# Block on a ramp

1. Sep 27, 2011

### 1MileCrash

A force P acts on a block weighing 45 N. The block is initially at rest on a plane inclined at angle 15 degrees. The positive direction of the x axis is up the plane. Coefficient of static friction is us = .5 and uk = 0.34. Find frictional force in unit vector notation, when P = -5i.

Normally, I'd call them liars and call the positive x direction to be, well, the positive x direction, and break that value for P into components. However, I can tell they REALLY want me to use a different plane.

I have no idea what it should look like. Everything I've tried doesn't work and I am getting a headache. I know I should break gravity down into it's components for this abnormal plane, and that P should run along x and normal force should run along y, but that gives me a diagram suggesting that the normal force is unaffected by p, but I KNOW that this isn't the case.

2. Sep 27, 2011

### Staff: Mentor

It's fairly common to use axes parallel and perpendicular to the incline. It often makes things easier.
Why would you think that a force parallel to the surface would affect the normal force?

The real thing to figure out is whether the force P is enough to start this block moving or not.

3. Sep 27, 2011

### 1MileCrash

P is pulling it at an angle down with gravity, shouldn't that increase normal force?

4. Sep 27, 2011

### Staff: Mentor

Only forces with components perpendicular to the surface can affect the normal force.

5. Sep 27, 2011

### Staff: Mentor

The choice of axes should actually help you. The force P is specified to act along the x-direction, which is parallel to the plane. So you won't have to split it into components, and it won't affect the normal force.

Can you take a stab at drawing the setup?

6. Sep 27, 2011

### 1MileCrash

Now I am extremely confused. My book requests for friction for several values of P. From what I am hearing, P shouldn't effect it..?

7. Sep 27, 2011

### 1MileCrash

I'm posting from a cell phone.

My setup looks like a cocked normal plane, axis paralell (Ive tried other ways) with 5 forces.

Fn up along y
Mgcos down along y
P down along x
Mgsin down along x
Friction up along x.

8. Sep 27, 2011

### Staff: Mentor

Why do you think that? (Hint: Static friction can vary, depending on the applied force.)

9. Sep 27, 2011

### Staff: Mentor

Looks good. As I mentioned in my first post, the real thing to figure out is whether the force P is enough to start this block moving or not. That will tell you whether you have static or kinetic friction.

10. Sep 27, 2011

### 1MileCrash

How could I possibly know that if I can't figure out what the normal force is?

According to my drawing, normal force must be mgcos along positive y, therefore it is the same regardless of p, therefore kinetic friction is the same regardless of p, therefore max static friction is the same regardless of p...

If I said any more untrue statements then Id have to attend confession. Solving with a normal plane clearly shows that normal force is a component of p + mg.

11. Sep 27, 2011

### Staff: Mentor

Why can't you? It doesn't depend on P.

Right.
Right--if the thing moves.
Right--the maximum value is the same, but the actual friction could be less than the maximum.
Huh?

12. Sep 27, 2011

### Staff: Mentor

Static friction exactly balances the plane-parallel component of the applied force up to the point where that force exceeds the maximum static frictional force. Then the block gets "unstuck", begins to move, and kinetic friction takes over.

13. Sep 27, 2011

### 1MileCrash

If fn = mgcos, then max static friction is 21.143 N.

Immediately, I know that is wrong because then the block would never move for the values of P I am asked to solve for.

And, according to the answer key, friction is greater for P at -8Ni than for -15Ni, which tells me it is moving at -15N.

14. Sep 27, 2011

### Staff: Mentor

The static friction will grow to its maximum value as the force trying to move the block grows. Until the block becomes unstuck, the static friction force will be equal and opposite to the applied force trying to move the block. So you certainly can expect the friction force to increase as P increases, even if the block doesn't move at all!

15. Sep 27, 2011

### 1MileCrash

I am well aware of that fact.

Let me repeat. According to the answer key, friction is LESS when p is 15 than when it is 8. That tells me that the block is STATIONARY when p is 8 and MOVING when p is 15.

That contradicts my calculation above which says that 15 newtons is insufficient force to overcome the maximum possible value for static friction.

16. Sep 27, 2011

### Staff: Mentor

When P is 15 the total downslope force (the sum of P and the downslope component of the block's weight) will exceed the maximum static friction force. So at that point dynamic friction will hold sway.

17. Sep 27, 2011

### 1MileCrash

So what I am doing wrong is comparing the applied force against usmax, when what I need to do is compare usmax to the applied force and the component of gravity that acts along the x axis of the plane?

18. Sep 27, 2011

### Staff: Mentor

Yup. When I do these sorts of problems I tend to think in terms of the total upslope forces and total downslope forces (and the normal forces, of course, which determine the frictional forces).

19. Oct 4, 2011

### theunloved

Ok, so I have the same problem with 1MileCrash, and I don't wanna start a new thread, so I'm going to show you how I did:
a: angle of the inclined plane

y direction:
(-mg)cosa + N = 0
N = (mg)cosa

x direction:
-P + fs - (mg)sina = 0 (1)
P = fs - (mg)sina
P = us(N) - (mg) sina
P = 10.08N

So If we have P > 10.08N, the block will start sliding down.

part a. Given vector P = (-5N)i ----> block will not move down

(1) -----> fs = P + (mg) sin a = 16.6N
Vector P = (16.6N)i

part b. Given vector P = (-8N)i ----> block not moving
(1)-----> fs = P + (mg) sin a = 19.6N
Vector P = (19.6N)i

part c. Given vector p = (-15N)i ---> block will start moving
This is right here that I got stuck, since the block starts moving down, it will have an acceleration, so the net force acting on block should be:
-P + fk - (mg) sina = ma

Question is: How do we find acceleration of the block ?

20. Oct 4, 2011

### Staff: Mentor

Your P is already negative (-15N; it carries the correct sign to indicate its direction in the given coordinate system). fk is positive, which means it's operating upslope, against the direction of motion. Good. You've got the downslope component of the block's weight also with a negative sign; Also good. But you're using variable name "a" for two different quantities. Call the angle something else, say θ. So your LHS should sum to the net force acting on the block, which is correct. Call that F. So force F is acting on block of mass m. F = ma. Solve for a.