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The question is in the photos (dangling separate weights by two separate ropes question)

The question is in the photos (dangling separate weights by two separate ropes question-example-1
The question is in the photos (dangling separate weights by two separate ropes question-example-1
The question is in the photos (dangling separate weights by two separate ropes question-example-2
User Itmuckel
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1 Answer

18 votes
18 votes

In order to calculate the acceleration of the system, let's consider both masses as one object, and use the Second Law of Newton (considering that the blocks are moving down):


\begin{gathered} \sum ^{}_{}F=m\cdot a \\ W-T=(m_1+m_2)\cdot a \\ (m_1+m_2)\cdot g-T=(m_1+m_2)\cdot a \\ 0.54\cdot9.8-6.6=0.54\cdot a \\ 5.29-6.6=0.54a \\ -1.31=0.54a \\ a=-(1.31)/(0.54)=-2.426\text{ m/s2} \end{gathered}

So the acceleration is 2.426 m/s² upwards.

The tension in the lower rope can be calculated using the Second law just in the lower block:


\begin{gathered} \sum ^{}_{}F=m\cdot a \\ T-W_2=m_2\cdot a \\ T-0.24\cdot9.8=0.24\cdot2.426 \\ T-2.352=0.582 \\ T=2.934\text{ N} \end{gathered}

The tension in the lower rope is equal to 2.934 N.

User Russ Egan
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2.7k points