Question

A clever device gently holds a sheep against a wall by compressing a spring, pushing the sheep against the wall, where it is held by static friction. The sheep has a mass of 31.3 kg, the coefficient of static friction is µ_s=0.97, and the spring has a spring constant of k=652 N/m. What is the minimum compression required to keep the sheep in place without falling downward?

Answer 1

solution:

We are given μ_s = 0.97

spring constant of k =652 N/m

The mass of the sheep m = 31.3 kg

then the weight of sheep W = mg = 31.3 x 9.8 = 306.74 N

The force of static friction is given by F = μ_sN.

The minimum force of friction required is equal to the weight of the sheep. so we can solve for N.

0.97 x N = 306.74

N = 316.2268 N

To push the sheep with 316.2268 N of force, the compression x must satisfy

k*x = 316.2268 N

652*x = 316.2268 N

minimum compression in the spring x = 0.485 m = 48.5 cm