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The 10.2kg block in FIGURE P7.37is held in place by a force applied to a rope passing over two massless, frictionless pulleys. Find the tensions T1toT5and the magnitude of force F→.

Short Answer

Expert verified

Tension in the rope,T1=100N,T2=T3=T5=F=50NandT4=150N.

Step by step solution

01

Given information

Mass of the block, m=10.2kg

Given the mass and pulleys system

02

Explanation

Consider the equilibrium block

T1=mg

localid="1649647716658" T1=10.2kg×9.81m/s2=100N

Consider the equilibrium of a small pulley

T1=T2+T3

Since T2 and T3 are attached to the same pulley, therefore the tension T2=T3

localid="1649647731493" 100N=2T2T2=T3=50N

Similarly, T2 and T5 are attached to the same pulley, therefore the tension T2=T5

T5=50N

Since the tension in the ropeT5would be equal to the forceF

F=T5=50N

Consider the equilibrium of a large pulley

localid="1649647749005" T4=T2+T3+T5T4=50N+50N+50N=150N

Tension in the rope,T1=100N,T2=T3=T5=F=50NandT4=150N.

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