Chapter 4, Problem 51P

To determine

The time taken by the wet bar of soap to slide down a ramp and reach the bottom of the ramp.

Answer to Problem 51P

Solution:

The bar of soap reaches the bottom of the ramp in 4.8 s.

Explanation of Solution

Given:

The length of the ramp $s=9.0\text{ m}$

The angle at which the ramp is inclined $\theta =8°$

The coefficient of kinetic friction ${\mu }_k=0.060$

Formula used:

The free body diagram of the soap sliding down the ramp is drawn as shown below.

  

Figure-1

The weight mg of the bar of soap acts vertically downwards. The normal force $F_N$ acts perpendicular to the ramp. The weight mg is resolved into two components, $mg\sin \theta \text{ and }mg\cos \theta$ parallel and perpendicular to the ramp. Since, the bar of soap slides downwards, the force of friction $F_{fr}$ acts upwards parallel to the ramp.

Since there is no motion perpendicular to the ramp,

  $F_N=mg\cos \theta$   ......(1)

The force of kinetic friction is related to the normal force as,

  $F_{fr}\text{=}{\mu }_k{F}_N$   ......(2)

Since the bar of soap slides down the ramp, there is a net force F that acts downwards. This is given by,

  $F=mg\sin \theta -F_{fr}$   ......(3)

Substitute equations (1) and (2) in (3).

  $\begin{array}{c}F=mg\sin \theta -{\mu }_{k}mg\cos \theta \\ =mg\left(\sin \theta -{\mu }_k\cos \theta \right)\end{array}$   .......(4)

From Newton’s second law,

  $F=ma$

Here, a is the resultant acceleration of the soap bar down the ramp.

Therefore,

  $\begin{array}{c}ma=mg\left(\sin \theta -{\mu }_k\cos \theta \right)\\ a=g\left(\sin \theta -{\mu }_k\cos \theta \right)\end{array}$   ......(5)

If the bar of soap slides down a distance s on the ramp in a time t , starting from rest, then

  $s=\frac{1}{2}a{t}^2$

Use the expression for acceleration from equation (5) and simplify for t .

  $\begin{array}{c}s=\frac{1}{2}a{t}^2\\ =\frac{1}{2}\left[g\left(\sin \theta -{\mu }_k\cos \theta \right)\right]t^2\\ t=\sqrt{\frac{2s}{g\left(\sin \theta -{\mu }_k\cos \theta \right)}}\end{array}$

Calculation:

Substitute the given values of s , $\theta$ , ${\mu }_k$ from the given values and 9.8 m/s² for g and solve for t .

  $\begin{array}{c}t=\sqrt{\frac{2s}{g\left(\sin \theta -{\mu }_k\cos \theta \right)}}\\ =\sqrt{\frac{2\left(9.0\text{ m}\right)}{\left(9.8{\text{ m/s}}^2\right)\left[\left(\sin 8°\right)-\left(0.060\right)\left(\cos 8°\right)\right]}}\\ =4.798\text{ s}\end{array}$

The time taken by the bar to reach the bottom of the ramp is 4.8 s.

Conclusion:

Using a free body diagram, the forces acting on the bar of soap is calculated. Whereas, using the equations of motion and Newton’s second law, the time taken to reach the bottom of the slope of length 9.0 m is calculated to be 4.8 s.