Chapter 4, Problem 152P

(a)

To determine

The magnitude of maximum tension when the cart rolls on a horizontal surface.

Answer to Problem 152P

The magnitude of maximum tension when the cart rolls on a horizontal surface is $\left(m_1+m_2\right){\mu }_{s}g$.

Explanation of Solution

The mass of the cart is $m_1$ and mass of the block is $m_2$.

Write the expression to calculate the acceleration of the system.

Here, a is the acceleration of the system.

Rewrite the above equation in terms of a.

$a=\frac{T}{\left(m_1+m_2\right)}$ (I)

Write the equilibrium condition of forces in vertical direction for the system.

$N-m_{2}g=0$ (II)

Here, N is the normal force and g is the acceleration due to gravity.

Rewrite the above equation in terms of N.

$N=m_{2}g$

The block is not sliding means the magnitude of frictional force and the force on the block would be equal.

Write the expression to calculate the equilibrium of force in horizontal direction.

${\mu }_{s}N=m_{2}a$

Substitute $N=m_{2}g$ in the above equation to rewrite.

${\mu }_s{m}_{2}g=m_{2}a$ (III)

Substitute equation (I) in the above equation to calculate T.

$\begin{array}{c}{\mu }_s{m}_{2}g=m_2\left(\frac{T}{\left(m_1+m_2\right)}\right)\\ \frac{T}{\left(m_1+m_2\right)}={\mu }_{s}g\\ T=\left(m_1+m_2\right){\mu }_{s}g\end{array}$

Conclusion:

Therefore, the magnitude of maximum tension when the cart rolls on a horizontal surface is $\left(m_1+m_2\right){\mu }_{s}g$.

(b)

To determine

The magnitude of maximum tension when the cart rolls on a ramp of angle $\theta$ above the horizontal.

Answer to Problem 152P

The magnitude of maximum tension when the cart rolls up on a ramp is $g\left(m_1+m_2\right)\left({\mu }_s\cos \theta +\sin \theta \right)$.

Explanation of Solution

Write the expression to calculate the acceleration of the system.

$T-\left(m_1+m_2\right)g\sin \theta =\left(m_1+m_2\right)a$

Here, a is the acceleration of the system.

Rewrite the above equation in terms of a.

$\begin{array}{c}T-\left(m_1+m_2\right)g\sin \theta =\left(m_1+m_2\right)a\\ a=\frac{T-\left(m_1+m_2\right)g\sin \theta }{\left(m_1+m_2\right)}\\ =\frac{T}{\left(m_1+m_2\right)}-g\sin \theta \end{array}$ (IV)

Write the equilibrium condition of forces in vertical direction for the system.

$N-m_{2}g\cos \theta =0$

Here, N is the normal force and g is the acceleration due to gravity.

Rewrite the above equation in terms of N.

$N=m_{2}g\cos \theta$ . (V)

The block is not sliding means the magnitude of frictional force and the force on the block would be equal.

Write the expression to calculate the equilibrium of force in horizontal direction.

${\mu }_{s}N=m_{2}a$

Substitute equation (IV) and (V) in the above equation to calculate T.

$\begin{array}{c}{\mu }_s\left(m_{2}g\cos \theta \right)=m_2\left(\frac{T}{\left(m_1+m_2\right)}-g\sin \theta \right)\\ \frac{T}{\left(m_1+m_2\right)}-g\sin \theta ={\mu }_{s}g\cos \theta \\ \frac{T}{\left(m_1+m_2\right)}=\left({\mu }_s\cos \theta +\sin \theta \right)g\\ T=g\left(m_1+m_2\right)\left({\mu }_s\cos \theta +\sin \theta \right)\end{array}$

Conclusion:

Therefore, the magnitude of maximum tension when the cart rolls on a horizontal surface is $g\left(m_1+m_2\right)\left({\mu }_s\cos \theta +\sin \theta \right)$.