Pulleys and weights - finding acceleration and tension

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Homework Help Overview

The problem involves a 24 kg block connected to a 6 kg block via a pulley system on a frictionless surface. The original poster seeks to find the acceleration of the 24 kg block and the tension in the connecting string after determining the horizontal distance traveled by the 24 kg block when the 6 kg block descends a specific distance.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the distances moved by the two blocks and the forces involved. There is an exploration of the implications of the pulley system on the forces acting on the blocks.

Discussion Status

Participants are actively engaging with the problem, questioning the relationships between forces and distances in the context of the pulley system. Some guidance has been offered regarding the force relationships, but there is no explicit consensus on the next steps to take.

Contextual Notes

The original poster has indicated uncertainty about how to approach the calculations for acceleration and tension, and there is an ongoing discussion about the mechanics of the system without providing definitive solutions.

iamjohnny56
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Homework Statement


A 24 kg block with a pulley attached slides along a frictionless surface. It is connected by a massless string to a 6 kg block via the arrangement shown.
The acceleration of gravity is 9.81 m/s2 .

The first part wanted you to find the horizontal distance traveled by the 24 kg block if the 6 kg blocked descended 10.1 cm. I found the answer, which is just half of that distance, 5.05 cm.

However, I don't know how to find the other two questions:

1. Find the acceleration of the 24 kg block. Answer in units of m/s2

2. Find the tension in the connecting string. Answer in units of N


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Homework Equations



F=ma

The Attempt at a Solution



I'm not asking for answers - but could someone point me in the direction I should take? I'm not sure about how to start this...
 
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If the 6 kg mass moves twice as far as the 24 kg mass, you get a multiplier of 2 in the force on the larger mass due to the weight of the smaller mass. Remember force times distance equals force times distance.

Realizing that, you can determine the force sliding the 24 kg mass on the table.
 
So if

(Smaller) = (Larger)
Fd = Fd
F(2x) = F(x)

Wouldn't the F cancel out? I'm a bit confused about what you said - could you perhaps elaborate a little bit more?
 
The F's are not the same. One F is twice the other...

This is how block and tackle works - I'm not talking football here.

It's F1 * d1 = F2 * d2. If you move twice as far, you double the force. However, the work done is the same.
 
Think of a little gear turning a larger gear. Torque gets multiplied by the ratio.

If you use a longer pry bar, you can lift more. It's force(1) times distance(1) equals force(2) times distance(2).
 
Must sign off now.
 

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