Question

A stubborn, 120 kg mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around the mule and pulls with his maximum force of 800 N. The coefficients of friction between the mule and the ground are μs=0.8 and μk=0.5. Is the farmer able to move the mule?

Answer 1

Answer:

No

Explanation:

In order for the farmer to be able to put the mule in motion, the force he applies must be larger than the maximum value of the force of static friction exerted by the ground on the mule.

The maximum value of the force of static friction on the mule is given by:

$F_f=\mu_s mg$

where:

$\mu_s=0.8$ is the coefficient of static friction

m = 120 kg is the mass of the mule

$g=9.8 m/s^2$ is the acceleration due to gravity

Substituting, we find:

$F_f=(0.8)(120)(9.8)=941 N$

Here in this problem, the force applied by the farmer through the rope is

F = 800 N

We see that this force is smaller than the value of the maximum force of friction: therefore, the farmer will not be able to move the mule.