# Motor in a pulley system with weight help! (Sort of a dynamics problem)

emzee1

## 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.

#### Attachments

• problem 1.png
13.8 KB · Views: 570

Homework Helper
Gold Member
Think a bit more about the acceleration of block A. Are you sure it's the same as the acceleration of point P?

emzee1
Well, reviewing some previous examples, it doesn't. How do I go about solving for the acceleration for block A, and does it matter for the overall problem? Because I can't solve the following equation:

2T-mg = ma; as I do not know T nor a
and I'm not entirely positive about the free-body diagram at point P other than Force of motor pointing downwards, and Tension pointing upwards (i.e what does that equal to?).

azizlwl
deleting my post.

Last edited:
Homework Helper
Gold Member
You can get the acceleration of block A in terms of the acceleration of point P. First consider displacements. If point P moves 1 m down, how far does block A go up?
[P.S. You won't need a free-body diagram for point P. You already have the correct equation for block A: 2T-mg = ma.]

emzee1
Let me try.

W=Fv
F-mg/2=ma
W=(ma+mg/2)v.

That is incorrect unfortunately. I've tried that also. The answer given is that the Power input should be 4803.75 W.

Homework Helper
Gold Member
Once you get the correct acceleration for block A, you will get the correct answer by repeating your original calculation.

emzee1
I'm really bad with multiple pulley-systems. Because there are 2 pulleys, is the acceleration just for block A just 1/2 of that at point P?

azizlwl
That is incorrect unfortunately. I've tried that also. The answer given is that the Power input should be 4803.75 W.

Yes i've done it wrongly by taking point P as reference.

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Homework Helper
Gold Member
Yes, block A has 1/2 the acceleration of P. Try to imagine what would happen if you used your hand to lift block A and its pulley by 0.5 m. That would leave 1.0 m of slack in the string due to there being string on both sides of the pulley connected to A. So, point P would need to travel 1.0 m down to take out the slack. So, P always moves twice as far as A. Thus P is always moving twice as fast as A. Hence, during any time interval, the change in speed of P is always twice the change in speed of A. Therefore, the acceleration of P is twice the acceleration of A.

emzee1
So the acceleration of block A would be 3 m/s^2 ?

Putting this into the equation: 2T-mg=ma; where m=50kg and a = 3m/s^2, solving for T gets me 320 N.

P_input = 320(12), and solving for P_ouput; ((320(12)/.8)=4800W.

Did I do everything correctly? Because the answer given is 4,803.75W...

Homework Helper
Gold Member
Try g = 9.81 m/s^2 rather than 9.8

emzee1
Try g = 9.81 m/s^2 rather than 9.8

Thank you so very much! Cleared up a lot of confusion for me!

Homework Helper
Gold Member
Good work, emzee1.