63. A crane's trolley at point P in Fig. 4-63 moves for a few seconds to...

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Answer 1

Answer #1

63.

TCos5 TSin5 mg

T = tension force in the rope

m = mass of the load = 870 kg

a = acceleration = ?

Along the vertical direction , force equation from the force diagram is given as

T Cos5 = mg eq-1

Along the horizontal direction, force equation from the force diagram is given as

T Sin5 = ma eq-2

dividing eq-2 by eq-1

T Sin5 /(T Cos5) = ma /(mg)

tan5 = a/g

a = g tan5

a = 9.8 tan5

a = 0.86 m/s²

Answer 2

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