Minimum friction coefficient for small block on sliding large block at 37°

[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?

 

[ehild]

It is zero. Do you see any difference between plus zero and minus zero?

 

[ehild]

friction force= -0?

It is zero. Do you see any difference between plus zero and minus zero?

 

[gracy]

Do you see any difference between plus zero and minus zero?

No.Not at all.Just edited.

 

[gracy]

But how I will get 0.75 more than or equal to coefficient of friction then?The answer should be 0.75.

 

[gracy]

The answer should be 0.75.

Just have a look at first question
http://www.educationguru4u.com/Pages/EasyPhysics.aspx

 

[ehild]

Just have a look at first question
http://www.educationguru4u.com/Pages/EasyPhysics.aspx

I see. Then my drawing was wrong, why did you say it was correct ?
Draw the FBD for the small block.

 

[gracy]

Then my drawing was wrong,

WHY?

 

[gracy]

Was the block M a wedge?

 

[gracy]

why did you say it was correct ?

I thought It is just 2 blocks on inclined surface as my title indicates.

 

[ehild]

In physics problems, block is meant that : http://www.wdlconcrete.co.uk/images/blockmain-dense-midi.jpg

 

[gracy]

So how that's going to change your answer?from zero to 0.75?please I have to finish this as soon as possible.I have many other problems to solve.

 

[gracy]

I see. Then my drawing was wrong, why did you say it was correct ?
Draw the FBD for the small block.

 

[BvU]

Picture looks awful still. m is lying on a horizontal surface. Picture in link is fine and under 1 in post #1 is hmmm.

So back to post #3 by Chet: three forces working on m have to result in an acceleration $g\sin\theta$ along the plane.
Draw the diagram and calculate the forces.

The $\tan\theta$ answer is correct.

 

[Chestermiller]

Gracy,

Are you saying that you see no difference in the figure ehild drew in post #12 and the figure in the link that you yourself presented in post #40, and that they should have the same answer?

Chet

 

[ehild]

View attachment 80831

I help with a new picture, as nothing can be seen from yours. Note that m is on a horizontal surface!

 

[gracy]

From where mg sin theta is coming?component of mg?

 

[ehild]

From where mg sin theta is coming?component of mg?

It is the resultant force on m. You know that it accelerates down the incline with a= gsin(37), so the resultant force is ma= mgsin(37).

 

[gracy]

I see. Then my drawing was wrong, why did you say it was correct ?

In physics problems, block is meant that : http://www.wdlconcrete.co.uk/images/blockmain-dense-midi.jpg

But the link I have provided i.e http://www.educationguru4u.com/Pages/EasyPhysics.aspx says M is also a block.

 

[ehild]

But the link I have provided says M is also a block.

Yes, but a wedge was drawn, and the solution they provided counts with the wedge.

 

[gracy]

Yes, but a wedge was drawn

You mean question is wrong?

 

[ehild]

In wider meaning, block can mean other shapes.
a. A solid piece of a hard substance, such as wood, having one or more flat sides.
b. Such a piece used as a construction member or as a support.
c. Such a piece upon which chopping or cutting is done: a butcher's block.
d. Such a piece upon which persons are beheaded.
e. One of a set of small wooden or plastic pieces, such as a cube, bar, or cylinder, used as a building toy.

So solve the problem according to the original figure. The top surface of M is horizontal, as in your second figure in the OP.

 

[gracy]

How your post 50 gives answer 0.75?Please answer.

 

[BvU]

In PF it's good habit that we help as much as we can, but do not do the exercise for the poster. You wouldn't let someone else train in your place if you want to run a marathon, would you ?

So back to post #3 by Chet: three forces working on m have to result in an acceleration gsinθ g\sin\theta along the plane.
Draw the diagram (that's done now) and calculate the forces.

 

[gracy]

I have done really hardwork on this particular question,that's why I was expecting that I will be understood that inspite of so much try she is not getting let's help her one step ahead by providing complete solution of the question..But no problem .

 

[gracy]

Actually I am running out of time.