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Force and Friction on an Inclined Plane

by wikidrox
Tags: force, friction, inclined, plane
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wikidrox
#1
Apr15-04, 05:58 PM
P: 44
This question is driving me insane please help.

A 25 kg block of ice slides down an incline 2.5 m long and 1.5 m high. A man pushes up the ice parallel to the incline so it slides down at a constant speed. The coefficient of friction between the ice and incline is 0.10. Find the force exerted by the man.
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Doc Al
#2
Apr15-04, 06:15 PM
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Quote Quote by wikidrox
A 25 kg block of ice slides down an incline 2.5 m long and 1.5 m high. A man pushes up the ice parallel to the incline so it slides down at a constant speed. The coefficient of friction between the ice and incline is 0.10. Find the force exerted by the man.
Sliding at constant speed implies equilibrium: the forces in any direction add to zero.

Start by identifying all the forces on the block of ice. I count four.
bullroar_86
#3
May12-05, 02:13 AM
P: 30
ok I have the same problem.

The four forces I have found are:

Fg = 147.1N (directly down)
FN = 195.9N (perpendicular to the slope)
Ff = 19.6N (up parallel the slope)
F = 147.1N (down parallel the slope)

just want to make sure here..

now I add the 4 vectors together and find out what Force I need to add (up parallel to the slope) in order to make the net force = 0

?


thanks

PhysicsinCalifornia
#4
May12-05, 02:53 AM
P: 58
Force and Friction on an Inclined Plane

Quote Quote by bullroar_86

The four forces I have found are:

Fg = 147.1N (directly down)
FN = 195.9N (perpendicular to the slope)
Ff = 19.6N (up parallel the slope)
F = 147.1N (down parallel the slope)

now I add the 4 vectors together and find out what Force I need to add (up parallel to the slope) in order to make the net force = 0
First, you have to draw your "free body diagram". (Did you do that bullroar?)
There are trig functions that should be applied here because the block is at an incline of [tex]\theta[/tex].
Once you draw it, you should have something like::

[tex] \Sigma F_{x} = F_{g}\sin\theta + F_{push} + f_{k} = ma[/tex]
[tex] \Sigma F_{y} = n - F_{g}\cos\theta = 0[/tex]
Doc Al
#5
May12-05, 08:23 AM
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Quote Quote by bullroar_86
ok I have the same problem.

The four forces I have found are:

Fg = 147.1N (directly down)
FN = 195.9N (perpendicular to the slope)
Ff = 19.6N (up parallel the slope)
F = 147.1N (down parallel the slope)
Not sure where you are getting these numbers. The four forces are:
(1) weight = mg, acting down. This will have components perpendicular and parallel to the incline. If you call the angle that the incline makes with the horizontal [itex]\theta[/itex], then [itex]W_{parallel} = mg \sin \theta[/itex] down the incline and [itex]W_{perpendicular} = mg \cos \theta[/itex] into the incline.

(2) normal force, N, which is the reaction to the forces pulling the block against the incline. [itex]N = mg \cos \theta[/itex], acting out of the plane. (This comes from setting the net force perpendicular to the plane equal to zero.)

(3) friction force = [itex]\mu N = \mu mg \cos \theta[/itex] acting up the plane.

(4) the applied force F that acts up the incline.
To find the applied force, do as PhysicsinCalifornia advised: set the net force parallel to the plane equal to zero: [itex]mg \sin \theta - \mu mg \cos \theta - F = 0[/itex].


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