Finding Applied Force from Coefficient of Friciton and mass

AI Thread Summary
To prevent the 3.0 kg box from sliding down a wall, the applied force must counteract the gravitational force acting on the box. The coefficient of static friction is 0.60, which means the frictional force can be calculated using the equation FF = uFn. The normal force (Fn) is equal to the weight of the box, calculated as Fn = mg, resulting in FF = 0.6 * (3 kg * 9.81 m/s²) = 17.658 N. The upward frictional force must equal the downward gravitational force for the box to remain stationary. Therefore, an applied force of at least 17.658 N is required to keep the box from sliding down.
jmcpherson82
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So I've been stuck on this type of question, please help!

You are pushing horizontally on a 3.0 kg box of wood, pressing against a wall. If the coefficient of static friction is 0.60, how much force must you exert on the block to prevent it from sliding down?


Homework Statement


What is the force applied to keep the box from sliding down?

m= 3.0kg
u= o.60
g=9.81


Homework Equations



FF=uFn
Fnet = ma


The Attempt at a Solution



I tried to use the equation FF=uFn:

FF = .6(3)(9.81)
FF=17.658 N


Am I doing this correct?
 
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Hi jmcpherson82. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif


The vertical forces must balance. The upward force due to friction = 0.6 Fn
 
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So how would I find Fnet without acceleration? Because then I can't do the equation Fnet = ma
 
jmcpherson82 said:
So how would I find Fnet without acceleration? Because then I can't do the equation Fnet = ma

Fnet=0
EDIT: Oh, N.O.'s talking about normal force not net force.
 
jmcpherson82 said:
So how would I find Fnet without acceleration? Because then I can't do the equation Fnet = ma
The upward force = the downward force

The upward force is due to friction.
 

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