Frictional Force in Unconventional Plane: Solving for P

[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.

 

[Doc Al]

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.

It's fairly common to use axes parallel and perpendicular to the incline. It often makes things easier.

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.

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.

 

[1MileCrash]

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

 

[Doc Al]

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

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

 

[gneill]

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?

 

[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..?

 

[1MileCrash]

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?

I'm posting from a cell phone.

My setup looks like a cocked normal plane, axis parallel (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.

 

[Doc Al]

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..?

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

 

[Doc Al]

I'm posting from a cell phone.

My setup looks like a cocked normal plane, axis parallel (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.

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.

 

[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.

 

[Doc Al]

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

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

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

Right.

therefore kinetic friction is the same regardless of p,

Right--if the thing moves.

therefore max static friction is the same regardless of p...

Right--the maximum value is the same, but the actual friction could be less than the maximum.

Solving with a normal plane clearly shows that normal force is a component of p + mg.

Huh?

 

[gneill]

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.

 

[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.

 

[gneill]

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.

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!

 

[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.

 

[gneill]

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.

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.

 

[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?

 

[gneill]

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?

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).

 

[theunloved]

Ok, so I have the same problem with 1MileCrash, and I don't want to 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 ?

 

[gneill]

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 ?

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.

 

[theunloved]

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.

Ok, so we use θ instead of a

we should have:
-P + fk - (mg)sinθ = ma

if you call F is the net force acting of the mass, we will have
F = -P + fk - (mg)sinθ = ma

The problem asks for the frictional force on the block from the plane which is fk, however, without knowing acceleration a, how can I solve for fk ?

 

[gneill]

You've already got fk. It's the normal force multiplied by the kinetic friction coefficient. It's constant as long as the block is in motion.

 

[theunloved]

You've already got fk. It's the normal force multiplied by the kinetic friction coefficient. It's constant as long as the block is in motion.

Ok, so static friction fs can vary, depending on the applied force and kinetic friction fk is constant as long as the block is in motion, and is independent with the applied force ? Is that right ?

 

[gneill]

Ok, so static friction fs can vary, depending on the applied force and kinetic friction fk is constant as long as the block is in motion, and is independent with the applied force ? Is that right ?

Right. You've got it. (assuming that the applied force is parallel to the surface)

 

[theunloved]

Right. You've got it.

Thanks for your help ;)