Calculating Tension: Pulley System with Different Masses | PhysicsForums

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In the discussion, a user seeks help calculating the tension in a pulley system with two different masses, 1.2 kg and 3.2 kg. Participants emphasize the importance of drawing force diagrams for each mass to understand the forces acting on them. The net force created by the difference in masses leads to acceleration, which is crucial for calculating tension. One user initially miscalculates tension by not considering acceleration, while others clarify that tension must be consistent throughout the system. Ultimately, the focus is on correctly applying the equations of motion to find the tension in the cord before one mass hits the ground.
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Hello PhysicsForums

I need help with a certain question which has been bugging me for a while.

a pulley is suspended by a cord with two masses attached. one mass is 1.2 kg. the second mass is 3.2 kg. the two masses are attached on different sides.

the question: determine the tension in this cord after the masses are released and before one hits the ground. ignore the mass of the pulley and cords.

Thanks
 
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Welcome to PF.

Draw a force diagram for each of the weights.

The heavier one will be showing a Tension and a net force from the weight greater than the tension. The opposite will be true for the lighter one.

From your 2 equations Tension must be the same. Solve for acceleration and plug back into one of the equations.
 
Thank you for replying, but the thing is for this question i don't think the acceleration is needed.. because were trying to find C is attached to the wall and the pulley. Then I thought of Fnet = t-mg and since Fnet = ma which has no acceleration. So I figure it would be 4.4*9.8 but I was wrong.. here is a picture of what I'm talking about.

Thanks again
 

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But there is a net force. 3.2 - 1.2 masses create a net force.

The system is accelerating.

If it was constant speed, the tension on one side would equal the weight and the tension on the other would equal that weight.
 
ok thanks again ... I figured for 17.12N for tension ... and what I am now suppose to do now? Do I make basically sum up what I know that Fnet = mg-t? where mg is (3.2+1.2)(9.8) and Fnet is is (3.2+1.2)(4.45)?
 
I thought tension was what you were solving for?
 
we are solving tension, but its not the tension of the string with the weights on it. we are solving the tension of the pulley carrying the string& weight. I believe i sent an image on post#3. Does this change how the concept of how we are solving it?
 

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