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Motor and Pulley System with weight help! (Kind of a Dynamics Problem)

  • Engineering
  • Thread starter emzee1
  • Start date
  • #1
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Homework Statement


At the instant shown, point P on the cable has a velocity 12 m/s, which is
increasing at the rate of 6m/s^2. Determine the power input of motor M at this
instant if it operates with an efficiency of 0.8. The mass of the block A is 50kg. I've attached a picture of the problem.


Homework Equations


F=ma
Power = F*v (F=force, v=velocity)


The Attempt at a Solution


All I did was make an attempt at drawing free-body diagrams. For the block A of 50kg, I did:

2T-mg=ma; 2T-50(g)=50(6); T= 395 N

Then for point P, I assumed 395 = force of the motor, and thus:
395(12)= 4,740
Input Power = (4,740/.8) = 5,925 W

This is wrong obviously, and was hoping if someone could help me out on where I'm going wrong.
 

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Answers and Replies

  • #2
Simon Bridge
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You can treat this as conservation of energy if you know how fast the weight is being raised.

You can think your way through this - i.e. does the top pulley matter? What are the speeds of the points on the other two sections of rope at the same height as P?
 
  • #3
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I'm not sure about the last point you made. Is the weight being pulled as fast and at the same rate as point P?
 
  • #4
Simon Bridge
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I'm not sure about the last point you made. Is the weight being pulled as fast and at the same rate as point P?
It's a pulley system - in your experience, is this generally true of pulleys?

You should have done problems involving pulleys by now.
If you don't have experience of pulleys or physics involving pulleys by now: get some!

Off that last point:
See that bit of rope from the ceiling going down to the pulley on the block?
Pick a point on that rope at the same height as point P.
Is that point moving?
 
  • #5
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So, after reviewing my notes. Block A isn't moving as fast/at the same rate as point P. How do I exactly solve for the acceleration for block A? Seeing I have these two equations then:

2T - mg = ma; where I don't know T or a (acceleration) of the block

At point P, the only two forces I can think of is T pointing upwards, and the Force of the motor pointing downwards. But what do I set this equation equal to?
 
  • #6
Simon Bridge
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So, after reviewing my notes. Block A isn't moving as fast/at the same rate as point P. How do I exactly solve for the acceleration for block A?
You'll have an example in your notes for just that.

Put a small mass at point P, which you can set to 0 later on.
So T-F=ma and 2T-Mg=Ma

That may help.

But - considering the pulley system and your notes about pulleys - you should be able to find a short-cut. If you pull on the motor end with speed v, how fast does the block rise?
 
  • #7
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Figured it out, block A has 1/2 the acceleration of point P
 
  • #8
Simon Bridge
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Cool - now you know the rate that the engine is doing work (lifting the block against gravity) and you know it's efficiency. You should be able to answer the question.
 

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