# Homework Help: Interesting Newtonian problem involving friction

1. Jun 28, 2013

### PsychonautQQ

1. The problem statement, all variables and given/known data
Two triangular wedges are placed next to each other on a flat floor. Both triangles have two 45 degree angles where they touch the floor and a 90 degree angle up in the air. The Triangular wedges have a mass of m. A cube of mass M is balanced on the wedges so it's weight is even distributed over both of them. The coefficient of static friction is less than one. What is the largest M that can be balanced without the wedges moving?

If this isn't making sense then it's problem #6 on this http://www.physics.ohio-state.edu/undergrad/greStuff/exam_GR9277.pdf

2. Relevant equations
Frictional force = Normal force * Coefficient of friction
Normal Force = mgcos(theta)
*cos(45) = sqrt(2)/2

3. The attempt at a solution
So my idea was to make the force of the big block onto the little block ((M/2)*g*(sqrt(2)/2)) equal the static friction force. The normal force acting up from the bottom of the small block i believe would be ((M/2)g + mg). So i set ((M/2)*g*(sqrt(2)/2)) = ((M/2)g + mg)μ. The answer i'm looking for is ((2μm) / (1-μ)) and i must have set it up wrong because I don't see that happening.

Last edited: Jun 28, 2013
2. Jun 28, 2013

### TSny

Hello, PsychonautQQ. Welcome to PF!

Can you explain how you got this expression? Does it represent the magnitude of the force of the cube on one of the little blocks, or does it represent the horizontal component of the force? Either way, I don't think it's correct.

Did you mean to include the coefficient of friction somewhere in this equation?

3. Jun 28, 2013

### PsychonautQQ

Yeah it was suppose to be the horizontal component M/2*g*cos(45)

Oops I meant to say ((M/2)g + mg)μ = (MgSqrt(2))/4. The term on the left was suppose to be the static friction force between one of the triangle wedges and the ground.

4. Jun 28, 2013

### TSny

OK, the left side looks good. Can you show how you got the right side?

5. Jun 28, 2013

### PsychonautQQ

Well, the force of the big block onto the small block would be M/2*g. I believe the force it exerts in the horizontal direction would be M/2*g*cos(45) with cos(45) = sqrt(2)/2. So it can be reduced to Mg*sqrt(2)/4

6. Jun 28, 2013

### TSny

What is the direction of the force of the big block on the little block? What is the direction of the force of the little block on the big block? Did you draw free body diagrams?

7. Jun 28, 2013

### PsychonautQQ

I did draw free body diagrams, though i'm not that great at it. I believe their is a normal force at 45 degree's above the horizontal on which the small block exerts onto the big block, and then in the opposite direction a (weight?) force that the big block exerts onto the small block. It seems natural to use cos(45) to get the horizontal component parallel to the floor?

8. Jun 28, 2013

### D H

Staff Emeritus
Much easier approach:
- What does the vertical component of the normal force have to be?
- Given that 45 degree angle, what does this mean the horizontal component of the normal force has to be?

9. Jun 28, 2013

### TSny

Yes. Good.

For the moment, let's stick with just the forces acting on the big block. You have already identified two of these forces, namely, the normal forces from the little blocks that act at 45o angles to the horizontal. There is one more force that acts on the big block. What is it and in what direction does it act?

10. Jun 28, 2013

### PsychonautQQ

The normal force between the small block and the big block? Wouldn't the vertical component of that normal force have to be Mg*sin(45) which again equals Mg*Sqrt(2)/2?
so the horizontal and vertical components are equal? Is that what you are getting at?

Or maybe you mean the Vertical component of the normal force has to equal the vertical component of the gravitational force??

11. Jun 28, 2013

### D H

Staff Emeritus
No.

The square block is not moving, so the net force must be zero. What does this mean as far as the vertical forces on the block? Draw a free body diagram.

Yes. That is what exactly I'm getting at. The vertical component of the normal force is easy. Because the horizontal and vertical components are equal thanks to that 45 degree angle, the horizontal component is also easy.

