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Figure 5-34 shows three blocks attached ...

Figure 5-34 shows three blocks attached by cords that loop over frictionless pulleys. Block B lies on a friction-less table, the masses are `m_(A)=6.00` kg, `m_(B)=8.00` kg, and `m_(C)=10.0` kg. When the blocks are released, what is the tension in the cord at the right?

Figure 5-34 Three blocks attached over two pulleys by a cord.

Text Solution

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First we analyze the entrie system with "clockwise" motion considered positive (that is, downward is positive for block C, rightward is positive for block B, and upward is positive for block A) and find acceleration. Then we analyze forces on block C only.
Calculation: The equation for the system is
`m_(C)g-m_(A)g=Ma" "("where M = mass of the system "=24.0 kg)`.
This yields an acceleration of
`a=g(m_(C)-m_(A))//M=1.63" m"//"s"^(2)`.
Next we analyze the forces just on block C : `m_(C)g-T=m_(C)a.`
Thus the tension is
`T=m_(C)g(2m_(A)+m_(B))//M=81.7" N".`
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