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2 years ago

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Suppose you have to move a heavy crate of weight 875 N by sliding it along a horizontal concrete floor. You push the crate to th

e right with a horizontal force of magnitude 300 N, but friction prevents the crate from sliding. What is the magnitude Fp of the minimum force you need to exert on the crate to make it start sliding along the floor? Let the coefficient of static friction ?s between the crate and the floor be 0.56 and that of kinetic friction, ?k, be 0.47.

1 answer:

AnnZ [28]2 years ago

7 0

Answer:

<u>490 N</u>

Explanation:

<u>1) Data:</u>

a) W = 875 N

b) Fx = 300 N (right direction = positive)

c) Fp = ?

d) μs = 0.56

e) μk = 0.47

<u>2) Physical principles and formulae</u>:

a) For sliding, Fp ≥ μs × N, where Fp = μs × N is the magnitud of the minimum force need to exert on the crate to make it start sliding

<u>3) Solution:</u>

<u>a) Free body diagram</u>

The balance of the vertical forces implies that the normal force (N) equals the weight (W) of the crate:

N = W = 875 N

b) Fp = μs × N = 0.56 × 875 N = <u>490 N ← answer</u>

<u>Remarks</u>:

Since the minimum force to make the crate start sliding along the floor is 490 N, when you push the crate to the right with a horizontal force (Fx) of magnitude 300N, the crate will not move.

You use the coefficient of static friction, μs, to determine the amount of force needed to make the crate start sliding. The coefficient of kinetic friction, μk, is used once the object is in motion, not before.

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