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KrisConnett

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?

  • one year ago
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  1. KrisConnett
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    |dw:1349917367041:dw|

    • one year ago
  2. maheshmeghwal9
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    |dw:1350795258686:dw|

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  3. maheshmeghwal9
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    |dw:1350795463231:dw|

    • one year ago
  4. maheshmeghwal9
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    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.

    • one year ago
  5. maheshmeghwal9
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    Here is Force diagram of 1kg block |dw:1350796181544:dw|

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  6. maheshmeghwal9
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    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 \[12-(T+m_2g \sin \theta)= m_2a \space \space \space \space \space ..........(ii)eq.\]\[m_2=1kg.\]

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  7. maheshmeghwal9
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    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:)

    • one year ago
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