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Three piñatas hang from the (stationary) assembly of massless pulleys and cords seen in Fig. 12-21. One long cord runs from the ceiling at the right to the lower pulley at the left, looping halfway around all the pulleys. Several shorter cords suspend pulleys from the ceiling or piñatas from the pulleys. The weights (in newtons) of two piñatas are given.

(a) What is the weight of the third piñata? (Hint:A cord that loops halfway around a pulley pulls on the pulley with a net force that is twice the tension in the cord.)

(b) What is the tension in the short cord labeled with T?

Short Answer

Expert verified

a) The weight of the third piñata is 15 N .

b) Tension in the short cord labelled is 10 N .

Step by step solution

01

The given data

The pulley system is in equilibrium, having various weights attached to it.

02

Understanding the concept of pulleys

We use the concept of balanced forces to determine the magnitudes of the various tensions and forces acting on the cords. We make use of free-body diagrams for pulleys.

Formulae:

The value of the net force at equilibrium, Fnet=0 (i)

03

a) Calculation of the weight of the third piñata

We will consider a free body diagram for each of the pulleys and balance the forces for each.

The tension in each part of the red rope is the same; this is as it wound continuously on the pulleys.

Let it be denoted as T'. Let, the tension in the green cords be denoted asF1,F2,F3 , and T as indicated in the diagram. Let the unknown weight be w.

We now consider the first pulley on the right side, to have weight 17 N hanging from it. The forces acting on this system are in each of the red ropesT'andF1in the green cord.

All three are directed upwards, whereas the weight is directed downwards.

Now, the net force equation for this pulley considering equation (i) can be given as:

F1+T'-17=0F1+2T'=17.................(i)

Similarly, for the next pulleys, we write the force equations using equation (i) as follows:

role="math" localid="1660977557551" 2T'=10...........(ii)F3-2T'=0............(iii)T-2T'=0.............(iv)3T'-w=0.............(v)

From equation (ii), we get,

T'=5N

And using this value of T’ in equation (v), we find,

w=15N

Hence, the value of the weight of the piñata is 15 N .

04

b) Calculation of the tension of the short labelled cord

Also, from equation (iv), we calculate the value of the tension using the value of

T'=5N can be calculated as:

T25N=0T=10N

Hence, the value of the tension of the short cord is 10 N .

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Most popular questions from this chapter

Figure 12-49ashows a vertical uniform beam of length Lthat is hinged at its lower end. A horizontal forceFa is applied tothe beam at distance yfrom the lower end. The beam remainsvertical because of a cable attached at the upper end, at angleθwith the horizontal. Figure12-49agives the tension Tin the cableas a function of the position of the applied force given as a fraction y/Lof the beam length. The scale of the Taxis is set byTs=600N.Figuregives the magnitude Fhof the horizontal force on thebeam from the hinge, also as a function of y/L. Evaluate (a) angleθand (b) the magnitude of Fa.

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Figure 12-81 shows a 300 kg cylinder that is horizontal. Three steel wires support the cylinder from a ceiling. Wires 1 and 3 are attached at the ends of the cylinder, and wire 2 is attached at the center. The wires each have a cross-sectional area of2.00×106m2 . Initially (before the cylinder was put in place) wires 1 and 3 were2.0000 m2 long and wire 2 was 6.00 mmlonger than that. Now (with the cylinder in place) all three wires have been stretched. What is the tension in (a) wire 1 and (b) wire 2?

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