Where Am I Going Wrong in Calculating Minimum Coefficient of Friction?

AI Thread Summary
The discussion revolves around calculating the minimum coefficient of friction required for a small block (m) to not slide on a larger block (M) on an incline. The key points include the need for static friction, as both blocks must accelerate together down the incline at g sin(θ). The user initially miscalculated the frictional forces and the role of gravity, leading to an incorrect conclusion about the coefficient of friction being less than 0.75 instead of the correct value. Participants emphasize the importance of drawing accurate free body diagrams to analyze the forces acting on the blocks. Ultimately, the correct approach involves recognizing that the force of static friction must balance the gravitational component acting on the smaller block along the incline.
gracy
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Homework Statement

:[/B]
upload_2015-3-23_1-48-25.png

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:
upload_2015-3-23_1-49-7.png

[/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]
 
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gracy said:
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.
 
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
 
I think my diagram was wrong.Here I have corrected myself.
 
upload_2015-3-23_7-19-13.png

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

Attachments

  • upload_2015-3-23_7-18-13.png
    upload_2015-3-23_7-18-13.png
    7.2 KB · Views: 611
will (m+M)g sin theta act on both the blocks?
 
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
 
Chestermiller said:
Your diagram is still awful.
please point out my mistakes.
Chestermiller said:
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
 
gracy said:
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
 
  • #10
Is not mass M also a block? If it is a wedge as it was drawn, what are its angles?
 
  • #11
ehild said:
Is not mass M also a block

Mass M is also a block.Sorry for my poor drawing.
 
  • #12
Is the set-up that shown in my figure?

twoblockswedge.jpg
 
  • #13
gracy said:
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.
 
  • #14
ehild said:
Is the set-up that shown in my figure?
Yes.Absolutely correct.
 
  • #15
And is the problem text as you wrote? Is zero friction between M and the incline?
 
  • #16
ehild said:
Is zero friction between M and the incline?
Yes.
 
  • #17
Then is friction needed that m accelerate the same as M?
 
  • #18
ehild said:
Then is friction needed that m accelerate the same as M?
Sorry.I did not understand.
 
  • #19
What force acts on m along the incline? What acceration does it cause without friction?
 
  • #20
ehild said:
What acceration does it cause without friction?
You mean excluding friction between the blocks what other forces are acting on m along the incline?
 
  • #21
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?
 
  • #22
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.
 
  • #23
ehild said:
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.
 
  • #24
gracy said:
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.
 
  • #25
gracy said:
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.
 
  • #26
gracy said:
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.
 
  • #27
ehild said:
gracy said:
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.
 
  • #28
gracy said:
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.
 
  • #29
I think I have solved the problem.May I show you how would I get 0.75 as coefficient of friction between blocks?
 
  • #30
gracy said:
I think I have solved the problem.May I show you how would I get 0.75 as coefficient of friction between blocks?
Show.
 
  • #31
Give me 5 minutes,.
 
  • #32
upload_2015-3-23_14-8-11.png

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
 
  • #33
gracy said:
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?
 
  • #34
ehild said:
Gravity acts on block m, and its component is mgsin(theta) along the incline.
This is what I have shown in image.
 
  • #35
ehild said:
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?
 
Last edited:
  • #36
It is zero. Do you see any difference between plus zero and minus zero?
 
  • #37
gracy said:
friction force= -0?

It is zero. Do you see any difference between plus zero and minus zero?
 
  • #38
ehild said:
Do you see any difference between plus zero and minus zero?
No.Not at all.Just edited.
 
  • #39
But how I will get 0.75 more than or equal to coefficient of friction then?The answer should be 0.75.
 
Last edited:
  • #40
gracy said:
The answer should be 0.75.
Just have a look at first question
http://www.educationguru4u.com/Pages/EasyPhysics.aspx
 
Last edited by a moderator:
  • #41
gracy said:
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.
 
Last edited by a moderator:
  • #42
ehild said:
Then my drawing was wrong,
WHY?
 
  • #43
Was the block M a wedge?
 
  • #44
ehild said:
why did you say it was correct ?
I thought It is just 2 blocks on inclined surface as my title indicates.
 
  • #45
In physics problems, block is meant that : http://www.wdlconcrete.co.uk/images/blockmain-dense-midi.jpg
 
Last edited by a moderator:
  • #46
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.
 
  • #47
ehild said:
I see. Then my drawing was wrong, why did you say it was correct ?
Draw the FBD for the small block.
upload_2015-3-23_15-43-22.png
 
  • #48
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.
 
  • #49
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
 
  • #50
gracy said:
View attachment 80831
I help with a new picture, as nothing can be seen from yours. Note that m is on a horizontal surface!
blocksonincline.JPG
 
Last edited:

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