Force and Friction on an Inclined Plane

[wikidrox]

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.

 

[Doc Al]

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]

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]

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

 

[Doc Al]

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.