What is the magnitude of P in a dynamics pulley problem?

[mwelly007]

Here is the problem;

The 40-lb block is moving downward with a speed of 3 ft/s at t=0 when constant forces P and 2P are applied through the ropes. Knowing that the block is moving upward with a speed of 2 ft/s when t=4 s, determine (a) the magnitude of P. (b) the time at which the speed is 0 ft/s. Neglect all friction.

**The picture is attached as a pdf, which consists of 4 pulleys.

 

[Doc Al]

Show your work and point out where you got stuck.

 

[mwelly007]

difficulties

I am having difficulties setting it up and getting the ball rolling. There are no examples of pulley problems in this chapter. Thus far, I have figured the length of the ropes to be from left to right:

L1=y1+3*y2
L2=y2+y3

where,
y1=length to P
y2=pulley to pulley/box
y3=length to 2P

From this we no that dL/dt=0 so
v1=y1'+3*y2'
v2=y2+y3

Then I can apply the Principle of Impulse:
mv1+Imp(1-->2)=mv2

I don't think I am setting it up correctly, but once I do, I think I could continue on the right track.

 

[mwelly007]

Any help or assistance would be greatly appreciated. Thank you.

 

[kamerling]

The only thing you really need to know about a pully, is that the tension in the ropes on both sides of the wheel is equal. The tension in the rope through the fixed end of the pully is then twice that value.
If you use this then it's easy to find the total upward force of all the ropes on the block.

 

[Doc Al]

I am having difficulties setting it up and getting the ball rolling. There are no examples of pulley problems in this chapter. Thus far, I have figured the length of the ropes to be from left to right:

L1=y1+3*y2
L2=y2+y3

where,
y1=length to P
y2=pulley to pulley/box
y3=length to 2P

From this we no that dL/dt=0 so
v1=y1'+3*y2'
v2=y2+y3

Although in many pulley problems you do have to worry about such constraints in order to solve for the tensions, in this particular problem they give you the tensions. So just follow kamerling's advice and find the net force on the block.

Then I can apply the Principle of Impulse:
mv1+Imp(1-->2)=mv2

You could use this principle, or you could just use Newton's 2nd law. (What's the acceleration?)

 

[mwelly007]

So, to clarify, my tensions from left to right are... P/4 (for all 4) and P for the last 2... correct? And the acceleration, using v=v0+at, is 5/4 ft/s^2 upward. Then, all the tensions forces are up, mg is down, and F=ma is up, and solve for P? Seems I'm missing something.

 

[Doc Al]

So, to clarify, my tensions from left to right are... P/4 (for all 4) and P for the last 2... correct?

No. The rope on the left is being pulled with a force P, so the tension in all its segments is P; similarly, the rope on the right has a tension of 2P.

And the acceleration, using v=v0+at, is 5/4 ft/s^2 upward. Then, all the tensions forces are up, mg is down, and F=ma is up, and solve for P? Seems I'm missing something.

That's all there is to it.

 

[mwelly007]

Makes sense, because otherwise adding pulleys would have no benefit. So then the equation is 5P+ma-mg=0, where mg=40 and m=40/32.3? Sorry, but for some reason, this problem is difficult for me to grasp.

 

[Doc Al]

So then the equation is 5P+ma-mg=0, where mg=40 and m=40/32.3?

I'd write it as: 5P - mg = ma

 

[JJBladester]

I know the original post was in 2008, but even two years later searching this forum has helped me understand pulley dynamics problems. Thanks kamerling and Doc Al!