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A 52.5-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.679 and 0.418, respectively. What horizontal pushing force is required to (a)just start the crate moving and (b) slide the crate across the dock at a constant speed?

Question

A 52.5-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.679 and 0.418, respectively. What horizontal pushing force is required to (a)just start the crate moving and (b) slide the crate across the dock at a constant speed?

check_circleAnswer
Step 1

The mass of the crate = 52.5 kg.

The coefficient of static friction = 0.679.

The coefficient of kinetic friction=0.418

(a)

Since the crate rests on level surface there are two forces acting on crate. One is normal force acting upwards, and other is the weight of the crate which is acting downwards.

Write the expression for force acting on crate

here
is the mass
F F-mg 0
T)
g is the acceleration due to gravity
F is the normal force
Fmg
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Image Transcriptionclose

here is the mass F F-mg 0 T) g is the acceleration due to gravity F is the normal force Fmg

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Step 2

The horizontal force needed to just start the crate moving is the static force.

Write the expression for kinetic force.

[substitute, mg for F]
here, 52.5kg for m,
9.80m/s for g
0.679 for u
=4mg
-(0.679)(52.5kg)(9.80m/s)
=349.3N
help_outline

Image Transcriptionclose

[substitute, mg for F] here, 52.5kg for m, 9.80m/s for g 0.679 for u =4mg -(0.679)(52.5kg)(9.80m/s) =349.3N

fullscreen
Step 3

(b)The horizontal force required to slide the crate is the kinetic for...

[substitute, mg for F
here, 52.5kg for m
9.80m/s for g
=4mg
0.418 for 4
(0.413)(52.5kg) (9.80m/s
= 215N
help_outline

Image Transcriptionclose

[substitute, mg for F here, 52.5kg for m 9.80m/s for g =4mg 0.418 for 4 (0.413)(52.5kg) (9.80m/s = 215N

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