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.

To determine

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

Answer to Problem 5.88AP

The expression for the mass $M$ is $3m\sin \theta$.

Explanation of Solution

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

Figure (1)

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

$\sum F_{\text{net}}=T_1-2mg\sin \theta$

Here,

$m$ is the mass of object.

$g$ is the acceleration due to gravity.

$\theta$ is the angle of inclined plane.

$T_1$ 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,

$\begin{array}{c}\sum F_{\text{net}}=\left(2m\right)a\\ =2ma\end{array}$

Here,

$a$ is the acceleration of the object.

Substitute $T_1-2mg\sin \theta$ for $\sum F_{\text{net}}$ in the above equation.

$2ma=T_1-2mg\sin \theta$ (1)

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

$\sum F_{\text{net}}=T_2-T_1-mg\sin \theta$

Here,

$T_2$ 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,

$\sum F_{\text{net}}=ma$

Substitute $T_2-T_1-mg\sin \theta$ for $\sum F_{\text{net}}$ in the above equation.

$ma=T_2-T_1-mg\sin \theta$ (2)

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

$\begin{array}{c}2ma+ma=T_1-2mg\sin \theta +\left(T_2-T_1-mg\sin \theta \right)\\ 3ma=T_2-3mg\sin \theta \\ T_2=3m\left(a+g\sin \theta \right)\end{array}$

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

$\sum F_{\text{net}}=Mg-T_2$

Here,

$M$ is the mass of hanging object.

Substitute $3m\left(a+g\sin \theta \right)$ for $T_2$ in the above equation.

$\sum F_{\text{net}}=Mg-3m\left(a+g\sin \theta \right)$

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

$\sum F_{\text{net}}=Ma$

Substitute $Mg-3m\left(a+g\sin \theta \right)$ for $\sum F_{\text{net}}$ in the above equation.

$\begin{array}{c}Mg-3m\left(a+g\sin \theta \right)=Ma\\ M\left(g-a\right)=3m\left(a+g\sin \theta \right)\\ M=\frac{3m\left(a+g\sin \theta \right)}{g-a}\end{array}$

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

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

$\begin{array}{c}M=\frac{3m\left(0+g\sin \theta \right)}{g-0}\\ =\frac{3mg\sin \theta }{g}\\ =3m\sin \theta \end{array}$

Conclusion:

Therefore, the expression for the mass $M$ is $3m\sin \theta$.

To determine

The expressions for tensions $T_1$ and $T_2$.

Answer to Problem 5.88AP

The expression for the tension $T_1$ is $2mg\sin \theta$ and the expression for the tension $T_2$ is $3mg\sin \theta$.

Explanation of Solution

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

$M=3m\sin \theta$

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

$\sum F_{\text{net}}=Mg-T_2$

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 $\sum F_{\text{net}}$ in the above equation.

$\begin{array}{c}Mg-T_2=0\\ T_2=Mg\end{array}$

Substitute $3m\sin \theta$ for $M$ in the above equation.

$\begin{array}{c}{T}_2=\left(3m\sin \theta \right)g\\ =3mg\sin \theta \end{array}$

Thus, the expression for the tension $T_2$ is $3mg\sin \theta$.

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

$ma=T_2-T_1-mg\sin \theta$

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

$\begin{array}{c}m\left(0\right)=T_2-T_1-mg\sin \theta \\ T_2-T_1=mg\sin \theta \end{array}$

Substitute $3mg\sin \theta$ for $T_2$ in the above equation.

$\begin{array}{c}3mg\sin \theta -T_1=mg\sin \theta \\ T_1=2mg\sin \theta \end{array}$

Thus, the expression for the tension $T_1$ is $2mg\sin \theta$.

Conclusion:

Therefore, the expression for the tension $T_1$ is $2mg\sin \theta$ and the expression for the tension $T_2$ is $3mg\sin \theta$.

To determine

The acceleration of each object.

Answer to Problem 5.88AP

The acceleration of each object is $\frac{g\sin \theta }{1+2\sin \theta }$.

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,

$\begin{array}{c}M=2\left(3m\sin \theta \right)\\ =6m\sin \theta \end{array}$

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

$2ma=T_1-2mg\sin \theta$

Rearrange the above equation.

$T_1=2ma+2mg\sin \theta$

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

$ma=T_2-T_1-mg\sin \theta$

Rearrange the above equation.

$T_2-T_1=ma+mg\sin \theta$

Substitute $2ma+2mg\sin \theta$ for $T_1$ in the above equation.

