Amakeshift swing is constructed by making a loop in one end of a rope and tying the other end to a tree limb. a child is sitting in the loop with the rope hanging vertically when the child's father pulls on the child with a horizontal force and displaces the child to one side. just before the child is released from rest, the rope makes an angle of 19o with the vertical and the tension in the rope is 280 n. (a) how much does the child weigh? (b) what is the magnitude of the (horizontal) force of the father on the child just before the child is released? (c) if the maximum horizontal force the father can exert on the child is 97 n, what is the maximum angle with the vertical the rope can make while the father is pulling horizontally?

a) 264.74 N

b) 91.15 N

c) 20.12°

Explanation:

Given:

Angle between the rope and the vertical, θ = 19°

Tension in the rope, T = 280 N

For the system to be in equilibrium,

The net force in vertical as well as in horizontal direction, should be zero

Therefore,

a) For Vertical direction

Weight of the child = vertical component of the tension

W = T cosθ (1)

or

W = 280 cos19° = 264.74 N          (a)

b) For horizontal

force on the child, F = T sinθ  (2)

or

F = 280 sin19° = 91.15 N

c) Now, on dividing (1) and (2), we have

W/F = Tcosθ/Tsinθ

or

tanθ = F/W

now, for F = 97 N

tanθ = 97/264.74 = 0.3663        (W from (a))

or

θ = tan⁻¹(0.3663)

or

θ = 20.12°

that would be because of both gravity and friction. so     c

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