How Do You Calculate Acceleration in a Two-Block System with Friction?

[mc8569]

Newton's Law of Motion problem! Help please!

Homework Statement

A small block of mass m1 is placed on block 2 of mass m2 this is then placed on the table. A string connecting block 2 to a hanging mass M passes over a pulley attached to one end of the table. The mass and friction of the pulley are negligible. The coefficients of friction between block 1 and 2 and block one and the tabletop are nonzero and are:
Coeff. between 1 & 2:
Static = (mus1) <-- mu sub static 1
Kinetic = (muk1) <-- mu sub kinetic 1

Coeff. between 2 & Table:
Static = (mus2)
Kinetic = (muk2)

Homework Equations

c) M is large enough that the hanging blocks descend when the block is released. Assume that blocks 1 and 2 are moving as a unit (no slippage) Determine the magnitude a of the acceleration.

d) Now suppose that M is large enough that as the hanging block desends, block 1 is slipping on block 2
Determine the following:
i. The magnitude a1 of the acceleration of block 1
ii. The magnitude a2 of the acceleration of block 2

The Attempt at a Solution

For c) I got:
a = g[M - (muk2)(m1 + m2)] / (m1 + m2 + M)
^^after factoring out g, correct?

Real concern is with d)
d) i. F = ma
fk1 = m1a1 = (muk1)n1
a1 = (muk1)m1g / m1
*m1 cancels out
a1 = (muk1)g
Is that right?

ii. was what confused me,
Should (ii) be the same answer as part (c)? Please help me on how to approach this if not. I am thinking that the reverse movement of m1 on m2 would decrease the acceleration, but I thought, as a system it shouldn't do anything because the friction on m1 is equal and opposite to m2. It isn't an external force. Help please?

 

[PhanthomJay]

Homework Statement

A small block of mass m1 is placed on block 2 of mass m2 this is then placed on the table. A string connecting block 2 to a hanging mass M passes over a pulley attached to one end of the table. The mass and friction of the pulley are negligible. The coefficients of friction between block 1 and 2 and block one and the tabletop are nonzero and are:
Coeff. between 1 & 2:
Static = (mus1) <-- mu sub static 1
Kinetic = (muk1) <-- mu sub kinetic 1

Coeff. between 2 & Table:
Static = (mus2)
Kinetic = (muk2)

Homework Equations

c) M is large enough that the hanging blocks descend when the block is released. Assume that blocks 1 and 2 are moving as a unit (no slippage) Determine the magnitude a of the acceleration.

d) Now suppose that M is large enough that as the hanging block desends, block 1 is slipping on block 2
Determine the following:
i. The magnitude a1 of the acceleration of block 1
ii. The magnitude a2 of the acceleration of block 2

The Attempt at a Solution

For c) I got:
a = g[M - (muk2)(m1 + m2)] / (m1 + m2 + M)
^^after factoring out g, correct?

looks good!

Real concern is with d)
d) i. F = ma
fk1 = m1a1 = (muk1)n1
a1 = (muk1)m1g / m1
*m1 cancels out
a1 = (muk1)g
Is that right?

Yes!

ii. was what confused me,
Should (ii) be the same answer as part (c)?

No.

Please help me on how to approach this if not. I am thinking that the reverse movement of m1 on m2 would decrease the acceleration, but I thought, as a system it shouldn't do anything because the friction on m1 is equal and opposite to m2. It isn't an external force. Help please?

You must isolate the blocks M and m2 in a free body diagram. The friction force and normal force on m2 from m1 act as an external force when you isolate the system this way, in accordance with Newton 3, but the mass m1 does not enter into the equation. The acceleration of the blocks are different.
Welcome to PF!

 

[mc8569]

Wow thanks so much! I'm understanding a much greater deal about this now.
But this rises another question now =P..

So force wise you are only looking at T (Mg) and the two kinetic frictional forces on m2 (top & bottom).
But for the "ma" side of F = ma, should you also only deal with the two masses for m in the equation? - that being F = (m2 + M)a. Because I feel like you should still include all the weights because m1 is still on top of m2 regardless, and it should be F = (m1 + m2 + M)a

And thanks for the welcome! Definitely going to come here more often now, so happy I found it.

 

[mc8569]

So if I solve using (m2 + M) for the system's mass, I have

T - fk2 - fk1 = (m2 + M)a
Mg - uk2(m1 + m2)g - uk1m1g = (m2 + M)a
a = g[M - uk2(m1 + m2) - uk1m1] / (m2 + M)

And if you're actually supposed to use m1 + m2 + M as the system's mass, then it's simply the top part over the bottom with an m1 added.

 

[PhanthomJay]

So force wise you are only looking at T (Mg) and the two kinetic frictional forces on m2 (top & bottom).
But for the "ma" side of F = ma, should you also only deal with the two masses for m in the equation? - that being F = (m2 + M)a. Because I feel like you should still include all the weights because m1 is still on top of m2 regardless, and it should be F = (m1 + m2 + M)a

Always draw free body diagrams. When you isolate M, you have Mg -T =Ma. When you isolate m2, you have T - uk2(mig +m2g) -uk1(m1g) = m2a (not (m1 +m2)a). Solve for the 2 equations with 2 unknowns.

 

[PhanthomJay]

So if I solve using (m2 + M) for the system's mass, I have

T - fk2 - fk1 = (m2 + M)a
Mg - uk2(m1 + m2)g - uk1m1g = (m2 + M)a
a = g[M - uk2(m1 + m2) - uk1m1] / (m2 + M)

You're working faster than I can think!This solution is correct, although you've combined a couple of steps.

And if you're actually supposed to use m1 + m2 + M as the system's mass, then it's simply the top part over the bottom with an m1 added.

But this would be wrong. You can't look at the entire system with Newton 2 to find the acceleration of each block with respect to the ground when they are different. If you did look at the entire system, the forces between the 2 blocks would be internal and not enter into the equation, and you'd end up with the acceleration of the center of mass of the system, which would confuse both of us for sure.

 

[mc8569]

Okay, I follow that and you can solve for T using the hanging mass M's equation.. but what I don't understand is:
T - uk2(m1g + m2g) - uk1(m1g) = m2a instead of = (m2 + M)a
I wrote (m2 + M) by the way, not (m1 + m2)
I feel like it should be that because that is the system's equation, right? The system is accelerating altogether, and therefore the mass of the whole system's movement would be (m2 + M). Can you please explain this conceptually to me please? Thanks SOOO much.

 

[mc8569]

I really like how this forum doesn't just give you the answer, it's very helpful. And that was non sarcastic.

 

[PhanthomJay]

Okay, I follow that and you can solve for T using the hanging mass M's equation.. but what I don't understand is:
T - uk2(m1g + m2g) - uk1(m1g) = m2a instead of = (m2 + M)a
I wrote (m2 + M) by the way, not (m1 + m2)
I feel like it should be that because that is the system's equation, right? The system is accelerating altogether, and therefore the mass of the whole system's movement would be (m2 + M). Can you please explain this conceptually to me please? Thanks SOOO much.

Well yes, you are correct, considering the system as the M and m2 part, and both my method (using 2 equations with 2 unknowns) and your method yield the same result. But it's best to isolate separately, to help understand what's going on, and you get into less trouble that way in the more complex problems.