma=T1−2mgsinθ−2mmsgcosθ$ (4)

Rearrange the above equation for $T1$ .

$T1=2ma+2mgsinθ+2mmsgcosθT1=2m(a+gsinθ+msgcosθ)$

Substitute $0$ for $a$ in the above equation.

$T1=2m(0+gsinθ+msgcosθ)=2mg(sinθ+mscosθ)$

From the Figure (2), the normal force on the object of mass $m$ is,

$n=mgcosθ$

Substitute $mgcosθ$ for $n$ in equation (3).

$fs=msmgcosθ$

From the Figure (1), the equilibrium forces acts on the object of mass $m$ is,

$ma=T2−T1−mgsinθ−fs$

Substitute $msmgcosθ$ for $fs$ in the above equation.

$ma=T2−T1−mgsinθ−msmgcosθ$

Add equation (4) with the above equation.

$2ma+ma=T1−2mgsinθ−2mmsgcosθ+T2−T1−mgsinθ−msmgcosθ3ma=−3mgsinθ−3mmsgcosθ+T2T2=3ma+3mgsinθ+3mmsgcosθT2=3m(a+gsinθ+msgcosθ)$

Substitute $0$ for $a$ in the above equation.

$T2=3m(0+gsinθ+msgcosθ)=3mg(sinθ+mscosθ)$

The equilibrium forces on the block of mass $M$ when mass is maximum represents as,

$T2=Mmaxg$

Substitute $3mg(sinθ+mscosθ)$ for $T2$ in the above equation.

$3mg(sinθ+mscosθ)=MmaxgMmax=3m(sinθ+mscosθ)$

Conclusion:

Therefore, the maximum value of $M$ is $3m(sinθ+mscosθ)$ .

(f)

Expert Solution

To determine

The minimum value of $M$ .

Answer to Problem 5.88AP

The minimum value of $M$ is $3m(sinθ−mscosθ)$ .

Explanation of Solution

From part (e), the expression of tension $T2$ for the minimum mass is,

$T2=3mg(sinθ−mscosθ)$

The equilibrium forces acts on block for minimum mass of $M$ is,

$T2=Mming$

Substitute $3mg(sinθ−mscosθ)$ for $T2$ in the above equation.

$3mg(sinθ−mscosθ)=MmingMmin=3m(sinθ−mscosθ)$

Conclusion:

Therefore, the minimum value of $M$ is $3m(sinθ−mscosθ)$ .

(g)

Expert Solution

To determine

The difference between the tension $T2$ for maximum and minimum mass.

Answer to Problem 5.88AP

The difference between the tension for maximum and minimum mass is $6msmgcosθ$ .

Explanation of Solution

From part (e), the expression for maximum mass is,

$Mmax=3m(sinθ+mscosθ)$

From part (f), the expression for the minimum mass is,

$Mmin=3m(sinθ−mscosθ)$

From part (e), the expression for the tension for maximum mass is,

$T2,max=Mmaxg$

From part (f), the expression for the tension for maximum mass is,

$T2,min=Mming$

Compare both the above equation.

$T2,max−T2,min=Mmaxg−Mming$

Substitute $3m(sinθ+mscosθ)$ for $Mmax$ and $3m(sinθ−mscosθ)$ for $Mmin$ in the above equation.

$T2,max−T2,min=(3m(sinθ+mscosθ))g−(3m(sinθ−mscosθ))g=6msmgcosθ$

Conclusion:

Therefore, the difference between the tension for maximum and minimum mass is $6msmgcosθ$ .

Answer 8

(a)

Expert Solution

To determine

The expression for the mass of $M$ of the object.

Answer to Problem 5.88AP

The expression for the mass $M$ is $3msinθ$ .

Explanation of Solution

The free body diagram of the three connected objects is shown in Figure below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 5, Problem 5.88AP , additional homework tip 1

Figure (1)

From Figure (1), the equilibrium forces acts on the object of mass $2m$ in inclined plane is,

$∑Fnet=T1−2mgsinθ$

Here,

$m$ is the mass of object.

$g$ is the acceleration due to gravity.

$θ$ is the angle of inclined plane.

$T1$ is the tension in the rope of mass $2m$ .

From the Newton’s second law of motion, the net force on the object of mass $2m$ is,

$∑Fnet=(2m)a=2ma$

Here,

$a$ is the acceleration of the object.

Substitute $T1−2mgsinθ$ for $∑Fnet$ in the above equation.

$2ma=T1−2mgsinθ$ (1)

From Figure (1), the equilibrium forces act on the object of mass $m$ in inclined plane is,

$∑Fnet=T2−T1−mgsinθ$

Here,

$T2$ is the tension in the rope of mass $m$ .

From the Newton’s second law of motion, the net force on the object of mass $m$ is,

$∑Fnet=ma$

Substitute $T2−T1−mgsinθ$ for $∑Fnet$ in the above equation.

$ma=T2−T1−mgsinθ$ (2)

Add the equation (1) with equation (2).

$2ma+ma=T1−2mgsinθ+(T2−T1−mgsinθ)3ma=T2−3mgsinθT2=3m(a+gsinθ)$

From Figure (1), the equilibrium forces act on the object of mass $M$ is,

$∑Fnet=Mg−T2$

Here,

$M$ is the mass of hanging object.

Substitute $3m(a+gsinθ)$ for $T2$ in the above equation.

$∑Fnet=Mg−3m(a+gsinθ)$

From the Newton’s second law of motion, the net force on the object of mass $M$ is,

$∑Fnet=Ma$

Substitute $Mg−3m(a+gsinθ)$ for $∑Fnet$ in the above equation.

$Mg−3m(a+gsinθ)=MaM(g−a)=3m(a+gsinθ)M=3m(a+gsinθ)g−a$

The system is in equilibrium so value of the acceleration is zero.

Substitute $0$ for $a$ in the above equation.

$M=3m(0+gsinθ)g−0=3mgsinθg=3msinθ$

Conclusion:

Therefore, the expression for the mass $M$ is $3msinθ$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The expression for the mass of $M$ of the object.

Answer 13

Answer to Problem 5.88AP

The expression for the mass $M$ is $3msinθ$ .

Answer 14

Explanation of Solution

The free body diagram of the three connected objects is shown in Figure below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 5, Problem 5.88AP , additional homework tip 1

Figure (1)

From Figure (1), the equilibrium forces acts on the object of mass $2m$ in inclined plane is,

$∑Fnet=T1−2mgsinθ$

Here,

$m$ is the mass of object.

$g$ is the acceleration due to gravity.

$θ$ is the angle of inclined plane.

$T1$ is the tension in the rope of mass $2m$ .

From the Newton’s second law of motion, the net force on the object of mass $2m$ is,

$∑Fnet=(2m)a=2ma$

Here,

$a$ is the acceleration of the object.

Substitute $T1−2mgsinθ$ for $∑Fnet$ in the above equation.

$2ma=T1−2mgsinθ$ (1)

From Figure (1), the equilibrium forces act on the object of mass $m$ in inclined plane is,

$∑Fnet=T2−T1−mgsinθ$

Here,

$T2$ is the tension in the rope of mass $m$ .

From the Newton’s second law of motion, the net force on the object of mass $m$ is,

$∑Fnet=ma$

Substitute $T2−T1−mgsinθ$ for $∑Fnet$ in the above equation.

$ma=T2−T1−mgsinθ$ (2)

Add the equation (1) with equation (2).

$2ma+ma=T1−2mgsinθ+(T2−T1−mgsinθ)3ma=T2−3mgsinθT2=3m(a+gsinθ)$

From Figure (1), the equilibrium forces act on the object of mass $M$ is,

$∑Fnet=Mg−T2$

Here,

$M$ is the mass of hanging object.

Substitute $3m(a+gsinθ)$ for $T2$ in the above equation.

$∑Fnet=Mg−3m(a+gsinθ)$

From the Newton’s second law of motion, the net force on the object of mass $M$ is,

$∑Fnet=Ma$

Substitute $Mg−3m(a+gsinθ)$ for $∑Fnet$ in the above equation.

$Mg−3m(a+gsinθ)=MaM(g−a)=3m(a+gsinθ)M=3m(a+gsinθ)g−a$

The system is in equilibrium so value of the acceleration is zero.

Substitute $0$ for $a$ in the above equation.

$M=3m(0+gsinθ)g−0=3mgsinθg=3msinθ$

Conclusion:

Therefore, the expression for the mass $M$ is $3msinθ$ .

Answer 15

(b)

Expert Solution

To determine

The expressions for tensions $T1$ and $T2$ .

