Friction problem for two block system on an inclined plane

In summary: Hmm.In summary, the conversation discussed a problem involving two blocks resting on an inclined plane and the conditions for when they will begin to slip. The equations for the limiting values of frictional forces were provided, and the attempts at solving the problem resulted in an inequality for both blocks that was equivalent to tanθ ≥ μ2. However, it was noted that there may have been a mistake in labeling the coefficients of friction and in the real world, only one block would slip first with no simultaneity.
  • #1
cooldudeachyut
5
0

Homework Statement



Two block M1 and M2 rest upon each other on an inclined plane. Coefficient of friction between surfaces are shown. If the angle θ is slowly increased, and M1<M2 then
b8c2269868c8412999129e2a6da0fad9.jpg

Options :
1- Block A slips first.
2- Block B slips first.
3- Both slip simultaneously.
4- Both remain at rest.

Homework Equations



Taking frictional force between the two blocks as f1 and between the block B and inclined plane as f2, the equations for limiting values f1max and f2max :

f1max = μ2M1gcosθ
f2max = μ2(M2+M1)gcosθ

The Attempt at a Solution



For Block A :
I only considered the case where f1 reaches its limiting value, hence I get this inequality as the condition when block A may start slipping,

M1gsinθ - μ2M1gcosθ ≥ 0

Which is equivalent to,
tanθ ≥ μ2

For Block B :
Again, I only considered the case where f2 reaches its limiting value but cannot figure out what's the magnitude/direction of f1 on this block. So I assumed f1max to act on this block in the direction up the slope as this basically provides least resistance and also complies with block A's case which may as well be the "limiting factor" for the case where block B slips, giving my inequality as,

M2gsinθ - μ2(M2+M1)gcosθ + μ2M1gcosθ ≥ 0

Which is again equivalent to,
tanθ ≥ μ2

I'm confused how to proceed now as I think both blocks should start slipping simultaneously however the answer provided is option 2, i.e., block B will slip first.
 

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  • #2
Suppose they both are on the threshold of slipping at the same angle as you've found. If both blocks begin to move, will there be any relative motion between them? How might you check for this?
 
  • #3
cooldudeachyut said:

Homework Statement



Two block M1 and M2 rest upon each other on an inclined plane. Coefficient of friction between surfaces are shown. If the angle θ is slowly increased, and M1<M2 then
View attachment 224984
Options :
1- Block A slips first.
2- Block B slips first.
3- Both slip simultaneously.
4- Both remain at rest.

Homework Equations



Taking frictional force between the two blocks as f1 and between the block B and inclined plane as f2, the equations for limiting values f1max and f2max :

f1max = μ2M1gcosθ
f2max = μ2(M2+M1)gcosθ

The Attempt at a Solution



For Block A :
I only considered the case where f1 reaches its limiting value, hence I get this inequality as the condition when block A may start slipping,

M1gsinθ - μ2M1gcosθ ≥ 0

Which is equivalent to,
tanθ ≥ μ2

For Block B :
Again, I only considered the case where f2 reaches its limiting value but cannot figure out what's the magnitude/direction of f1 on this block. So I assumed f1max to act on this block in the direction up the slope as this basically provides least resistance and also complies with block A's case which may as well be the "limiting factor" for the case where block B slips, giving my inequality as,

M2gsinθ - μ2(M2+M1)gcosθ + μ2M1gcosθ ≥ 0

Which is again equivalent to,
tanθ ≥ μ2

I'm confused how to proceed now as I think both blocks should start slipping simultaneously however the answer provided is option 2, i.e., block B will slip first.
I get the same answer you get.
 
  • #4
gneill said:
Suppose they both are on the threshold of slipping at the same angle as you've found. If both blocks begin to move, will there be any relative motion between them? How might you check for this?
At that angle there should be no relative motion between the blocks initially as they both receive same acceleration but I don't know how to calculate for later instants as f1's direction/magnitude bothers me.

So does that mean block A actually doesn't slip first because slipping is only considered relative to block B? Or does that mean both start slipping at the same time?

Chestermiller said:
I get the same answer you get.
I see, option 2 is wrong after all.
 
  • #5
cooldudeachyut said:
I see, option 2 is wrong after all.
It is suspicious that both coefficients are labelled μ2. Looks like a cut-and-paste error in the diagram, and one of them should have had a different value.
 
  • #6
haruspex said:
It is suspicious that both coefficients are labelled μ2. Looks like a cut-and-paste error in the diagram, and one of them should have had a different value.
That makes sense, there must've been a different μ1 coefficient between both blocks in the original problem.
 
  • #7
cooldudeachyut said:
So does that mean block A actually doesn't slip first because slipping is only considered relative to block B?
That would be my interpretation of "slipping", yes.
 
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  • #8
gneill said:
That would be my interpretation of "slipping", yes.
Ok, but I don't think that corresponds to reality.
In the real world, kinetic friction is always less than static, and there is no simultaneity. One will slip first, and as soon as that happens there is less tendency to slip at the other interface. So only one block will slip, but if the coefficients are the same we cannot say which.
 

Related to Friction problem for two block system on an inclined plane

1. What is friction?

Friction is a force that occurs when two surfaces are in contact and moving relative to each other. It acts in the opposite direction to the direction of motion and can cause objects to slow down or stop.

2. How does friction affect a two block system on an inclined plane?

In a two block system on an inclined plane, friction can cause the blocks to resist motion and make it more difficult for them to slide down the plane. This can be seen as an opposing force to the force of gravity pulling the blocks down the incline.

3. How is the coefficient of friction calculated for a two block system on an inclined plane?

The coefficient of friction is calculated by dividing the force of friction by the normal force between the two surfaces. In the case of a two block system on an inclined plane, the normal force is equal to the weight of the upper block acting perpendicular to the surface of the incline.

4. What factors can affect the amount of friction in a two block system on an inclined plane?

The amount of friction in a two block system on an inclined plane can be affected by the type of surfaces in contact, the weight of the blocks, and the angle of the incline. Rougher surfaces and heavier blocks will typically have a higher coefficient of friction, while a steeper incline will result in a larger component of the force of gravity acting parallel to the surface, increasing the force of friction.

5. How can friction be reduced in a two block system on an inclined plane?

Friction can be reduced in a two block system on an inclined plane by using smoother surfaces or by applying a lubricant between the two surfaces. Additionally, reducing the weight of the blocks or decreasing the angle of the incline can also decrease the amount of friction.

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