$\begin{array}{c}{T}_2-\left(2ma+2mg\sin \theta \right)=ma+mg\sin \theta \\ T_2=3ma+3mg\sin \theta \end{array}$ (3)

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

$\begin{array}{c}Ma=Mg-T_2\\ T_2=M\left(g-a\right)\end{array}$

Substitute $6m\sin \theta$ for $M$ in the above equation.

$T_2=6m\sin \theta \left(g-a\right)$ (4)

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

$\begin{array}{c}{T}_2-T_2=6m\sin \theta \left(g-a\right)-\left(3ma+3mg\sin \theta \right)\\ 6ma\sin \theta +3ma=6mg\sin \theta -3mg\sin \theta \\ a\left(1+2\sin \theta \right)=g\sin \theta \\ a=\frac{g\sin \theta }{1+2\sin \theta }\end{array}$

Conclusion:

Therefore, the acceleration of each object is $\frac{g\sin \theta }{1+2\sin \theta }$.

To determine

The expressions for tensions $T_1$ and $T_2$.

Answer to Problem 5.88AP

The expression for the tension $T_1$ is $4mg\sin \theta \left(\frac{1+\sin \theta }{1+2\sin \theta }\right)$ and the expression for the tension $T_2$ is $6mg\sin \theta \left(\frac{1+\sin \theta }{1+2\sin \theta }\right)$.

Explanation of Solution

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

$a=\frac{g\sin \theta }{1+2\sin \theta }$

From part (c), the equation for tension $T_1$ is,

$T_1=2ma+2mg\sin \theta$

Substitute $\frac{g\sin \theta }{1+2\sin \theta }$ for $a$ in the above equation.

$\begin{array}{c}{T}_1=2m\left(\frac{g\sin \theta }{1+2\sin \theta }\right)+2mg\sin \theta \\ =2mg\sin \theta \left(\frac{1}{1+2\sin \theta }+1\right)\\ =2mg\sin \theta \left(\frac{2+2\sin \theta }{1+2\sin \theta }\right)\\ =4mg\sin \theta \left(\frac{1+\sin \theta }{1+2\sin \theta }\right)\end{array}$

Thus, the expression for tension $T_1$ is $4mg\sin \theta \left(\frac{1+\sin \theta }{1+2\sin \theta }\right)$.

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

$T_2=3ma+3mg\sin \theta$

Substitute $\frac{g\sin \theta }{1+2\sin \theta }$ for $a$ in the above equation.

$\begin{array}{c}{T}_2=3m\left(\frac{g\sin \theta }{1+2\sin \theta }\right)+3mg\sin \theta \\ =3mg\sin \theta \left(\frac{1}{1+2\sin \theta }+1\right)\\ =3mg\sin \theta \left(\frac{1+1+2\sin \theta }{1+2\sin \theta }\right)\\ =6mg\sin \theta \left(\frac{1+\sin \theta }{1+2\sin \theta }\right)\end{array}$

Thus, the expression for tension $T_2$ is $6mg\sin \theta \left(\frac{1+\sin \theta }{1+2\sin \theta }\right)$.

Conclusion:

Therefore, the expression for the tension $T_1$ is $4mg\sin \theta \left(\frac{1+\sin \theta }{1+2\sin \theta }\right)$ and the expression for the tension $T_2$ is $6mg\sin \theta \left(\frac{1+\sin \theta }{1+2\sin \theta }\right)$.

To determine

The maximum value of $M$.

Answer to Problem 5.88AP

The maximum value of $M$ is $3m\left(\sin \theta +m_{\text{s}}\cos \theta \right)$.

Explanation of Solution

Given info: The coefficient of static friction between mass $m$, $2m$ and the inclined plane is $m_{\text{s}}$.

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

Figure (2)

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

$n=2mg\cos \theta$

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

$f_{\text{s}}=m_{\text{s}}n$ (3)

Here,

$m_{\text{s}}$ is the coefficient of static friction.

$n$ is the normal force.

Substitute $2mg\cos \theta$ for $n$ in the above equation.

$\begin{array}{c}{f}_{\text{s}}=m_{\text{s}}\left(2mg\cos \theta \right)\\ =2m{m}_{\text{s}}g\cos \theta \end{array}$

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

$2ma=T_1-2mg\sin \theta -f_{\text{s}}$

Substitute $2m{m}_{\text{s}}g\cos \theta$ for $f_{\text{s}}$ in the above equation.

$2ma=T_1-2mg\sin \theta -2m{m}_{\text{s}}g\cos \theta$ (4)

Rearrange the above equation for $T_1$.

