Understanding the Effects of Force on a Pulley System

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SUMMARY

The discussion focuses on calculating the acceleration of a pulley system involving two blocks of mass m1 and m2 connected by a massless string over a frictionless pulley of mass mp. The applied force F affects the system's dynamics, specifically the tension in the ropes and the acceleration of the pulley. Key equations derived include F = mp * ap + 2T, where T represents the tension in the ropes. The relationship between the accelerations of the blocks and the pulley is critical for solving the problem accurately.

PREREQUISITES
  • Newton's Second Law (F = ma)
  • Understanding of tension in a pulley system
  • Basic knowledge of kinematics and acceleration
  • Frictionless dynamics in physics
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  • Learn about the relationship between different masses' accelerations in a pulley system
  • Explore advanced applications of Newton's laws in multi-body systems
  • Investigate the effects of friction on pulley systems
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funnyboy1000
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Homework Statement



Two blocks of mass m1 and m2, resting on a frictionless tabletop, are connected by a massless string passing through a frictionless pulley of mass mp. The figure shows a top view of the arrangement. If a force of magnitude F is applied to the pulley in the direction indicated, what is the acceleration ap of the pulley?

Homework Equations


F=ma?

The Attempt at a Solution


I honestly have no idea how to go about this problem. I know the tension in the strings will be F=ma, but I don't understand how the force of the pulley being pulled acts on the rope. Rather than giving me the answers, I would appreciate if someone helped me understand the effects of everything in the problem. I understand that ap cannot equal (m1+m2+mp)/F because there is tension in the ropes, but I don't understand the effect of the force applied to the pulley and how it effects the ropes.
 

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welcome to pf!

hi funnyboy1000! welcome to pf! :wink:

call the tension T, and the lengths of the two strings L1 and L2

(obviously, (L1 and L2)'' = 0)

then apply good ol' Newton's second law (F = ma) three times, to each of the three masses …

what do you get? :smile:
 
Would you add the forces together? And I don't think the lengths of the string in this problem actually matter. I also don't understand why the length of the two strings are 0. Also, how can I figure out the acceleration of the two blocks if I don't know the acceleration of the pulley? Would the force being exerted on the pulley also effect the tension in the ropes? Sorry, I am kind of a idiot. Thanks a lot for the help though.
 
funnyboy1000 said:
… I also don't understand why the length of the two strings are 0.

no, the (single and) double derivative of their sum is 0

just write all the F = ma equations, and solve them
 
Tension in the top rope = m1*ap = T1
Tension in the bottom rope = m2*ap = T2
F= mp*ap + T1 + T2
This is what I got so far... is that correct? If so then... F= ap(m1+m2+mp)... and the only given hint in the problem states that ap =/= F/(m1+m2+mp)
 
(try using the X2 button just above the Reply box :wink:)
funnyboy1000 said:
Tension in the top rope = m1*ap = T1
Tension in the bottom rope = m2*ap = T2

no
F= mp*ap + T1 + T2

yes, but T1 and T2 are the same, so just use "T"
 
Then you would get ap=F/(mp+2m)... which is essentially the same thing because T=2(ap)m. I am so confused, sorry about this.
 
funnyboy1000 said:
Tension in the top rope = m1*ap = T1
Tension in the bottom rope = m2*ap = T2

no..
 
So umm.. how do I find the tension in the rope? Does the force exerted on the pulley have a effect on the tension in the rope?
 
  • #10
funnyboy1000 said:
Tension in the top rope = m1*ap = T1
Tension in the bottom rope = m2*ap = T2

you need to use the acceleration of the individual masses
 
  • #11
So... acceleration of the individual masses= F/m1? But I can't see how that could be if the acceleration of the pulley is dragging the blocks at the same acceleration as the pulley, due to the string being attached to the blocks.
 
  • #12
funnyboy1000 said:
But I can't see how that could be if the acceleration of the pulley is dragging the blocks at the same acceleration as the pulley …

it isn't!

the heavier mass is lagging behind (accelerating less than the lighter mass)

start again! :smile:
 
  • #13
Actually.. that was a stupid statement. If there is tension in the rope, the acceleration of the mass being dragged has to be less than the acceleration of the pulley. I don't understand how to find the acceleration of the mass being dragged though.
 
  • #14
correct these two :rolleyes:
funnyboy1000 said:
Tension in the top rope = m1*ap = T1
Tension in the bottom rope = m2*ap = T2
 
  • #15
Wait... i feel stupid now... I just sort of worked it out and (F-2T)/mp= ap. I had F=mp*ap+2T written the entire time, but it didn't dawn upon me to try and solve for ap. Is my thinking correct? If so, how do I solve for T?
 
  • #16
funnyboy1000 said:
(F-2T)/mp= ap.

yes, but T is unknown, so you still need the other two equations
 
  • #17
Which other two equations? I am back to being confused... :( I tried 2T=-mp *ap+F but I just over complicated things... and I don't think I was on the right track. I also know that T=T1=T2=m1*a1=m2*a2, but that doesn't help much...
 
  • #18
funnyboy1000 said:
Which other two equations?

T=m1*a1=m2*a2
 
  • #19
How do I find a1? Or do I just leave it as is? Thanks for putting up with me so far...
 
  • #20
what is the geometric equation relating a1 a2 and ap ?
 
  • #21
a1=F/m1? I don't understand how ap and a1 are related... Could you hint something please?
 
  • #22
call the position of the pulley xp, and the lengths of the the two parts of the string x1 and x2

then differentiate :smile:
 

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