Two Blocks and Pulley (Rotational Statics & Dynamics)

AI Thread Summary
The discussion revolves around a physics problem involving two blocks and a pulley, focusing on calculating the acceleration of the falling mass and the tensions in the strings. The original poster initially calculated an acceleration of 4.01 m/s², which was incorrect due to overlooking the coefficient of friction affecting the block on the table. Participants highlighted the importance of accounting for the static friction between the block and the table, as well as the correct application of forces and torques on the pulley. Clarifications were made regarding the role of the pulley's mass, emphasizing that it affects energy distribution when the string does not slip. The conversation concluded with the poster acknowledging the misinterpretation of the problem and the need to adjust their calculations accordingly.
Pat2666
Messages
33
Reaction score
0
Here's the problem:

http://img187.imageshack.us/img187/5958/showmenq5.gif

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
 
Last edited by a moderator:
Physics news on Phys.org
I'll assume there's no coefficient of friction to oversimplify the problem a little.
Apply Newton's 2nd law:
(1) \sum F_{x}=T_{1}=m_{1}a
(2) \tau_{p}=(T_{3}-T{1})r=I_{p}\alpha *I'll keep the second tension as T_3 to keep the same notations
(3) F_{x}=m_{3}g-T_{2}=m{3}a

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

Now, if there was a coefficient of friction, how would a change?
*I think the problem with your equation is that you put a \mu m_{3}; however, the coefficient of friction is only between the ledge and the block m_{1}
 
Last edited:
Hi Pat2666! :smile:

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

At height h, KE1 + KE2 + KE3 = ghm3 - work. :smile:
 
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?
 
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.
 
Alright thanks guys! I misread the problem and was thinking that the friction was occurring 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!
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top