Chapter 5, Problem 77QAP

To determine

(a)

If the crate resting on the bed of a decelerating truck would slide during the braking period.

Answer to Problem 77QAP

The crate will not slide, since the acceleration it experiences due to the truck's deceleration is less than the acceleration it experiences due to the force of static friction.

Explanation of Solution

Given:

The mass of the crate

  $m=150\text{ kg}$

Initial velocity of the truck

  $u=50.0\text{ km/h}$

Final velocity of the truck

  $v=0\text{ m/s}$

Time during which the truck comes to a stop

  $\Delta t=12.0\text{ s}$

Coefficient of static friction between the truck and the crate

  ${\mu }_s=0.655$

Formula used:

To find if the crate would slide or not, a free body diagram is drawn for the crate and the force equations for equilibrium and for motion of the crate are determined.

The free body diagram for the crate is shown below:

  

The weight of the block $\overrightarrow{w}$ acts downwards, the normal force $\overrightarrow{n}$ is exerted on the crate by the bed of the truck. The truck moves along the +x direction initially, before the brakes are applied. The decoration of the truck exerts a force $\overrightarrow{F}$ in the −x direction of the crate. The crate would tend to slip along the −x direction and hence, the force of friction ${\overrightarrow{f}}_s$ acts along the +x direction.

The block is in equilibrium along the vertical ( y ) direction.

Therefore,

  ${\displaystyle \sum F_y}=n-w=0$

Since, $w=mg$, where, g is the acceleration of free fall with a value 9.80 m/s².

Therefore,

  $n=w=mg$

If ${\mu }_s$ is the coefficient of static friction between the blocks, then

  $f_s={\mu }_{s}n={\mu }_{s}mg$......(1)

If the force of friction produces an acceleration, $a_f$, then by Newton's second law,

  $f_s=m{a}_f$......(2)

Write an expression for the acceleration due to the force of friction using equations (1) and (2).

  $\begin{array}{c}m{a}_f={\mu }_{s}mg\\ a_f={\mu }_{s}g\end{array}$......(3)

The expression for the acceleration produced by the truck is given by,

  $a=\frac{v-u}{\Delta t}$......(4)

Calculation:

Calculate the acceleration $a_f$ by substituting the values of the variables in equation (3).

  $\begin{array}{c}{a}_f={\mu }_{s}g\\ =\left(0.655\right)\left(9.80{\text{ m/s}}^2\right)\\ =6.419{\text{ m/s}}^2\end{array}$

Express the initial velocity of the crate in m/s.

  $\begin{array}{c}u=50.0\text{ km/h}\times \frac{1000\text{ m}}{1\text{ km}}\times \frac{1\text{ h}}{3600\text{ s}}\\ =13.89\text{ m/s}\end{array}$

Calculate the acceleration a experienced by the crate due to the truck's decoration by substituting the values of the variables in equation (4).

  $\begin{array}{c}a=\frac{v-u}{\Delta t}\\ =\frac{\left(0\text{ m/s}\right)-\left(13.89\text{ m/s}\right)}{12.0\text{ s}}\\ =-1.158{\text{ m/s}}^2\end{array}$

This acceleration acts along the −x direction.

The crate slides if a >a_f.

Conclusion:

It can be seen that the acceleration in the +x direction is 6.419 m/s² and that along the −x direction is 1.158 m/s². Since $a<a_f$, the crate will not slide.

To determine

(b)

The minimum stopping time for the truck, that prevents the crate from sliding.

Answer to Problem 77QAP

For the crate to not slide on the bed of the truck, the minimum stopping time of the truck is 2.16 s.

Explanation of Solution

Given:

Acceleration produced by the force of friction

  $a_f=6.419{\text{ m/s}}^2$

Initial velocity of the truck

  $u=13.89\text{ m/s}$

Final velocity of the truck

  $v=0\text{ m/s}$

Formula used:

The crate will just remain in equilibrium, if the acceleration acting on it along the −x direction due to the truck's deceleration is equal to the acceleration along the +x direction due to the force of friction.

  $a=-a_f=\frac{v-u}{\Delta t}$

Calculation:

Rewrite the expression for $\Delta t$.

  $\Delta t=\frac{v-u}{-a_f}$

Substitute the values of the variables in the equation and solve for $\Delta t$.

  $\begin{array}{c}\Delta t=\frac{v-u}{-a_f}\\ =\frac{\left(0\text{ m/s}\right)-\left(13.89\text{ m/s}\right)}{-6.419{\text{ m/s}}^2}\\ =2.16\text{ s}\end{array}$

Conclusion:

Thus, for the crate to not slide on the bed of the truck, the minimum stopping time of the truck is 2.16 s.