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You must push a crate across a floor to a docking bay. The crate weighs 165 N. The coefficient of static friction between crate and floor is 0.510, and the coefficient of kinetic friction is 0.32. Your force on the crate is directed horizontally. (a) What magnitude of your push puts the crate on the verge of sliding? (b) With what magnitude must you then push to keep the crate moving at a constant velocity? (c) If, instead, you then push with the same magnitude as the answer to (a), what is the magnitude of the crate’s acceleration?

Short Answer

Expert verified

a)Fapplied=84.15N

b)Fapplied=52.8N

c)a=1.87m/s2

Step by step solution

01

Given

Weight of the crate:W=mg=165N

Coefficient of static friction: μs=0.51

Coefficient of kinetic friction:μk=0.32

02

Understanding the concept

The problem deals with the Newton’s laws of motion which describe the relations between the forces acting on a body and the motion of the body. Also it involves friction force. Draw a free body diagram for the given situation and then apply Newton’s first and second law.

Formula:

Frictional force is given by,

fk=μkFN

03

Draw the free body diagram

04

 Calculate the magnitude of your push puts the crate on the verge of sliding

(a)

The push or the Fappliedto get the crate just moving the applied force must be as big as the maximum force of static friction = fs.max=μsFNand here FN=W=165N

Hence, required push,

Fapplied=fs.max=μsFNFapplied=0.51×165NFapplied=84.15N

05

Step 5: Find out with what magnitude the crate must be pushed to keep it moving at a constant velocity

(b)

While in motion, constant velocity (zero acceleration) is maintained if the push is equal to the kinetic friction force

Fapplied=fk=μsFNFapplied=0.32×165NFapplied=52.8N

06

Step 6: If the crate is pushed with the same magnitude as the answer to (a), calculate the magnitude of the crate’s acceleration

(c)

Mass of the crate,

m=165N9.8m/s2=16.84kg

The acceleration, using the push from part (a), is:

a=84.15N-52.8N16.84kg

a=1.87m/s2

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