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_{s}**N**.

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**

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You are on an amusement park ride with your back against the
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(c) If the radius of the cylinder...

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