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A 60 kg package rests on the level floor of a warehouse.

1.) A 60 kg package rests on the level floor of a warehouse. If the coefficients of static and kinetic frictions are 0.56 and 0.37 respectively, what horizontal pushing force is required to
(a) start the package’s motion?
(b) slide the package across the floor at constant velocity?


60kg at rest
Find all the forces on the object
There is gravity…and Normal force pushing upward
Since the normal force opposes gravity and the object is not acclerating, the normal force = the force of gravity = mg

mg = 60kg x 9.8m/s^2 = 588N

mu = coefficient of friction
Ff=Force of friction
Fn = Force normal

mu = Ff/Fn

mu= 0.56
Fn= 588N

0.56 = Ff / 588N
0.56 x 588N = Ff
Ff = 329.28
The force of friction holding the object in place is 329.28N. NOTE that there is no force of friction YET because the object is at REST. But that’s how much resistance friction can apply
SO you need a force GREATER THAN 329.28N to START its motion

mu = Ff/Fn
0.37 = Ff/588N
0.37 x 588N = Ff
Ff = 217.56N
The maximum force of friction that can be applied at motion is 217.56N
To keep at a constant velocity, the force has to = 217.56

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