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The force F→ in Fig. 12-70 keeps the 6.40 k²µ block and the pulleys in equilibrium. The pulleys have negligible mass and friction. Calculate the tension Tin the upper cable. (Hint:When a cable wraps halfway around a pulley as here, the magnitude of its net force on the pulley is twice the tension in the cable.)

Short Answer

Expert verified

Tension T in the upper cable is 71.68 N.

Step by step solution

01

Understanding the given information

Mass of the block, m=6.40 k²µ

02

Concept and formula used in the given question

Using the equilibrium conditions, you can write the equations for force for the lowest pulley and find the tension in terms of applied force. You can do it for the remaining two pulleys and find the tension in the string holding the topmost pulley. The formula used is given below.

∑Fnet=0

03

Calculation for the tension T in the upper cable

From the figure, we can conclude that

T1=FT2=F+T1T2=2FT3=2F+T2T3=4FT=4F+T3T=8F

In addition, we know the value of force,F=62.72 N

We can write,

T1+T2+T3=mg7F=mgF=mg7F=62.72 N7F=8.96 N

We already found that,

T=8F

So,

T=(8.96 N)(8)=71.68 N

Thus, the tension in the upper cable is 71.68 N.

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Most popular questions from this chapter

In Fig. 12-73, a uniform beam with a weight of 60 Nand a length of 3.2″¾ is hinged at its lower end, and a horizontal force of magnitude 50 N acts at its upper end. The beam is held vertical by a cable that makes angle θ=25°with the ground and is attached to the beam at height h=2.0″¾ . What are (a) the tension in the cable and (b) the force on the beam from the hinge in unit-vector notation?

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