Chapter 5, Problem 88AP

(a)

To determine

The mass M in terms of m, g , and $\theta$

Answer to Problem 88AP

The mass M in terms of m, g , and $\theta$ is $3m\sin \theta$

Explanation of Solution

The system is in equilibrium, so force on either sides of the pulley must be the same.

Write the expression for the equilibrium.

  $Mg=mg\sin \theta +2mg\sin \theta$

Here, $M$ is the mass of the hanging from the pulley, $m$ is the mass on the inclined plane, $g$ is the acceleration due to gravity, and $\theta$ is the angle of the incline.

Conclusion:

Simplify the above equation.

    $\begin{array}{c}Mg=3mg\sin \theta \\ M=3m\sin \theta \end{array}$

Therefore, the mass M in terms of m, g, and $\theta$ is $3m\sin \theta$

(b)

To determine

The tension $T_1$ and $T_2$ .

Answer to Problem 88AP

The tension $T_1$ and $T_2$ is $T_1=2mg\sin \theta$ and $T_2=3mg\sin \theta$

Explanation of Solution

Below figure shows the forces and tension acting on the string.

$T_2$ supports both the below masses.

From the figure,

  $\begin{array}{c}{T}_2=mg\sin \theta +2mg\sin \theta \\ =3mg\sin \theta \end{array}$

$T_1$ supports one mass.

From the figure,

  $T_1=2mg\sin \theta$

Conclusion:

Therefore, the tension $T_1$ and $T_2$ is $T_1=2mg\sin \theta$ and $T_2=3mg\sin \theta$

(c)

To determine

The acceleration of each object

Answer to Problem 88AP

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

Explanation of Solution

Below figure shows the forces and tension acting on the string.

Use Newton’s second law to write the expression for tension $T_1$ .

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

Use Newton’s second law to write the expression for tension $T_2$ .

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

Substitute $6m\sin \theta$ for $M$ in the above equation for $T_2$ and simplify.

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

Use Newton’s second law to write the expression for tension $T_2-T_1$ .

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

Conclusion:

Substitute equation (I) and (II) in (III) and simplify.

    $\begin{array}{l}6m\sin \theta \left(g-a\right)-2m\left(g\sin \theta +a\right)=m\left(g\sin \theta +a\right)\\ 6mg\sin \theta -6ma\sin \theta -2mg\sin \theta -2ma=mg\sin \theta +ma\\ ma+2ma+6ma\sin \theta =4mg\sin \theta -mg\sin \theta \\ 3ma\left(1+2\sin \theta \right)=3mg\sin \theta \\ a=\frac{g\sin \theta }{1+2\sin \theta }\end{array}$

All the objects are connected to a single string and hence they have same acceleration.

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

(d)

To determine

The tension $T_1$ and $T_2$

Answer to Problem 88AP

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

Explanation of Solution

Rewrite (I).

  $T_1=2m\left(g\sin \theta +a\right)$

Rewrite (II).

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

Conclusion:

Substitute $\frac{g\sin \theta }{1+2\sin \theta }$ for $a$ in (I) and simplify.

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

Substitute $\frac{g\sin \theta }{1+2\sin \theta }$ for $a$ in (II) and simplify.

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

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

(e)

To determine

The maximum value of $M$ .

Answer to Problem 88AP

The maximum value of $M$ is $M_{\max }=3m\left(\sin \theta +{\mu }_s\cos \theta \right)$

Explanation of Solution

Below figure shows the forces acting on the system (including frictional force).

From the figure

Write the expression for friction force acting on $2m$.

    $f_1=2{\mu }_{s}mg\cos \theta$

Here, $f_1$ is the frictional force and ${\mu }_s$ is the coefficient of static friction.

Write the expression for friction force acting on $m$.

    $f_2={\mu }_{s}mg\cos \theta ,$

Write the expression for tension $T_1$.

    $F_{\text{net}}=T_1-2mg\sin \theta -f_1$                                                                             (III)

The system is in equilibrium, $a=0$, $F_{\text{net}}=0$.

Substitute 0 for $F_{\text{net}}$ and $2{\mu }_{s}mg\cos \theta$ for $f_1$ in (III) and simplify.