$\begin{array}{c}{T}_1=2ma+2mg\sin \theta +2m{m}_{\text{s}}g\cos \theta \\ T_1=2m\left(a+g\sin \theta +m_{\text{s}}g\cos \theta \right)\end{array}$

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

$\begin{array}{c}{T}_1=2m\left(0+g\sin \theta +m_{\text{s}}g\cos \theta \right)\\ =2mg\left(\sin \theta +m_{\text{s}}\cos \theta \right)\end{array}$

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

$n=mg\cos \theta$

Substitute $mg\cos \theta$ for $n$ in equation (3).

$f_{\text{s}}=m_{\text{s}}mg\cos \theta$

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

$ma=T_2-T_1-mg\sin \theta -f_{\text{s}}$

Substitute $m_{\text{s}}mg\cos \theta$ for $f_{\text{s}}$ in the above equation.

$ma=T_2-T_1-mg\sin \theta -m_{\text{s}}mg\cos \theta$

Add equation (4) with the above equation.

$\begin{array}{l}2ma+ma=T_1-2mg\sin \theta -2m{m}_{\text{s}}g\cos \theta +T_2-T_1-mg\sin \theta -m_{\text{s}}mg\cos \theta \\ 3ma=-3mg\sin \theta -3m{m}_{\text{s}}g\cos \theta +T_2\\ T_2=3ma+3mg\sin \theta +3m{m}_{\text{s}}g\cos \theta \\ T_2=3m\left(a+g\sin \theta +m_{\text{s}}g\cos \theta \right)\end{array}$

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

$\begin{array}{c}{T}_2=3m\left(0+g\sin \theta +m_{\text{s}}g\cos \theta \right)\\ =3mg\left(\sin \theta +m_{\text{s}}\cos \theta \right)\end{array}$

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

$T_2=M_{\text{max}}g$

Substitute $3mg\left(\sin \theta +m_{\text{s}}\cos \theta \right)$ for $T_2$ in the above equation.

$\begin{array}{c}3mg\left(\sin \theta +m_{\text{s}}\cos \theta \right)=M_{\text{max}}g\\ M_{\text{max}}=3m\left(\sin \theta +m_{\text{s}}\cos \theta \right)\end{array}$

Conclusion:

Therefore, the maximum value of $M$ is $3m\left(\sin \theta +m_{\text{s}}\cos \theta \right)$.

To determine

The minimum value of $M$.

Answer to Problem 5.88AP

The minimum value of $M$ is $3m\left(\sin \theta -m_{\text{s}}\cos \theta \right)$.

Explanation of Solution

From part (e), the expression of tension $T_2$ for the minimum mass is,

$T_2=3mg\left(\sin \theta -m_{\text{s}}\cos \theta \right)$

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

$T_2=M_{\text{min}}g$

Substitute $3mg\left(\sin \theta -m_{\text{s}}\cos \theta \right)$ for $T_2$ in the above equation.

$\begin{array}{c}3mg\left(\sin \theta -m_{\text{s}}\cos \theta \right)=M_{\text{min}}g\\ M_{\text{min}}=3m\left(\sin \theta -m_{\text{s}}\cos \theta \right)\end{array}$

Conclusion:

Therefore, the minimum value of $M$ is $3m\left(\sin \theta -m_{\text{s}}\cos \theta \right)$.

To determine

The difference between the tension $T_2$ for maximum and minimum mass.

Answer to Problem 5.88AP

The difference between the tension for maximum and minimum mass is $6m_{\text{s}}mg\cos \theta$.

Explanation of Solution

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

$M_{\text{max}}=3m\left(\sin \theta +m_{\text{s}}\cos \theta \right)$

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

$M_{\text{min}}=3m\left(\sin \theta -m_{\text{s}}\cos \theta \right)$

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

$T_{2,\text{max}}=M_{\text{max}}g$

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

$T_{2,\text{min}}=M_{\text{min}}g$

Compare both the above equation.

$T_{2,\text{max}}-T_{2,\text{min}}=M_{\text{max}}g-M_{\text{min}}g$

Substitute $3m\left(\sin \theta +m_{\text{s}}\cos \theta \right)$ for $M_{\text{max}}$ and $3m\left(\sin \theta -m_{\text{s}}\cos \theta \right)$ for $M_{\text{min}}$ in the above equation.

$\begin{array}{c}{T}_{2,\text{max}}-T_{2,\text{min}}=\left(3m\left(\sin \theta +m_{\text{s}}\cos \theta \right)\right)g-\left(3m\left(\sin \theta -m_{\text{s}}\cos \theta \right)\right)g\\ =6m_{\text{s}}mg\cos \theta \end{array}$

Conclusion:

Therefore, the difference between the tension for maximum and minimum mass is $6m_{\text{s}}mg\cos \theta$.