# Friction on a Ramp

1. Feb 18, 2009

### mikey555

1. The problem statement, all variables and given/known data

A small block of mass m rests on the rough, sloping side of a triangular block of mass M which itself rests on a horizontal frictionless table as shown in the figure.

http://s7.photobucket.com/albums/y252/mikey555/?action=view&current=GIANCOLIch05p033.jpg
(edit: image doesn't appear. it's here:http://s7.photobucket.com/albums/y252/mikey555/?action=view&current=GIANCOLIch05p033.jpg)

If the coefficient of static friction is $$\mu$$ determine the minimum horizontal force F applied to M that will cause the small block m to start moving up the incline.

2. Relevant equations

Newton's 2nd law for each individual block.

3. The attempt at a solution

So first I drew FBDs for both blocks.
- For the small block, there's a frictional force to the right, its weight downward, and a normal force on M on m upward.
- For the wedge, its weight downward, the normal force of the ground on M upward, an applied horizontal force towards the right, and the normal force from m perpendicular to the ramp of the wedge.

I think there might be a force on each that I'm missing. Is there a frictional force back on the wedge from the small block? Also, is there a force pushing the small block up the ramp?

Where I get confused is how to form the equations for x- and y-directions. I need to make an F=ma equation for the horizontal direction of the small block to find out how much external force on the wedge is required for it to move up the ramp, except I don't know which way acceleration is going. And if I'm trying to find out how much force is required to break m's static friction, then won't it not be accelerating in either direction (instantaneously still)?

So basically I'm stuck after making my FBDs. Obviously there's a slough of questions here and I don't expect anyone to answers them all. But I would appreciate it very much if someone could give me some direction on where to go after I make FBDs (and if I have any mistakes on mine!).

Thank you!

Last edited by a moderator: Apr 24, 2017
2. Feb 18, 2009

### Tom Mattson

Staff Emeritus

No, the frictional force always points in the opposite direction of the block's motion relative to the incline. So if the block is to move up the incline, the frictional force points in which direction? Likewise, the normal force always points normal to the surface of the incline (that's why it's called that!). So it isn't pointing upward.

3. Feb 18, 2009

### mikey555

Oops, I've said my directions completely wrong!
The small block has a frictional force down the ramp, a normal force up and perpendicular to the ramp, and weight downward.

4. Feb 20, 2009

### mikey555

I just realized that my last response makes it sound like I solved the problem. I actually drew the directions correctly on my FBDs, but I wrote them out wrong on the forum.

Thanks for the corrections, any more help is appreciated :)

5. Feb 20, 2009

### Delphi51

Wow, this is a mind boggler!
Perhaps it would make sense to play Einstein and think of the acceleration as being like gravity - combine the mg down with the ma to the right and treat the combo as we would normally treat gravity.

Maybe makes more sense to say gravity is the same as accelerating upward at g. Add to that acceleration to the right at a.

Seems to be promising. Combined force on the m is m*sqrt(a^2 + g^2) tilted by angle
A = arctan(a/g) away from the old vertical to the LEFT.

Now it is possible to get expressions for the normal and up-ramp forces on the mass.

Last edited: Feb 20, 2009
6. Feb 20, 2009

### lanedance

So assuming your FBDs are correct...

write separately for both wedge & block
sum (Fx) = max (horiz)
sum (Fy) = may = 0 (vert)

Knowing that
F = mu.N at slip point
ax is same for both wedge & block
F & N are equal & opposite for both wedge & block

should get you pretty close ...

7. Feb 20, 2009

### Delphi51

I would sure be interested in seeing that done lanedance's way.
Combining the acceleration of gravity with the horizontal acceleration, I ended up with
F = (M+m)*g*(u+tan(A))/(1-u*tan(A)) where A = ramp angle, u = coefficient of friction

8. Feb 22, 2009

### mikey555

Thanks for the help, guys!