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Forces and Newton's Laws of Motion

  • Thread starter kaspis245
  • Start date
  • #1
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Homework Statement


The drawing shows a large cube (mass = 25 kg) being accelerated across a horizontal frictionless surface by a horizontal force P. A small cube (msmall = 4 kg) is in contact with the front surface of the large cube and will slide downward unless P is sufficiently large. The coefficient of friction between the cubes is f = 0.71 . What is the smallest magnitude that P can have in order to keep the small cube from sliding down?

image.jpg


Homework Equations


Ffriction= f*mg

The Attempt at a Solution


Ffriction=msmall*g
f*P=msmall*g
P=56,34N

But the answer must be P = 400 N.
 

Answers and Replies

  • #2
34,056
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P has to accelerate both the small and the large block, that will take more force than the force between the two blocks.
 
  • #3
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Great answer but still not clear.
 
  • #4
Doc Al
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Ffriction=msmall*g
f*P=msmall*g
P=56,34N
What you found is the force that must push against the small mass, not P, which is the force that pulls the large mass. Use what you found to figure out what P must be.
 
  • #5
186
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I kinda figured that out. Can somebody show me how to do it?
 
  • #6
Doc Al
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Here's a hint: Consider the acceleration.
 
  • #7
186
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You mean P = ma ?
 
  • #8
Doc Al
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44,892
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You mean P = ma ?
Yes. Well, F = ma. Consider the acceleration of the small mass first.
 
  • #9
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The acceleration of the small mass is equal to the large mass.
 
  • #10
Doc Al
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The acceleration of the small mass is equal to the large mass.
Exactly! What's the acceleration of the small mass?
 
  • #11
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The acceleration of the small mass a = P/msmall
 
  • #12
186
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Got it. asmall = 14 m/s2

Plarge = (m + msmall)*a = 400 N
 
  • #13
Doc Al
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The acceleration of the small mass a = P/msmall
No. It's the force acting on the small mass divided by the small mass.
 
  • #14
Doc Al
Mentor
44,892
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Got it. asmall = 14 m/s2

Plarge = (m + msmall)*a = 400 N
Good!
 

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