Posted by **diane** on Monday, December 11, 2006 at 10:36am.

A 160-N child sits on a light swing and is pulled back and held with a horizontal force of 100N. The tension force of each of the two supporting ropes is:

a) 60N

b) 94N

c) 120N

d) 190N

e) 260N

work:

weight is down, the pull is back, so the tension must act at an angle to counter both of these. The vertical component of the tension is 160N, the horizontal component is 100. Then Im lost on how to solve it.

I think the equation for vertical components is:

Ta + Tb= 160N

but then how to I handle an angle in this case? I could use some help of how to solve this one. Thanks

The angle A must be such that

T sin A = 100 (horizontal balance)

T cos A = 160 (vertical balance)

sin A/cos A= tan A = 100/160 = 5/8

A = 32.0 degrees

T is the combined rope tension force.

T = 100/sin A = 188 N

T/2 = 94 N is the tension per rope.

<<T/2 = 94 N is the tension per rope.>>

It is assumed that she sits in the middle of the swing, midway between the ropes. Off-center seating would result in unequal rope tension.

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