What are the tension and acceleration in this pulleys and rope problem?

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Homework Help Overview

The discussion revolves around a problem involving pulleys and ropes, specifically focusing on calculating the tension in the rope and the acceleration of two masses. The context includes two masses, one heavier than the other, with the assumption of massless pulleys and no friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the tension in the rope and the weights of the masses, with some questioning the original poster's reasoning about the tension being half the weight of mass 2. Others suggest analyzing the forces acting on mass 1 to understand the dynamics better.

Discussion Status

The discussion is active, with participants providing insights and corrections to the original poster's reasoning. Some have offered guidance on how to approach the problem by focusing on the forces acting on the masses, and there is an acknowledgment of a potential correct reasoning path by one participant.

Contextual Notes

There is an ongoing examination of the assumptions regarding the motion of the masses and the distribution of tension in the rope, with participants noting that mass 2 is not stationary and that its acceleration is related to that of mass 1.

Teegvin
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Mass 1 = 10.0 kg
Mass 2 = 3.00 kg

The pulleys and rope are massless, and there is no friction.

What is the tension in the rope?

What is the acceleration of mass 1?

If my thinking is correct, the tension is half the weight of mass 2 (14.7 N), and the acceleration of mass 1 is 1.47 m/s^2.

My textbook gives the solutions as 13.7 N and 1.37 m/s^2.
 
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Teegvin said:
If my thinking is correct, the tension is half the weight of mass 2...
What makes you think that? Show your work and we can see what happened.
 
You're right about the tension being evenly distributed on both sides of mass 2, but mass 2 isn't stationary. As mass 1 slides across the table, mass 2 is moving downward, so the net is less than weight of mass 2.

Try thinking about mass 1 first. The force required (10kg * a) equals the tension on either side of mass 2.
 
BobG said:
Try thinking about mass 1 first. The force required (10kg * a) equals the tension on either side of mass 2.

I figured it out after I read this, and after I realized that the acceleration of mass 2 is half that of mass 1, because it goes half the distance in the same time.

m2(a1)/2 = m2g - 2T
= m2g - 2(10a1)
m2(a1)/2 + 20a1 = m2g
(3/2)a1 + 20a1 = 3g
(43/2)a1 = 3(9.80)
a1 = 1.37

Is this reasoning correct?
 
Looks good to me. (As a matter of style, I would solve for the acceleration in terms of m1 and m2. I wouldn't plug in the actual masses until the last step. But you've got it!)
 

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