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KrisConnett
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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?
 one year ago
 one year ago
KrisConnett Group Title
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
 one year ago

This Question is Open

KrisConnett Group TitleBest ResponseYou've already chosen the best response.0
dw:1349917367041:dw
 one year ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
dw:1350795258686:dw
 one year ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
dw:1350795463231:dw
 one year ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
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 ydirection. But there is acceleration in xdirection so we have made above equation.
 one year ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
Here is Force diagram of 1kg block dw:1350796181544:dw
 one year ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
here also normal & m2g cos theta would balance each other. Therefore no need of them. we need only xdirection forces because acceleration is only in xdirection. So our next equation is \[12(T+m_2g \sin \theta)= m_2a \space \space \space \space \space ..........(ii)eq.\]\[m_2=1kg.\]
 one year ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
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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