12. Jun 28, 2013

### PsychonautQQ

Gravitational force acting straight down on the block which is Mg. Since the block isn't moving that means that Mg - Fn(1y) - (Fn(2y) = 0 (Sorry if this is bad notation i'm still new to this, 1y = the first blocks y component normal.) But this reduces down to Mg = Mg(Sqrt(2)/2) which is not correct ;-/. (Since Fn(1y) = (M/2)g*sin45 and same for Fn(2y)) Ahhh what am i doing wrong ;-/

13. Jun 28, 2013

### TSny

Your assumption that Fn(1y) = (M/2)g*sin45 is wrong. Let the equation Mg - Fn(1y) - (Fn(2y) = 0 tell you the value of Fn(1y).

14. Jun 28, 2013

### PsychonautQQ

Fn(1y) + Fn(2y) = Mg so Fn(1y) and Fn(2y) = (Mg/2).
Since the vertical and horizontal components are the same, that means that the Fn(x) = (Mg/2)

So the big block puts Mg/2 force onto the small block in the horizontal direction. That force has to equal the static friction force from the small block against the floor.

Mg/2 = ((M/2)*g + mg)μ
M/2 = (Mμ)/2 + mμ
M/2 - (Mμ)/2 = mμ
M-Mμ = 2mμ
M = 2mμ / (1-μ)

Thank you so much!!!
A little confused conceptually though, the big block put's Mg/2 force ALL into the horizontal direction for the small block?? The block M's force due to gravity is 100% in the x component of the block under it? In my misguided brain it seems like (M/2) * g * cos(45) = Fn(1x). Any idea where this misguided idea stems from or something? anyway thanks a ton i appreciate it :D

15. Jun 28, 2013

### D H

Staff Emeritus
You apparently assumed that the normal force had to have a magnitude of Mg. Don't do that.

The normal force is an example of a constraint force, which means it's constrained by the geometry of the problem. Let the geometry tell you what those constraint forces must be, assuming the geometry is correct. In this case, the geometry wouldn't be correct if the mass of the square block exceeds 2mμ/(1-μ). The triangular blocks would slide out of the way. But so long as the constraints are satisfied, the geometry (rather than guesses) will tell you what those constraint forces have to be.

16. Jun 28, 2013

### TSny

Good work!

The big block exerts a horizontal force of Mg/2 on each little block and a vertical force of Mg/2 on each little block. Edited: So, the magnitude of the total force that the big block exerts on each little block is Mg/√2 which is greater than half the weight of the block. So, each little block receives a force greater than half the weight of the big block

Maybe the confusion stems from believing that the big block cannot exert a force greater than half its weight on each little block.

It might help to consider the attached figure where a block is supported by two ropes which are almost horizontal. In this case, you can show that the force which the block exerts on each rope is much greater than the weight of the block.

It's important to keep in mind that the weight of the block is just the force of gravity acting on the block. The weight of the block only acts on the block, not on any other object. If an external object exerts a force on the block, then the block will exert the same magnitude of force on the external object (no matter how large). In the case of the ropes, the ropes must pull hard on the block to provide enough vertical component of force on the block to cancel the weight acting on the block. If a rope pulls hard on the block, then the block pulls back hard on the rope.

#### Attached Files:

• ###### Hanging Block.png
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17. Jun 28, 2013

### PsychonautQQ

Okay wow thanks for helping so much!! I'm starting to understand a bit more, but still when i think of like a human pyramid with one person on top and two on the bottom, each person doesn't get the full weight of the person above them, they share the load. How are these situations different?

18. Jun 28, 2013

### TSny

The human pyramid is essentially the same problem. Suppose the top person's legs are spread to form a right angle and assume that the top person's feet exert forces on the shoulders of the bottom people in the direction of the legs of the top person (i.e., at 45o to the horizontal.) Then each foot of the top person will exert a force on a shoulder of a bottom person that is greater than half the weight of the the top person.

See attachment (where angle between legs is less than 90o)

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