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A mass of 1kg on an incline of 37 degrees is being pulled with a force of 12 N upward that incline. The mass is attached to a string going to the bottom of the incline, which is attached to a pulley, which is attached to a string attached to a mass of 3kg on the ground. Assuming there is no friction and the mass of the string and pulley are negligible, what is the tension of the string?

MIT 8.01 Physics I Classical Mechanics, Fall 1999
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Other answers:

For 3kg block \[T = m_1a \space \space \space \space \space \space ..........(i)eq.\]\[m_1=3kg.\]We would ignore Normal reaction & its weight because they balance each other. So no acceleration in y-direction. But there is acceleration in x-direction so we have made above equation.
Here is Force diagram of 1kg block |dw:1350796181544:dw|
here also normal & m2g cos theta would balance each other. Therefore no need of them. we need only x-direction forces because acceleration is only in x-direction. So our next equation is \[12-(T+m_2g \sin \theta)= m_2a \space \space \space \space \space ..........(ii)eq.\]\[m_2=1kg.\]
Now u know all the values because they are given in the question itself. So solve eq. (i) & (ii). U will get T which is the tension in the rope. But \[\LARGE{\color{red}{\text{Please read slowly all the stuff.}}}\] Thanx:)

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