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: $mg \sin \theta - \mu mg \cos \theta - F = 0$.