Two blocks on inclined surface

[gracy]

Homework Statement

:[/B]

block M slides down on frictionless incline .Find the minimum co efficient of friction so that m does not slide with respect to M.

Homework Equations

:acceleration of the system=total driving force/total mass:

[/B]

The Attempt at a Solution

m does not slide with respect to M means there should not be kinetic friction rather static friction.both small and big block will have same acceleration.

acceleration of the system=total driving force/total mass:total driving force=(m+M)g sin theta

acceleration of the system=(m+M)g sin theta/m+M

=g sin theta
force responsible for acceleration of small block would be force of static friction between two blocks.

Hence force of static friction between two blocks=mass of small block i.e "m" multiplied by g sin theta

but this force of static friction between two blocks is also equal to or smaller than μ multiplied by normal force on small block i. mg

force of static friction ≤ μmg force of static friction/mg≤ μ

m× g sin theta= force of static friction

Hence m× g sin theta/mg ≤ μ

theta= 37 degrees

sin theta ≤ μ
sin 37 degrees ≤ μ
0.601 ≤ μ
but the correct answer should be 0.75 ≤ μ
where am I going wrong?[/B]

 

[TSny]

force responsible for acceleration of small block would be force of static friction between two blocks.

There are other forces acting on the small block besides friction. What is the direction of the friction force acting on m?
Draw a good free body diagram for the small mass m.

 

[Chestermiller]

You correctly obtained the acceleration of both masses, which is oriented in the same direction as the incline. Can you resolve this acceleration into components in the horizontal and vertical directions? These are the horizontal and vertical components of the acceleration of the smaller mass m. Once you know these, can you do force balances on m in the horizontal and vertical directions? What do you obtain?

Chet

 

[gracy]

I think my diagram was wrong.Here I have corrected myself.

 

[gracy]

I think my diagram was wrong.Here I have corrected myself.

 

[gracy]

will (m+M)g sin theta act on both the blocks?

 

[Chestermiller]

Your diagram is still awful. I thought I understood your original diagram. I'm still trying to figure out how to draw figures with whiteboard that can have straight lines. Meanwhile, please see if you can answer my questions in post #3.

I still don't know from your figure whether θ = 37° is the angle of the incline with the horizontal, or with the vertical. I assumed it was with the horizontal.

Chet

 

[gracy]

Your diagram is still awful.

please point out my mistakes.

θ = 37° is the angle of the incline with the horizontal, or with the vertica

θ = 37° is the angle of the incline with the horizontal

 

[Chestermiller]

please point out my mistakes.

θ = 37° is the angle of the incline with the horizontal

OK. That's all I needed to know. So now, back to post #3.

Chet

 

[ehild]

Is not mass M also a block? If it is a wedge as it was drawn, what are its angles?

 

[gracy]

Is not mass M also a block

Mass M is also a block.Sorry for my poor drawing.

 

[ehild]

Is the set-up that shown in my figure?

 

[gracy]

please point out my mistakes.

I want to solve this as soon as I can.I don't have much time.I have a test tomorrow.

 

[gracy]

Is the set-up that shown in my figure?

Yes.Absolutely correct.

 

[ehild]

And is the problem text as you wrote? Is zero friction between M and the incline?

 

[gracy]

Is zero friction between M and the incline?

Yes.

 

[ehild]

Then is friction needed that m accelerate the same as M?

 

[gracy]

Then is friction needed that m accelerate the same as M?

Sorry.I did not understand.

 

[ehild]

What force acts on m along the incline? What acceration does it cause without friction?

 

[gracy]

What acceration does it cause without friction?

You mean excluding friction between the blocks what other forces are acting on m along the incline?

 

[ehild]

The two blocks have to move together, with the same acceleration. You have shown that the acceleration is gsin(37°).
If you suppose zero friction between the blocks, what would be the acceleration of block m?

 

[gracy]

Please tell me is my free body diagram correct in post 5?I don't think so.Please point out my mistakes in post 5.

 

[gracy]

If you suppose zero friction between the blocks, what would be the acceleration of block m

Zero,as I think that it is the only net force acting on block m.

 

[ehild]

Please tell me is my free body diagram correct in post 5?I don't think so.Please point out my mistakes in post 5.

It is correct for the blocks as a single system, when they move together. To get the force of friction necessary acting on m you need to draw separate FBD-s for both blocks.

 

[ehild]

Zero,as I think that it is the only net force acting on block m.

The net force acting on m along the incline is mgsin(37)-Friction. But you want m accelerate with a=gsin(37°), so you are right about the friction.

 

[gracy]

as I think that it is the only net force acting on block m.

I meant to say Friction is the only net force acting on block m.Excluding fiction net force is zero.

 

[gracy]

Zero,
meant to say Friction is the only net force acting on block m.

If you suppose zero friction between the blocks

Excluding fiction net force is zero.

 

[ehild]

meant to say Friction is the only net force acting on block m.Excluding fiction net force is zero.

You can not switch off gravity. Gravity and normal force act on block m anyway.

 

[gracy]

I think I have solved the problem.May I show you how would I get 0.75 as coefficient of friction between blocks?

 

[ehild]

I think I have solved the problem.May I show you how would I get 0.75 as coefficient of friction between blocks?

Show.

 

[gracy]

Give me 5 minutes,.

 

[gracy]

as mg sin theta is the net force friction opposes it.
m does not slide with respect to M means there should not be kinetic friction rather static friction.acceleration of the system=(m+M)g sin theta/m+M

=g sin theta
force responsible for acceleration of small block would be force of static friction between two blocks.

Hence force of static friction between two blocks= m× g sin theta
but Static frictional force=≤ μ Normal force
But Normal force=mg cos theta
so,Static frictional force=≤ μ mg cos theta
m× g sin theta=≤ μ mg cos theta
mg cancels out
sin theta/ cos theta =≤ μ
tan theta =≤ μ
theta is 37 degrees
tan 37≤ μ
0.75 ≤μ

 

[ehild]

View attachment 80830
as mg sin theta is the net force friction opposes it.
m does not slide with respect to M means there should not be kinetic friction rather static friction.acceleration of the system=(m+M)g sin theta/m+M

=g sin theta
force responsible for acceleration of small block would be force of static friction between two blocks.

No this is not right. You can not switch of gravity. Gravity acts on block m, and its component is mgsin(theta) along the incline.
The acceleration of the block is determined by the sum of forces acting on it (in the direction of the incline). That sum is mgsin(theta)-friction.
So ma= mgsin(theta) -friction.
But you determined that the acceleration of both blocks is a=gsin(theta). Plug in this a into the previous equation. What do you get for the friction force?

 

[gracy]

Gravity acts on block m, and its component is mgsin(theta) along the incline.

This is what I have shown in image.

 

[gracy]

ma= mgsin(theta) -friction.
But you determined that the acceleration of both blocks is a=gsin(theta). Plug in this a into the previous equation. What do you get for the friction force?

friction force= 0?