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Two Blocks and Pulley (Rotational Statics & Dynamics)

  • Thread starter Pat2666
  • Start date
  • #1
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Here's the problem:

http://img187.imageshack.us/img187/5958/showmenq5.gif [Broken]

A block of mass m1 = 1 kg rests on a table with which it has a coefficient of friction µ = 0.52. A string attached to the block passes over a pulley to a block of mass m3 = 3 kg. The pulley is a uniform disk of mass m2 = 0.5 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley.


--------------------------------------------------------------------------------
a) With what acceleration does the mass m3 fall?
a =

--------------------------------------------------------------------------------
b) What is the tension in the horizontal string, T1?
T1 =

--------------------------------------------------------------------------------
c) What is the tension in the vertical string, T3?
T3 =

Well I set up free body diagrams for all three parts, found the tension equations for the masses and torque equation for the pulley and attempted to solve for a (translational acceleration). I got an answer of 4.01m/s^2, but it was incorrect. Where could I have gone wrong?

My work :

http://img187.imageshack.us/img187/6857/workow7.jpg [Broken]
 
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Answers and Replies

  • #2
238
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I'll assume there's no coefficient of friction to oversimplify the problem a little.
Apply Newton's 2nd law:
(1) [tex]\sum F_{x}=T_{1}=m_{1}a[/tex]
(2) [tex]\tau_{p}=(T_{3}-T{1})r=I_{p}\alpha[/tex] *I'll keep the second tension as T_3 to keep the same notations
(3) [tex]F_{x}=m_{3}g-T_{2}=m{3}a[/tex]

Then, by substituting we get [tex]T_{3}-T_{1}=\frac{1}{2} m_{2}a[/tex]
Thus, [tex]a=\frac{m_{3}g}{m_{3}+m_{1}+\frac{1}{2}m_2}[/tex]

Now, if there was a coefficient of friction, how would a change?
*I think the problem with your equation is that you put a [tex]\mu m_{3}[/tex]; however, the coefficient of friction is only between the ledge and the block [tex]m_{1}[/tex]
 
Last edited:
  • #3
tiny-tim
Science Advisor
Homework Helper
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Hi Pat2666! :smile:

I wouldn't use torque, I'd use energy …

At height h, KE1 + KE2 + KE3 = ghm3 - work. :smile:
 
  • #4
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I am kinda new to this but does it make a difference if the pulley has mass or not because we assume that it is completely smooth?
 
  • #5
41
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Hi this is my first post, but I'll try to answer the questions.

First to answer Ed Aboud's question, if we assume that the pulley is completely smooth than it doesn't matter if it has mass because the rope would just slide over it. However the problem states, "the string does not slip on the pulley." In that case the pulley spins with the rope, meaning there is a minuscule amount of static friction that spins the pulley. If the rope doesn't slide, then the pulley's mass now has meaning because some of the potential energy goes into spinning the pulley instead of into the falling mass.

Regarding Pat2666's original problem.

I think I see two issues in your solution. The first is that the sum of forces on mass one does not take into account the stipulated friction. The second is that the sum of torques on the pulley are only the tensions, not the reaction forces you have labeled as F1 and F2. Those forces act on the ropes not the pulley. If you fix those two problems it should work out.
 
  • #6
33
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Alright thanks guys! I misread the problem and was thinking that the friction was occuring on the pulley and not the block on the table. Which would also make sense as why I was confused that it said the rope didn't slip and thought it said there was friction! lol

The additional part I need to add was -umg on the denominator to the equation for acceleration :) At least I was somewhat on track with that!

Thanks again everyone!
 

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