Answer to Problem 5.88AP

The expression for the tension $T1$ is $2mgsinθ$ and the expression for the tension $T2$ is $3mgsinθ$ .

Explanation of Solution

From part (a), the expression for the mass of $M$ is,

$M=3msinθ$

From part (a), the equilibrium forces act on the object is,

$∑Fnet=Mg−T2$

The system is in equilibrium, the value of acceleration is zero so the net force acts on the system is also zero.

Substitute $0$ for $∑Fnet$ in the above equation.

$Mg−T2=0T2=Mg$

Substitute $3msinθ$ for $M$ in the above equation.

$T2=(3msinθ)g=3mgsinθ$

Thus, the expression for the tension $T2$ is $3mgsinθ$ .

From part (a), the equation (2) is,

$ma=T2−T1−mgsinθ$

Substitute $0$ for $a$ in the above equation.

$m(0)=T2−T1−mgsinθT2−T1=mgsinθ$

Substitute $3mgsinθ$ for $T2$ in the above equation.

$3mgsinθ−T1=mgsinθT1=2mgsinθ$

Thus, the expression for the tension $T1$ is $2mgsinθ$ .

Conclusion:

Therefore, the expression for the tension $T1$ is $2mgsinθ$ and the expression for the tension $T2$ is $3mgsinθ$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The expressions for tensions $T1$ and $T2$ .

Answer 20

Answer to Problem 5.88AP

The expression for the tension $T1$ is $2mgsinθ$ and the expression for the tension $T2$ is $3mgsinθ$ .

Answer 21

Explanation of Solution

From part (a), the expression for the mass of $M$ is,

$M=3msinθ$

From part (a), the equilibrium forces act on the object is,

$∑Fnet=Mg−T2$

The system is in equilibrium, the value of acceleration is zero so the net force acts on the system is also zero.

Substitute $0$ for $∑Fnet$ in the above equation.

$Mg−T2=0T2=Mg$

Substitute $3msinθ$ for $M$ in the above equation.

$T2=(3msinθ)g=3mgsinθ$

Thus, the expression for the tension $T2$ is $3mgsinθ$ .

From part (a), the equation (2) is,

$ma=T2−T1−mgsinθ$

Substitute $0$ for $a$ in the above equation.

$m(0)=T2−T1−mgsinθT2−T1=mgsinθ$

Substitute $3mgsinθ$ for $T2$ in the above equation.

$3mgsinθ−T1=mgsinθT1=2mgsinθ$

Thus, the expression for the tension $T1$ is $2mgsinθ$ .

Conclusion:

Therefore, the expression for the tension $T1$ is $2mgsinθ$ and the expression for the tension $T2$ is $3mgsinθ$ .

Answer 22

(c)

Expert Solution

To determine

The acceleration of each object.

Answer to Problem 5.88AP

The acceleration of each object is $gsinθ1+2sinθ$ .

Explanation of Solution

Given info: The value of mass $M$ is double.

From part (a), the expression for the mass when it is double represents as,

$M=2(3msinθ)=6msinθ$

From part (a), the equation (1) is,

$2ma=T1−2mgsinθ$

Rearrange the above equation.

$T1=2ma+2mgsinθ$

From part (a), the equation (2) is,

$ma=T2−T1−mgsinθ$

Rearrange the above equation.

$T2−T1=ma+mgsinθ$

Substitute $2ma+2mgsinθ$ for $T1$ in the above equation.

$T2−(2ma+2mgsinθ)=ma+mgsinθT2=3ma+3mgsinθ$ (3)

From Figure (1), the equilibrium forces act on the object of mass $M$ is,

$Ma=Mg−T2T2=M(g−a)$

Substitute $6msinθ$ for $M$ in the above equation.

$T2=6msinθ(g−a)$ (4)

Subtract the equation (3) from equation (4).

$T2−T2=6msinθ(g−a)−(3ma+3mgsinθ)6masinθ+3ma=6mgsinθ−3mgsinθa(1+2sinθ)=gsinθa=gsinθ1+2sinθ$

Conclusion:

Therefore, the acceleration of each object is $gsinθ1+2sinθ$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The acceleration of each object.

Answer 27

Answer to Problem 5.88AP

The acceleration of each object is $gsinθ1+2sinθ$ .

Answer 28

Explanation of Solution

Given info: The value of mass $M$ is double.

From part (a), the expression for the mass when it is double represents as,

$M=2(3msinθ)=6msinθ$

From part (a), the equation (1) is,

$2ma=T1−2mgsinθ$

Rearrange the above equation.

