# Friction and moving blocks on inclines

1. Jul 23, 2008

### WarPhalange

This is a GRE question, but it's so basic I thought I'd just put it here. Here it is with the solution:

http://grephysics.net/ans/9277/6 The answer is "D" by the way.

So my problem is with the way the weight of the block is distributed.

I get that each triangular block only "feels" half of the square block, so M/2 of mass. But then they assume that's the horizontal force via some weird trig that seems pointless (I can't figure out where they get a square-root 2 from).

My understanding was that you get half of the block's weight pushing down on the triangular block, and half pushing it horizontally. So you'd have g*M/4 pushing straight down and (m+M/2)*g*u pushing horizontally, for each block.

Why am I wrong?

2. Jul 23, 2008

### Staff: Mentor

Since the triangles are frictionless, they can only exert forces perpendicular to their surface. It's the vertical component of that normal force that must equal half the weight of the square block. (The weight of the block does not act directly on the triangles; the only force between block and triangle is the normal force.)

3. Jul 23, 2008

### WarPhalange

So if the Square Block (SB) is only exerting force downwards due to gravity, F = g*M/2, right?

Then what does the triangular block exert back? g*M/2 in the Normal direction, or g*M in the Normal direction in order to get g*M/2 straight up? If it's the former, it would seem like horizontal force is just g*M/4 in that case. If it's the latter, how come it is "pushing back" with more force than the block is pushing with on it?

4. Jul 23, 2008

### Staff: Mentor

No. That's just the vertical component of the force it exerts on the triangles. Mg is the force of gravity on the block, not the force that the block exerts on the triangles.

It's certainly true that the force the block exerts on the triangle must be equal and opposite to the force that the triangle exerts on the block.

5. Jul 23, 2008

### WarPhalange

Hold on, there are TWO triangular blocks here. Doesn't each only get g*M/2 of Force exerted on it? Then a further half of that goes towards the horizontal component?

6. Jul 23, 2008

### Staff: Mentor

The vertical component of the force exerted on each triangle is g*M/2.
What do you mean "half"? It's true that the vertical and horizontal components must be equal, since the triangles are at 45 degrees.

7. Jul 23, 2008

### WarPhalange

Sure. Each triangle gets half of the weight distributed on to it due to symmetry, right?

Okay, so if the Square Block exerts M/2*g force onto a Triangle Block, the Triangle Block exerts M/2*g force back onto the Square Block vertically? Where does the horizontal component come in?

8. Jul 23, 2008

### Staff: Mentor

As I said before, the vertical component of the force exerted by the block on each triangle equals M*g/2.
The force between block and triangle is a normal force which acts perpendicular to the surface of the triangle. It has both vertical and horizontal components.

I suggest drawing a freebody diagram of the triangles.

9. Jul 23, 2008

### WarPhalange

I have, but I don't know what values to put where.

If the vertical component = M*g/2, then the horizontal is the same, because of the 45 degree incline? So because both the left and right blocks both exert M*g/2 force horizontally, but in opposite directions, the net force is 0 horizontally and only M*g vertically from both blocks, right?

So that much makes sense (if I explained it correctly). I *think* I'm understanding it...

Thanks a lot Doc Al. You really stuck with me even though it looked like I was a lost cause. I appreciate it.

10. Jul 23, 2008

### Staff: Mentor

Right. Sounds like you've got it now.

11. Jul 23, 2008

### WarPhalange

I should hope so, I graduate with a physics BS this coming year. :tongue: