Two Masses Hanging on a Massive Pulley problem

In summary: The first sentence should be: "If the system were static it would certainly be true that m1g + m2g + mpg is equal to T3, but the system is not static."In summary, a pulley with a mass Mp and a radius Rp attached to the ceiling in a gravity field of 9.81 m/s2 rotates with no friction about its pivot. The mass M2 is larger than mass m1 and the quantities Tn and g are magnitudes. From this, we can determine that T2 is greater than T1 since the pulley needs a bigger torque to rotate in that direction. Additionally, m1g is less than T1, as T1 = m1g
  • #1
MissBisson
24
0

Homework Statement


A pulley with mass Mp and a radius Rp is attached to the ceiling, in a gravity field of 9.81 m/s2 and rotates with no friction about its pivot. Mass M2 is larger than mass m1. The quantities Tn and gare magnitudes.

Homework Equations



(Select T-True, F-False, G-Greater than, L-Less than, E-Equal to. For example, if the first is T, the second L and the rest E, enter TLEEEE).
A) T2 is ... T1.
B) m1g is ... T1.
C) m1g + M2g + Mpg is ... T3.
D) The center of mass of Mp+M1+M2 accelerates.
E) T1 + T2 is ... T3.
F) The magnitude of the acceleration of M2 is ... that of m1.

The Attempt at a Solution



A) T2 is greater than T1 since the pulley needs a bigger torque to rotate in that direction
B) m1g is less than T1. T1 = m1g + ma hence T1 is greater
C) m1g + m2g + mpg is equal to T3 because it holds up the whole system
D) I believe it is true since the whole system is accelerating but i am not sure
E) T1 +T2 is less than T3 because T3 holds up the whole system while T1 and T2 only take into account the masses and accelerations
F) The magnitude of the acceleration of M2 is equal that of m1 because the are all accelerating at same magnitude just different directions

I tried many other combinations but i can't seem to get it right..

Help would be appreciated :)
 

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  • #2
MissBisson said:
C) m1g + m2g + mpg is equal to T3 because it holds up the whole system

Is the acceleration of the centre of mass zero?

MissBisson said:
D) I believe it is true since the whole system is accelerating but i am not sure

How would this affect the answer to the previous question?

MissBisson said:
E) T1 +T2 is less than T3 because T3 holds up the whole system while T1 and T2 only take into account the masses and accelerations

The answer is correct, but I would here rather just consider the free body diagram of the pulley.
 
  • #3
Orodruin said:
Is the acceleration of the centre of mass zero?

But how does the acceleration of the mass affect T3? T3 just holds up the whole system so isn't T3 just related to the mass of the system and not the acceleration?
 
  • #4
T3 and the total gravitational force are the only external forces acting on the system. What is the acceleration of the CoM of the whole system?
 
  • #5
Orodruin said:
T3 and the total gravitational force are the only external forces acting on the system. What is the acceleration of the CoM of the whole system?

It is accelerating in the counterclockwise direction making m2 go down and m1 go up so the negative direction..?
 
  • #6
So based on this, what can you say about (m1+m2+mp)g - T3?

Edit: It also helps to think about special cases. What should happen when m1 = mp = 0?
 
  • #7
Based on (m1+m2+mp)g - T3, T3 would be equal to (m1+m2+mp)g. but i don't understand what you mean by m1=mp=0 :S
 
  • #8
As I said, you are implicitly making the assumption that the center of mass does not accelerate. This is not true in general. If m2 is not equal to m1 and the system accelerates, then the center of mass accelerates. You must take this into account in your force equation.

Regarding puttin masses to zero: It often helps intuition to think of what would happen in extreme cases, such as when some masses are very small in comparison, which essentially corresponds to them being zero.
 
  • #9
A) T2 is greater than T1 since the pulley needs a bigger torque to rotate in that direction
B) m1g is less than T1. T1 = m1g + ma hence T1 is greater
C) m1g + m2g + mpg is greater than T3
D) true since the whole system is accelerating evern the pulley
E) T1 +T2 is less than T3 because T3 holds up the whole system
F) The magnitude of the acceleration of M2 is equal that of m1 because the are all accelerating at same magnitude just different directions

This showed as incorrect... Can you spot where i am wrong?
 
  • #10
MissBisson said:
B) m1g is less than T1. T1 = m1g + ma hence T1 is greater
.. and because m1 is accelerating upwards.
MissBisson said:
D) true since the whole system is accelerating evern the pulley
Right answer, wrong reasoning.
The question is in regard to the mass centre of the system. Each component contributes to that in respect of the acceleration of its own mass centre. Is the pulley's mass centre accelerating?
The fact that the mass centres of certain component are accelerating does not imply that the mass centre of the system is accelerating. The accelerations of the components might cancel each other. You need to explain why they do not.
MissBisson said:
E) T1 +T2 is less than T3 because T3 holds up the whole system
Consider the net vertical force on the pulley, and the vertical acceleration of the pulley.
MissBisson said:
F) The magnitude of the acceleration of M2 is equal that of m1 because the are all accelerating at same magnitude just different directions
Yes, but the reason you give is just restating the answer. Why are the magnitudes the same?
 
  • #11
Thank you for clarifying the explanations! From what you have said it seems that it is on E) that i am wrong. when i consider the net force and acceleratin on the pulley, the vertical accelerations cancel out since the both masses are accelerating at the same magnitude but in opposite directions. for the forces we have two tensions acting down and one tension acting up on the pulley. so wouldn't that make T3 greater since it has to account for the 2 tensions acting down?
 
  • #12
from the last combination i posted, is there anywhere else where i am wrong?
 
  • #13
MissBisson said:
From what you have said it seems that it is on E) that i am wrong
No - I somehow missed out my comment on C:
If the system were static it would certainly be true that m1g + m2g + mpg is equal to T3, but the system is moving. Consider the system below T3 as a unit. What is the net force on this? Is its mass centre accelerating?
MissBisson said:
on E) ... when i consider the net force and acceleratin on the pulley, the vertical accelerations cancel out
No, again it's just the explanation that was unconvincing. When you consider the individual components of an assembly you need to focus on each in turn. The pulley doesn't 'know' anything about the suspended masses and hence nothing about their accelerations. Its only connections with the outside world are gravity and the tensions.
 
  • #14
Ohh alright i understand what you mean! however since the pulley is not moving it would be considered a system at equilibrium for that specific case, with a tension force on both sides of the pulley acting down and one acting up. so wouldn't that make T3 = T1+T2?
 
  • #15
MissBisson said:
Ohh alright i understand what you mean! however since the pulley is not moving it would be considered a system at equilibrium for that specific case, with a tension force on both sides of the pulley acting down and one acting up. so wouldn't that make T3 = T1+T2?
You're forgetting the force of gravity on the pulley.
 
  • #16
ah right that makes sense so T3 is greater than T1+T2

Thank you for clarifying that :) your help is very much appreciated!

i went to talk to my prof and there was a glitch in the system, my last post of answers was correct, the system wasnt reading the lower case letters.
so the final answer is

A) T2 is greater than T1
B) m1g is less than T1.
C) m1g + m2g + mpg is greater than T3
D) true
E) T1 +T2 is less than T3
F) The magnitude of the acceleration of M2 is equal that of m1

Thanks again for everything :)
 

1. What is a "Two Masses Hanging on a Massive Pulley" problem?

A "Two Masses Hanging on a Massive Pulley" problem is a physics problem that involves two masses hanging on opposite sides of a pulley, with the pulley itself being a massive object. It is commonly used to teach concepts of force, acceleration, and Newton's laws of motion.

2. How do you solve a "Two Masses Hanging on a Massive Pulley" problem?

To solve this type of problem, you will need to use principles of Newton's laws of motion and the equations of motion. You will also need to consider the forces acting on each mass, including tension, weight, and any external forces. Then, you can use algebraic equations to find the acceleration, velocity, and displacement of the masses.

3. What is the significance of the pulley being a massive object in this problem?

The fact that the pulley is a massive object means that it has its own weight and inertia, which will affect the overall dynamics of the system. This adds an extra level of complexity to the problem as compared to a simple pulley system, where the pulley is assumed to be massless.

4. What are some common assumptions made when solving a "Two Masses Hanging on a Massive Pulley" problem?

Some common assumptions include neglecting air resistance, assuming the rope or string is massless and inextensible, and assuming the pulley is frictionless. These assumptions make the problem easier to solve, but may not accurately reflect real-world scenarios.

5. How can this problem be applied in real-life situations?

This type of problem can be applied in various real-life situations, such as determining the tension in a cable system, analyzing the dynamics of elevators or cranes, or understanding the mechanics of a pulley system in a weightlifting machine. It can also be used to understand the principles behind simple machines and their applications in everyday life.

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