$T1=2ma+2mgsinθ$

From part (a), the equation (2) is,

$ma=T2−T1−mgsinθ$

Rearrange the above equation.

$T2−T1=ma+mgsinθ$

Substitute $2ma+2mgsinθ$ for $T1$ in the above equation.

$T2−(2ma+2mgsinθ)=ma+mgsinθT2=3ma+3mgsinθ$ (3)

From Figure (1), the equilibrium forces act on the object of mass $M$ is,

$Ma=Mg−T2T2=M(g−a)$

Substitute $6msinθ$ for $M$ in the above equation.

$T2=6msinθ(g−a)$ (4)

Subtract the equation (3) from equation (4).

$T2−T2=6msinθ(g−a)−(3ma+3mgsinθ)6masinθ+3ma=6mgsinθ−3mgsinθa(1+2sinθ)=gsinθa=gsinθ1+2sinθ$

Conclusion:

Therefore, the acceleration of each object is $gsinθ1+2sinθ$ .

Answer 29

(d)

Expert Solution

To determine

The expressions for tensions $T1$ and $T2$ .

Answer to Problem 5.88AP

The expression for the tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ and the expression for the tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

Explanation of Solution

From part (c), the expression for the acceleration is,

$a=gsinθ1+2sinθ$

From part (c), the equation for tension $T1$ is,

$T1=2ma+2mgsinθ$

Substitute $gsinθ1+2sinθ$ for $a$ in the above equation.

$T1=2m(gsinθ1+2sinθ)+2mgsinθ=2mgsinθ(11+2sinθ+1)=2mgsinθ(2+2sinθ1+2sinθ)=4mgsinθ(1+sinθ1+2sinθ)$

Thus, the expression for tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ .

From part (c), the equation (3) is,

$T2=3ma+3mgsinθ$

Substitute $gsinθ1+2sinθ$ for $a$ in the above equation.

$T2=3m(gsinθ1+2sinθ)+3mgsinθ=3mgsinθ(11+2sinθ+1)=3mgsinθ(1+1+2sinθ1+2sinθ)=6mgsinθ(1+sinθ1+2sinθ)$

Thus, the expression for tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

Conclusion:

Therefore, the expression for the tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ and the expression for the tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The expressions for tensions $T1$ and $T2$ .

Answer 34

Answer to Problem 5.88AP

The expression for the tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ and the expression for the tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

Answer 35

Explanation of Solution

From part (c), the expression for the acceleration is,

$a=gsinθ1+2sinθ$

From part (c), the equation for tension $T1$ is,

$T1=2ma+2mgsinθ$

Substitute $gsinθ1+2sinθ$ for $a$ in the above equation.

$T1=2m(gsinθ1+2sinθ)+2mgsinθ=2mgsinθ(11+2sinθ+1)=2mgsinθ(2+2sinθ1+2sinθ)=4mgsinθ(1+sinθ1+2sinθ)$

Thus, the expression for tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ .

From part (c), the equation (3) is,

$T2=3ma+3mgsinθ$

Substitute $gsinθ1+2sinθ$ for $a$ in the above equation.

$T2=3m(gsinθ1+2sinθ)+3mgsinθ=3mgsinθ(11+2sinθ+1)=3mgsinθ(1+1+2sinθ1+2sinθ)=6mgsinθ(1+sinθ1+2sinθ)$

Thus, the expression for tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

Conclusion:

Therefore, the expression for the tension $T1$ is $4mgsinθ(1+sinθ1+2sinθ)$ and the expression for the tension $T2$ is $6mgsinθ(1+sinθ1+2sinθ)$ .

Answer 36

(e)

Expert Solution

To determine

The maximum value of $M$ .

Answer to Problem 5.88AP

The maximum value of $M$ is $3m(sinθ+mscosθ)$ .

Explanation of Solution

Given info: The coefficient of static friction between mass $m$ , $2m$ and the inclined plane is $ms$ .

The static friction forces on the masses $m$ and $2m$ is shown in Figure below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 5, Problem 5.88AP , additional homework tip 2

Figure (2)

From the Figure (2), the normal force on the object of mass $2m$ is,

$n=2mgcosθ$

The expression for the static friction force on the object of mass $2m$ and mass $m$ is,

$fs=msn$ (3)

Here,

$ms$ is the coefficient of static friction.

