How to Solve a Coupled Motion Pulley Problem with Two Masses of Equal Weight?

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Homework Help Overview

The discussion revolves around a physics problem involving a weightless, frictionless pulley system with two masses of equal weight, specifically focusing on calculating the tension in the rope when one mass is accelerated upward.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the masses, including tension, gravitational force, and the applied force from a hand. There are attempts to set up equations based on Newton's second law, with some participants questioning the initial assumptions about the system's acceleration and the role of the second mass.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have provided calculations and reasoning, while others express confusion about the correct approach and the assumptions made regarding the system's acceleration.

Contextual Notes

Participants note discrepancies between their calculations and the book's answers, indicating potential misunderstandings about the setup of the problem and the forces involved. There is a recognition of the need to clarify the relationship between the two masses and the overall acceleration of the system.

johnsonandrew
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This has been frustrating me:

Problem:

The pulley in the figure is essentially weightless and frictionless. Suppose someone holds on to m=10.0 kg and accelerates it upward at 4.905 m/s/s. What will then be the tension in the rope given that M=10.0 kg?

Attempt:

The sum of the forces on m = Ftension + Fhand - Fgravity = ma
The sum of the forces on M = Ftension - Fgravity = Ma

I said Fhand= m * upward acceleration

I tried solving for 'a' on both equations, then set them equal to each other to solve for Ftension. This gave me a ridiculous answer, and I'm pretty sure I didn't screw up my math. I must have set it up wrong? Please help!
 

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I would guess you calculate the tension in the string for when the Weight is completely motionless, this is equal to:

[tex]F=mg = 10*9.8 = 98N[/tex]

With this tension in mind, this will always be persistent through your calculations, now the mass is being accelerated upwards at a acceleration of 4.905m/s^2 so:

[tex]F=ma = 4.905*10 = 49.05N[/tex]

Total reactive tensile force in the rope is equal to the force of gravity pulling downwards which causes a tension upwards and opposite to it plus the force applied to the weight to accelerate it upwards these tension forces are in the same direction, imagine, the tension force that is reactive to gravity, being larger then the force of gravity, things on pullies would magically float upwards, so it follows that the reactive tension force is equal and opposite to gravity, and the upwards accelerative force is in the same direction of the tension force that prevents gravity otherwise this accelerative force would be pushing the weight downwards, which is counterintuitive:

[tex]F_{total} = F_{gravity}+F_{acceleration}=49.05+98=147.05N[/tex]
 
Last edited:
The second answer, 49.1 N, is the answer the book gives. But I'm confused as to why you calculate the tension when the weight is motionless first. And aren't you neglecting the force the other mass, M, in calculating the tensile force? The answer 147.05 N is incorrect, says the book. :confused:
 
Ohhh you know what? I think the acceleration given becomes the acceleration of the system. So a = 4.905. Why didn't I think of that before! I'm making it more complicated than it is, making two separate accelerations for some reason. That is the problem isn't it?
 
But then I get 147 N... Damn .. What am I doing wrong..
 

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