Question

What is the difference between two problems?

Ex 14.9 has -mgh and F14-13 has +mgh...

14.6 CONSERVATION OF E XAMPLE 14.9 The gantry struch airplane during of & Me is cable AC is release the plane just befon the maximum tens motion? Neglect suructure in the photo is used to test the response of an ng a crash. As shown in Fig. 14-21a, the plane, having a is hoisted back until - 60, and then the pull-back eleased when the plane is at rest. Determine the speed of before it crashes into the ground, 15". Also, what is tension developed in the supporting cable during the elect the size of the airplane and the effect of lift caused the wings during the motion Datum C 20 m SOLUTION Since the force of the cable does no work on the plane, it must be obtained using the equation of motion. First, however, we must determine the plane's speed at B. Potential Energy. For convenience, the datum has been established at the top of the gantry, Fig 14-21a. Conservation of Enemy TA + VA = TB + VB 0 - 8000 kg (9.81 m/s?)(20 cos 60°m) = (8000 kg)ux – 8000 kg (9.81 m/s?)(20 cos 15°m) Up = 13.52 m/s = 13.5 m/s Ans.

Answer 1

Solution:

Reason:

1.) In the Example 14.9 A is not the highest position of the plane, its at position when $\theta$ = 60^o. But In Problem F4.13 A is the highest position of the bob with $\theta$ = 90^o

2.) In the Example 14.9  B is not the lowest position of the plane, its at position when $\theta$ = 15^o.But In Problem F4.13 B is the lowest position of the bob with $\theta$ = 0^o.

As in both problems, the angles at which they both start and end are completely different. So These problems are some what different .

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