$n$ is the normal force.

Substitute $2mgcosθ$ for $n$ in the above equation.

$fs=ms(2mgcosθ)=2mmsgcosθ$

From the Figure (1), the equilibrium forces acts on the object of mass $2m$ is,

$2ma=T1−2mgsinθ−fs$

Substitute $2mmsgcosθ$ for $fs$ in the above equation.

$2ma=T1−2mgsinθ−2mmsgcosθ$ (4)

Rearrange the above equation for $T1$ .

$T1=2ma+2mgsinθ+2mmsgcosθT1=2m(a+gsinθ+msgcosθ)$

Substitute $0$ for $a$ in the above equation.

$T1=2m(0+gsinθ+msgcosθ)=2mg(sinθ+mscosθ)$

From the Figure (2), the normal force on the object of mass $m$ is,

$n=mgcosθ$

Substitute $mgcosθ$ for $n$ in equation (3).

$fs=msmgcosθ$

From the Figure (1), the equilibrium forces acts on the object of mass $m$ is,

$ma=T2−T1−mgsinθ−fs$

Substitute $msmgcosθ$ for $fs$ in the above equation.

$ma=T2−T1−mgsinθ−msmgcosθ$

Add equation (4) with the above equation.

$2ma+ma=T1−2mgsinθ−2mmsgcosθ+T2−T1−mgsinθ−msmgcosθ3ma=−3mgsinθ−3mmsgcosθ+T2T2=3ma+3mgsinθ+3mmsgcosθT2=3m(a+gsinθ+msgcosθ)$

Substitute $0$ for $a$ in the above equation.

$T2=3m(0+gsinθ+msgcosθ)=3mg(sinθ+mscosθ)$

The equilibrium forces on the block of mass $M$ when mass is maximum represents as,

$T2=Mmaxg$

Substitute $3mg(sinθ+mscosθ)$ for $T2$ in the above equation.

$3mg(sinθ+mscosθ)=MmaxgMmax=3m(sinθ+mscosθ)$

Conclusion:

Therefore, the maximum value of $M$ is $3m(sinθ+mscosθ)$ .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

To determine

The maximum value of $M$ .

Answer 41

Answer to Problem 5.88AP

The maximum value of $M$ is $3m(sinθ+mscosθ)$ .

Answer 42

Explanation of Solution

Given info: The coefficient of static friction between mass $m$ , $2m$ and the inclined plane is $ms$ .

The static friction forces on the masses $m$ and $2m$ is shown in Figure below,

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 5, Problem 5.88AP , additional homework tip 2

Figure (2)

From the Figure (2), the normal force on the object of mass $2m$ is,

$n=2mgcosθ$

The expression for the static friction force on the object of mass $2m$ and mass $m$ is,

$fs=msn$ (3)

Here,

$ms$ is the coefficient of static friction.

$n$ is the normal force.

Substitute $2mgcosθ$ for $n$ in the above equation.

$fs=ms(2mgcosθ)=2mmsgcosθ$

From the Figure (1), the equilibrium forces acts on the object of mass $2m$ is,

$2ma=T1−2mgsinθ−fs$

Substitute $2mmsgcosθ$ for $fs$ in the above equation.

$2ma=T1−2mgsinθ−2mmsgcosθ$ (4)

Rearrange the above equation for $T1$ .

$T1=2ma+2mgsinθ+2mmsgcosθT1=2m(a+gsinθ+msgcosθ)$

Substitute $0$ for $a$ in the above equation.

$T1=2m(0+gsinθ+msgcosθ)=2mg(sinθ+mscosθ)$

From the Figure (2), the normal force on the object of mass $m$ is,

$n=mgcosθ$

Substitute $mgcosθ$ for $n$ in equation (3).

$fs=msmgcosθ$

From the Figure (1), the equilibrium forces acts on the object of mass $m$ is,

$ma=T2−T1−mgsinθ−fs$

Substitute $msmgcosθ$ for $fs$ in the above equation.

$ma=T2−T1−mgsinθ−msmgcosθ$

Add equation (4) with the above equation.

$2ma+ma=T1−2mgsinθ−2mmsgcosθ+T2−T1−mgsinθ−msmgcosθ3ma=−3mgsinθ−3mmsgcosθ+T2T2=3ma+3mgsinθ+3mmsgcosθT2=3m(a+gsinθ+msgcosθ)$

Substitute $0$ for $a$ in the above equation.