    $\begin{array}{c}0=T_1-2mg\sin \theta -2{\mu }_{s}mg\cos \theta \\ T_1=2mg\sin \theta +2mg{\mu }_s\cos \theta \\ =2mg\left(\sin \theta +{\mu }_s\cos \theta \right)\end{array}$

Similarly, Write the expression for tension $T_2$.

    $F_{\text{net}}=T_2-M_{\max }g$                                                                                         (IV)

Substitute 0 for $F_{\text{net}}$ in (IV) and simplify.

    $\begin{array}{c}0=T_2-M_{\max }g\\ T_2=M_{\max }g\end{array}$                                                                                             (V)

Write the expression for $T_2-T_1$.

    $F_{\text{net}}=T_2-T_1-mg\sin \theta -{\mu }_{s}mg\cos \theta$                                                           (VI)

Substitute 0 for $F_{\text{net}}$ in (VI) and simplify.

    $\begin{array}{c}0=T_2-T_1-mg\sin \theta -{\mu }_{s}mg\cos \theta \\ T_2-T_1=mg\sin \theta +{\mu }_{s}mg\cos \theta \\ =mg\left(\sin \theta +{\mu }_s\cos \theta \right)\end{array}$                                                     (VII)

Substitute $2mg\left(\sin \theta +{\mu }_s\cos \theta \right)$ for $T_1$ in (VII) and simplify.

    $\begin{array}{c}{T}_2=2mg\left(\sin \theta +{\mu }_s\cos \theta \right)+mg\left(\sin \theta +{\mu }_s\cos \theta \right)\\ =3mg\left(\sin \theta +{\mu }_s\cos \theta \right)\end{array}$

Conclusion:

Equate (V) and (VII).

    $\begin{array}{c}{M}_{\max }g=3mg\left(\sin \theta +{\mu }_s\cos \theta \right)\\ M_{\max }=3m\left(\sin \theta +{\mu }_s\cos \theta \right)\end{array}$                                                                    (VIII)

Therefore, the maximum value of $M$ is $M_{\max }=3m\left(\sin \theta +{\mu }_s\cos \theta \right)$

(f)

To determine

The minimum value of $M$ .

Answer to Problem 88AP

The minimum value of $M$ is $M_{\min }=3m\left(\sin \theta -{\mu }_s\cos \theta \right)$

Explanation of Solution

For minimum value of $M_{\min }$, the masses $m$ and $2m$ moves down the incline and $M$ moves up. Only the direction of the frictional force changes in the expression and other remains same as $M_{\max }$.

Below figure shows the forces acting on the system (including frictional force).

Therefore,

  $T_1=2mg\left(\sin \theta -{\mu }_s\cos \theta \right)$

  $T_2-T_1=mg\left(\sin \theta -{\mu }_s\cos \theta \right)$

  $T_2=M_{\min }g$

  $T_2=3mg\left(\sin \theta -{\mu }_s\cos \theta \right)$

Conclusion:

Equate both the above expression for $T_2$.

    $\begin{array}{c}{M}_{\min }g=3mg\left(\sin \theta -{\mu }_s\cos \theta \right)\\ M_{\min }=3m\left(\sin \theta -{\mu }_s\cos \theta \right)\end{array}$                                                                 (IX)

Therefore, the minimum value of $M$ is $M_{\min }=3m\left(\sin \theta -{\mu }_s\cos \theta \right)$

(g)

To determine

Comparison the values of $T_2$ when $M$ has its minimum and maximum values.

Answer to Problem 88AP

Comparison the values of $T_2$ when $M$ has its minimum and maximum values is $6{\mu }_{s}mg\cos \theta$

Explanation of Solution

Compare (IX) and (VIII).

    $\begin{array}{c}{M}_{\max }g-M_{\min }g=3mg\left(\sin \theta +{\mu }_s\cos \theta \right)-\left[3mg\left(\sin \theta -{\mu }_s\cos \theta \right)\right]\\ =6{\mu }_{s}mg\cos \theta \end{array}$

Conclusion:

Therefore, the Comparison of the values of $T_2$ when $M$ has its minimum and maximum values gives $6{\mu }_{s}mg\cos \theta$