Tension with a with a massless rope

In summary, the conversation discusses a physics problem involving a massless pulley, ropes, and boxes with different masses. The first question considers the system in equilibrium with friction, and the second question looks at the system with negligible friction. The summary also includes a link to a diagram of the situation and a follow-up question about the suggested solution.
  • #1
kbka
25
0
Hi...

I have a little assignment where I'm not quite sure, what would be the correct answer...

A massless pulley, is mounted to the ceiling with a massless rope. Over the pulley runs another massles rope, with a box with mass m in one end and a box with mass 2*m in the other.

Q1) In this first question, the friction holds the whole system in equilibrium.
What is the tension T0 between pulley and ceiling, T1 between pulley and box with mass m, T2 between pulley and box with mass 2*m

A1: T1=m*g, T2=2*m*g and T0=(m*2m)*g=3*m*g

Q2) Now, friction is negligible, and the system starts to move... What is the tension T0, T1 and T2
A2: I have calculated T1=T2=4/3*mg and T0=3*m*g (still)

Is the last answer for T0 correct? My assumption is that this part of the system (pulley+rope) is still in equilibrium, and therefore Fres=m*g-T=0
 
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  • #2


kbka said:
Hi...

I have a little assignment where I'm not quite sure, what would be the correct answer...

A massless pulley, is mounted to the ceiling with a massless rope. Over the pulley runs another massles rope, with a box with mass m in one end and a box with mass 2*m in the other.

Q1) In this first question, the friction holds the whole system in equilibrium.
What is the tension T0 between pulley and ceiling, T1 between pulley and box with mass m, T2 between pulley and box with mass 2*m

A1: T1=m*g, T2=2*m*g and T0=(m*2m)*g=3*m*g

Q2) Now, friction is negligible, and the system starts to move... What is the tension T0, T1 and T2
A2: I have calculated T1=T2=4/3*mg and T0=3*m*g (still)

Is the last answer for T0 correct? My assumption is that this part of the system (pulley+rope) is still in equilibrium, and therefore Fres=m*g-T=0

If T1 + T2 = T0, then isn't T0 = 8/3mg ?
 
  • #3
LowlyPion said:
If T1 + T2 = T0, then isn't T0 = 8/3mg ?

Thank you for the reply, but i highly doubt that's the case... I have drawn the situation to make it more clear...

http://peecee.dk/upload/view/136892
 
  • #4
kbka said:
Thank you for the reply, but i highly doubt that's the case... I have drawn the situation to make it more clear...

http://peecee.dk/upload/view/136892

Is it your suggestion then that T1 + T2 ≠ T0 ?

Looks to me like T1 = m*(g + 1/3g) = T2 = 2m*(g - 1/3g)

2*T1 = 2*T2 = 8/3*m*g = T0
 

Related to Tension with a with a massless rope

1. What is tension with a massless rope?

Tension with a massless rope refers to the force that is exerted on the rope when it is pulled or stretched. In other words, it is the resistance of the rope to being pulled apart.

2. Is it possible for a rope to be massless?

In theory, a rope can be considered massless if its weight is negligible compared to the forces acting upon it. However, in reality, all ropes have some amount of mass, no matter how small it may be.

3. How does the mass of the rope affect the tension?

The mass of the rope does not directly affect the tension. The tension is determined by the forces acting upon the rope, such as the force applied to one end of the rope and the force of gravity acting on the object at the other end of the rope.

4. Does the length of the rope affect the tension?

Yes, the length of the rope can affect the tension. A longer rope will have a greater tension than a shorter rope if the same amount of force is applied to both. This is because the longer rope has more surface area to distribute the force, resulting in a higher tension.

5. How can tension with a massless rope be calculated?

Tension with a massless rope can be calculated using the formula T = F/a, where T is tension, F is the applied force, and a is the acceleration of the object at the end of the rope. This formula assumes that the rope has no mass, and therefore, all of the applied force is transferred to the object at the end of the rope.

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