$T2=3m(0+gsinθ+msgcosθ)=3mg(sinθ+mscosθ)$

The equilibrium forces on the block of mass $M$ when mass is maximum represents as,

$T2=Mmaxg$

Substitute $3mg(sinθ+mscosθ)$ for $T2$ in the above equation.

$3mg(sinθ+mscosθ)=MmaxgMmax=3m(sinθ+mscosθ)$

Conclusion:

Therefore, the maximum value of $M$ is $3m(sinθ+mscosθ)$ .

Answer 43

(f)

Expert Solution

To determine

The minimum value of $M$ .

Answer to Problem 5.88AP

The minimum value of $M$ is $3m(sinθ−mscosθ)$ .

Explanation of Solution

From part (e), the expression of tension $T2$ for the minimum mass is,

$T2=3mg(sinθ−mscosθ)$

The equilibrium forces acts on block for minimum mass of $M$ is,

$T2=Mming$

Substitute $3mg(sinθ−mscosθ)$ for $T2$ in the above equation.

$3mg(sinθ−mscosθ)=MmingMmin=3m(sinθ−mscosθ)$

Conclusion:

Therefore, the minimum value of $M$ is $3m(sinθ−mscosθ)$ .

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

To determine

The minimum value of $M$ .

Answer 48

Answer to Problem 5.88AP

The minimum value of $M$ is $3m(sinθ−mscosθ)$ .

Answer 49

Explanation of Solution

From part (e), the expression of tension $T2$ for the minimum mass is,

$T2=3mg(sinθ−mscosθ)$

The equilibrium forces acts on block for minimum mass of $M$ is,

$T2=Mming$

Substitute $3mg(sinθ−mscosθ)$ for $T2$ in the above equation.

$3mg(sinθ−mscosθ)=MmingMmin=3m(sinθ−mscosθ)$

Conclusion:

Therefore, the minimum value of $M$ is $3m(sinθ−mscosθ)$ .

Answer 50

(g)

Expert Solution

To determine

The difference between the tension $T2$ for maximum and minimum mass.

Answer to Problem 5.88AP

The difference between the tension for maximum and minimum mass is $6msmgcosθ$ .

Explanation of Solution

From part (e), the expression for maximum mass is,

$Mmax=3m(sinθ+mscosθ)$

From part (f), the expression for the minimum mass is,

$Mmin=3m(sinθ−mscosθ)$

From part (e), the expression for the tension for maximum mass is,

$T2,max=Mmaxg$

From part (f), the expression for the tension for maximum mass is,

$T2,min=Mming$

Compare both the above equation.

$T2,max−T2,min=Mmaxg−Mming$

Substitute $3m(sinθ+mscosθ)$ for $Mmax$ and $3m(sinθ−mscosθ)$ for $Mmin$ in the above equation.

$T2,max−T2,min=(3m(sinθ+mscosθ))g−(3m(sinθ−mscosθ))g=6msmgcosθ$

Conclusion:

Therefore, the difference between the tension for maximum and minimum mass is $6msmgcosθ$ .

Answer 51

(g)

Expert Solution

Answer 52

(g)

Expert Solution

Answer 53

Expert Solution

Answer 54

To determine

The difference between the tension $T2$ for maximum and minimum mass.

Answer 55

Answer to Problem 5.88AP

The difference between the tension for maximum and minimum mass is $6msmgcosθ$ .

Answer 56

Explanation of Solution

From part (e), the expression for maximum mass is,

$Mmax=3m(sinθ+mscosθ)$

From part (f), the expression for the minimum mass is,

$Mmin=3m(sinθ−mscosθ)$

From part (e), the expression for the tension for maximum mass is,

$T2,max=Mmaxg$

From part (f), the expression for the tension for maximum mass is,

$T2,min=Mming$

Compare both the above equation.

$T2,max−T2,min=Mmaxg−Mming$

Substitute $3m(sinθ+mscosθ)$ for $Mmax$ and $3m(sinθ−mscosθ)$ for $Mmin$ in the above equation.

$T2,max−T2,min=(3m(sinθ+mscosθ))g−(3m(sinθ−mscosθ))g=6msmgcosθ$

Conclusion:

Therefore, the difference between the tension for maximum and minimum mass is $6msmgcosθ$ .

Answer 57

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Answer to Problem 5.88AP

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