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Chapter 12: Q61P (page 351)

The force F→ in Fig. 12-70 keeps the 6.40 kg 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 kg

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

Question: In Fig. 12-26, a uniform sphere of mass m = 0.85 m and radius r = 4.2 mis held in place by a massless rope attached to a frictionless wall a distance L = 8.0 cm above the center of the sphere. Find (a) the tension in the rope and (b) the force on the sphere from the wall.

A gymnast with mass46.0 kg stands on the end of a uniform balance beam as shown in Fig. 12-80. The beam is 5.00 m long and has a mass of250 kg (excluding the mass of the two supports). Each support is0.540m  from its end of the beam. In unit-vector notation, what are the forces on the beam due to (a) support 1 and (b) support 2?

Question: In Fig.12-32, a horizontal scaffold, of length 2.00 m and uniform mass 50 kg , is suspended from a building by two cables. The scaffold has dozens of paint cans stacked on it at various points. The total mass of the paint cans is 75 kg . The tension in the cable at the right is 722 N . How far horizontally from thatcable is the center of mass of the system of paint cans?

A particle is acted on by forces given, in newtons, by F1→=8.40i^−5.70j^and F2→=16.0i^+4.10j^ . (a) What are the xcomponentF3→ and (b) ycomponent of the force Fthat balances the sum of these forces? (c) What angle does F3→have relative to the+xaxis ?

In Fig 12-66, asphere is supported on a frictionless plane inclined at angle θ=45°from the horizontal. Angle ϕis 25°. Calculate the tension in the cable.
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