What am I missing here? Mechanics

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  • #26
tiny-tim
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Now I have the last part of the question where they tell me that Object C's mass is 4kg and I need to find the acceleration of all three objects.
Now the friction is not static, how does it influence the system? I know that object B's acceleration is probably different than Object A and C (which I think is the same) but I cant realy explaion why, maybe you can help me understand?

Yes, A and C have the same speed and the same acceleration, because the distance between them is constant.

(and the distance between A and B isn't constant if B is sliding)

You know that the friction force will be the normal force on B times µk.

Use Newton's second law on A. :smile:
 
  • #27
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Yes, A and C have the same speed and the same acceleration, because the distance between them is constant.

(and the distance between A and B isn't constant if B is sliding)

You know that the friction force will be the normal force on B times µk.

Use Newton's second law on A. :smile:

m_C*g - T = m_C*a
so that 28 = 10m_C - 2m_C
hmm.. still trying to understand why it accelerates down. isnt object A going up the slope?

"You know that the friction force will be the normal force on B times µk."
so its the same friction as in the question before?

has anything changed in the equations above now that C's mass is 4kg, so that I know for sure that all three object are not accelerating together?
I know I'm asking realy basic questions but I haven't touched this subject in 3 years since I've been drafted to the army.
 
  • #28
tiny-tim
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m_C*g - T = m_C*a
so that 28 = 10m_C - 2m_C
hmm.. still trying to understand why it accelerates down. isnt object A going up the slope?

Yes … A goes up the slope, and C is attached to A, so C goes vertically down. :smile:
so its the same friction as in the question before?

Yes, if µk = µs.
has anything changed in the equations above now that C's mass is 4kg, so that I know for sure that all three object are not accelerating together?

Yes … for mC ≤ 3.5 kg, the force needed to exactly balance the other forces on B will always be ≤ µsN, and so that can be supplied by friction.

But, for mC > 3.5 kg, the force needed to exactly balance the other forces on B will always be > µsN, and so that cannot be supplied by friction, and so B will slip.
 
  • #29
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ok thanks a lot! I finally got it all figured out :~)
I have another question (that is much easier and simpler to explain/understand) but ill save it for tommorow.
Thanks again, everyone!
 

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