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Newton Force Problem #2

1. Homework Statement
A block of Mass 0.5Kg rests on the inclined surface of a wedge of mass 2Kg. The wedge is acted on by a horizontal force that slides on a frictionless surface.

note that the force is towards the box on the incline and not away from it.

a) If the coefficient of static friction between the wedge and the block is u=0.8 and the angle of the incline is 35 degrees, find the maximum and minimum values for F for which the block does not slip.


3. The Attempt at a Solution

I set up the i and j vectors for the block and the ramp. I solved the general equations of friction between the two. Not sure what to do from there.

Both these questions have been eating away at me today.
 

Answers and Replies

Doc Al
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I set up the i and j vectors for the block and the ramp. I solved the general equations of friction between the two. Not sure what to do from there.
Show exactly what you did.

I would start by analyzing the forces acting on the block. Hint: Try to find the acceleration of block.
 
Mass 1:
i = m1*g*sin theta - (F-friction)=m1*a
j = (F-normal) - m1*g*cos Theta = 0

mass 2

F = m2*a

j = (F-normal)-m1*g - m2*g = 0

I then solved for the general equation of F-normal and pluged in u to get the friction equations.

I don't know where to go next. Does the force in F=ma the same force that will be exerted in the negative x direction>?

thanks
 
Doc Al
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Several problems here:
Mass 1:
i = m1*g*sin theta - (F-friction)=m1*a
j = (F-normal) - m1*g*cos Theta = 0
Seems like you are trying to treat the block as if it were accelerating down the wedge--but it's not sliding down the wedge. Further, you are trying to measure acceleration of the block with respect to the wedge, but the wedge itself is accelerating.
mass 2

F = m2*a

j = (F-normal)-m1*g - m2*g = 0
How does the acceleration of the wedge relate to the acceleration of the block?

Answer this question: Since the block is not sliding down the wedge, what direction must the wedge and ramp be accelerating?

Hint: Stick to analyzing the forces on the block as viewed in the inertial frame of the floor. (Instead of using coordinates parallel and perpendicular to the wedge surface, use horizontal and vertical.)
 
Are you saying that the block should be accelerating along the horizon at the same magnitude as the ramp?

thanks
 
Doc Al
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Are you saying that the block should be accelerating along the horizon at the same magnitude as the ramp?
Yes. The block is not slipping down the ramp, so they must move together.
 
a=F/(m+M) towards Earth for m+M
 

Attachments

Thanks a lot you guys. I understand it now.
 
you are welcome:smile:
 

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