How to Calculate Tension and Acceleration in a Frictionless Pulley System?

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To calculate tension and acceleration in a frictionless pulley system, start by identifying the forces acting on each mass, including their weights and the tension in the rope. Apply Newton's second law to both masses to derive two equations that will help solve for the unknowns. The weight of the 10.0-kg block is 98 N, while the 3.0-kg block's weight is 29.4 N, and the larger mass moves twice as far as the smaller mass. The normal force only applies to the mass on the table, balancing its weight. Understanding these principles will lead to the correct answers of 13.7 N for tension and 1.37 m/s² for acceleration.
Miss1nik2
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Tension question. PLEASE HELP! I have a test tmw!

In the drawing, the rope and the pulleys are massless, and there is no friction. Find (a) the tension in the rope and (b) the acceleration of the 10.0-kg block. (Hint: The larger mass moves twice as far as the smaller mass.)I know the answers are 13.7 and 1.37, I just don't know how to get them. PLEASE EXPLAIN IT TO ME!So far I have found that W1 = 98 and W2 = 29.4. And I know the given information M1= 10 and M2= 3. I am completely stuck after that. I have tried in ever way I can think of and I cannot get to the right answer.

Thank you VERY much!
 

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Start by indentifying the forces on each mass. Apply Newton's 2nd law to each mass. You'll get two equations, which will allow you to solve for the unknown tension and acceleration.

You might find it helpful to review some of these sample problems: Standard Newton's Laws Problems
 
The forces on each of on each are their weights, and the Normal Force. Which cancel each other out and = 0. Right?
 
Miss1nik2 said:
The forces on each of on each are their weights, and the Normal Force. Which cancel each other out and = 0. Right?
While gravity acts on both masses, only the one on the table will have a normal force. But yes, for that mass the normal force will equal the weight and cancel out.

Don't forget the tension in the rope, which acts on both masses.
 
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