# Force and Friction on an Inclined Plane

by wikidrox
Tags: force, friction, inclined, plane
 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.
Mentor
P: 41,579
 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.
 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
P: 58
Force and Friction on an Inclined Plane

 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 $$\theta$$.
Once you draw it, you should have something like::

$$\Sigma F_{x} = F_{g}\sin\theta + F_{push} + f_{k} = ma$$
$$\Sigma F_{y} = n - F_{g}\cos\theta = 0$$
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P: 41,579
 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 $\theta$, then $W_{parallel} = mg \sin \theta$ down the incline and $W_{perpendicular} = mg \cos \theta$ into the incline.

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

(3) friction force = $\mu N = \mu mg \cos \theta$ 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: $mg \sin \theta - \mu mg \cos \theta - F = 0$.

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