Another pulley problem, but on an incline

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SUMMARY

The discussion focuses on calculating the tension in a pulley system involving two boxes, one on a horizontal surface and the other on an inclined plane. The key equations derived include f - t = ma for the horizontal box and t - m'gsinx = m'a for the inclined box, where 'f' is the applied force, 't' is the tension, 'm' is the mass of the horizontal box, 'm'' is the mass of the inclined box, 'g' is the acceleration due to gravity, and 'x' is the angle of the incline. By combining these equations, users can solve for acceleration 'a' and subsequently determine the tension 't' in the system.

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murphy
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There is this question that I feel like it should be easy, but I can't get the right answer. There is a box sitting on a horizontal surface attached to another box that is sitting on a "decline" The horizontal box is bigger than the other one. a horizontal force is applied to the horizontal box. how do i find the tension in the system? (all surfaces, strings, pulleys, etc, are ideal). I understand that the tension is the same in both strings and that the acceleration (if there is any) is the same, but I don't know how to use that because I can only do the pulley systems that are 90 degrees to each other!what happens when the acceleration and tension are in components?
 
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What i understan in your problem is one block is placed on the horizontal table and other one is down the decline plane of it, if it is so then it could be easily solved as bellow, otherwise give me a daigram of it.
Let 'f' be the force applied horizontly, 't' be the tension, m be mass of block on horizontal and (m') is of the other
f - t= ma .. 1 'a' is the acceleration due to the force applied
t - m'gsinx = m'a 'x' is the angle of the inclined plane
adding
f - m'gsinx = (m + m')a
calculate 'a', substitute its value in (1)
 
Thank your so much! It finally makes sense!
 

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