Block-wedge problem.

1. Jun 23, 2013

judas_priest

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

What is the maximum magnitude of a horizontal Force F that can be applied without causing the block to move up the incline?

2. Relevant equations

3. The attempt at a solution

Resolving F into it's components and writing down the equations

For x:

$Fcos(theta) + μMgcos(theta) = Mg sin(theta)$

For y:

$Fsin(theta) + mgcos(theta) = N$

Are the two equations correct? Because solving for F from this doesn't give me the correct answer.
Where am I going wrong?

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2. Jun 23, 2013

judas_priest

Is the direction of Friction correct?

3. Jun 23, 2013

dreamLord

You should check the direction and magnitude of friction that you have used. Remember, friction requires the total normal force applied on the body.

4. Jun 23, 2013

judas_priest

Why is the friction force acting in the direction you think it's acting in? I've added normal force in my equation. Please reason your answers.

5. Jun 23, 2013

dreamLord

Static friction is written as $\mu$N, where N is the total normal force on the body. You have written it as $\mu$Mgcos(theta).

The direction is correct - I don't know why I asked you check the direction. I meant for you to check the magnitude. Also, does the question say anything about the surface of the wedge? It is possible that they meant for it to be frictionless.

6. Jun 23, 2013

judas_priest

Thanks, you made it clear. Actually, this is the second part of a problem It talks about friction in the first part.

Thank you!

7. Jun 23, 2013

PeterO

I think your original expression had the friction in the wrong direction.

To clairify your thinking for me, if the block is about to move up the slope - ie you are at the maximum force asked for - what direction will friction be acting?

8. Jun 23, 2013

dreamLord

If we go on increasing the F from 0 to it's correct value, then the OP's direction of friction is correct in my opinion.

9. Jun 23, 2013

PeterO

"What is the maximum magnitude of a horizontal Force F that can be applied without causing the block to move up the incline?"
means what force can be applied when the block is about to move up the slope.
What direction do you think friction will be acting in that case?

If F = 0, the block will (tend to) slide down the slope - friction acting up.

As we increase F, the friction force acting up the slope is reduced until we reach the situation where the block would stay there even on a frictionless slope.

As we increase F further, friction acts down the slope, preventing the block from slipping up the slope.

eventually F will become large enough that the friction is no longer sufficient, and the block will slide up the slope.

The question asks for the maximum F can reach before that slipping happens.

I thus think you are wrong.

10. Jun 23, 2013

PeterO

The applied Force F has a component parallel to the slope (your direction x) which tends to make the block move up the slope, it also has a component perpendicular to the slope (your direction y or -y) which increases the Normal force over the usual Mgcos(θ). and that friction acts in a direction opposed to motion, so in the opposite direction for Fcos(θ) so should have the opposite sign to that term.

11. Jun 23, 2013

ehild

PeterO is right, the static friction opposes the effect of the applied force, to push the block up on the incline. Also, the maximum static friction is μN.

ehild

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12. Jun 23, 2013

judas_priest

Yup, you're right!
Silly mistake.

13. Jun 24, 2013

dreamLord

I am having trouble understanding this - should the friction not increase due to increased normal force?

14. Jun 24, 2013

judas_priest

I too was doing it that way. But you're looking at it wrong. The force applied is trying to push the block up. They asked for the maximum force that can be applied before the box starts to move UP. Hence, the frictional force to oppose it from going up, points down.

Needs a little logic. Think for a while. You'll get there.

15. Jun 24, 2013

PeterO

You are confusing the maximum possible friction, μN, and the actual friction force, which is only ever as big as it needs to be.

eg: 2kg block on a flat surface, μ = 0.3, and lets take g=10 for simplicity here.

μN = 0.3 x 2 x 10 = 6N. ie maximum possible friction is 6N.

If we apply a lateral force of 2N, friction will only be 2N and the block won't move.

Increase the lateral force to 4N, and friction will increase to 4N and the block still won't move

Increase the lateral force to 6N and the friction force increases to 6N and it still doesn't move.

Increase the lateral force further and finally you get some action.

On the slope of your problem, lets suppose the maximum friction available was just enough to stop the block sliding down the slope. The block will thus not move.

once we apply some horizontal force, there will be a component of that force acting up the slope, so although the maximum available friction force will have increased, the magnitude of friction needed will be less, so the friction will reduce.

If the applied force gets big enough, the component of applied Force acting up the slope matches the component of the weight force down the slope so no friction force is needed at all - despite the maximum available friction force being even larger than anything I have written about yet.

If the applied force gets even bigger, the component of applied force acting up the slope will exceed the component of the weight force, so the block will tend to slide up the slope, but the available friction may prevent that slip (by now the available friction is getting quite large)

eventually, if we continue increasing the applied force, the now greater friction force is still not able to prevent the block sliding, so away it goes.

16. Jun 24, 2013

dreamLord

Ah I get it now. Thank you for the great explanation Peter. Silly thinking on my part - apologies if I caused you any problems OP.

17. Jan 24, 2015

Poetria

This is a corrected version with a minus sign:

"Resolving F into it's components and writing down the equations

For x:

Fcos(theta)-μMgcos(theta)=Mgsin(theta)

For y:

Fsin(theta)+mgcos(theta)=N"

May I ask how you can solve for F if you also have N? Two unknowns to my mind. I didn't get it. :(

18. Jan 24, 2015

haruspex

As explained, the frictional force is not μMgcos(theta). It is μN. N is greater than Mgcos(theta) because F contributes to it, as shown in the second equation.

19. Jan 25, 2015

Poetria

OK. I got it. Many thanks. :) :) :)