Swing Set Support Forces

## Question:

A 40 kg child is swinging on a swing supported by a pair of A
frames with a beam connecting them at the top from which the
swing is hung. The A frame angle is 60 degrees, each post being
tipped in 30 degrees from the vertical. If she is swinging to a
maximum angle of 60 degrees from the vertical, determine the
force developed along each of the supporting posts as a result of
her swinging, at the instant the swing passes through the
vertical. The length of the swing is two meters.
## Answer:

The gravitational potential energy of the child at the top of a
swing is m*g*h where h is 2 m minus 2*cos(60) m, h = 1 m. So
PE=40*9.8*1= 392 Joules. When the swing
passes through the vertical, the potential energy is zero so the
kinetic energy is 392 Joules =
1/2*m*v^{2} = 1/2*40*v^{2}. So v^{2}=392/20 = 19.6
m^{2}/s^{2}, v=4.43
m/s. The tension in the swing is then the sum of the force
providing the centripetal acceleration and the child's
weight. Her weight is 40*9.8 N = 392 N.
The centripetal acceleration is v^{2}/r
= 19.6/2 = 9.8 m/s^{2}. The force is her mass
times this acceleration or 392 N. This total force of 784 N is
acting straight downward on the beam holding the swing. The four
A frame legs each acting at an angle of 30 degrees from the
vertical must provide the counter force. If they were vertical
each would support 784/4 N. Because they are at an angle, each
must support 784/4/cos(30) = 227 N.
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