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

Question

**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?**

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General Physics
4 years
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## Answer ( 1 )

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