A 35.0-kg child swings in a swing supported by two chains, each 2.94 m long. The

Question

A 35.0-kg child swings in a swing supported by two chains, each 2.94 m long. The tension in each chain at the lowest point is 416 N. (a) Find the child's speed at the lowest point. Consider all the vertical components of force acting on the swing when it is at its lowest point and relate them to the acceleration of the swing at that instant. m/s (b) Find the force exerted by the seat on the child at the lowest point. (Ignore the mass of the seat.) N (upward)

Please include the formulas. Diagrams are always helpful too! Thank you!

Explanation / Answer

F(n) = 2*T - m*g = m*a

where T is the tension (since there is two chains, we multiply it by two), m is the mass, g = 9.81 m/s^2, and a is acceleration in the normal direction. We know that by relationships found in any fundamental dynamics textbook that:

a = v^2 / r

where v is the speed we are looking for, a is acceleration in the normal direction, and r is the radius. So substituting this equation into our original equation we have:

2*T - m*g = m*(v^2 / r)

Now plugging in what we have and solving for v, we get:

2*(416) - (35)*(9.81) = (35)*(v^2 / 2.94)

v = 6.4067 m/s

For Part B) we already have the answer. Since the tension in the chain at the lowest point is 350 N, we know that since there are two chains, the force exerted by the seat is simply:

F = 2*(416)

F = 832 N

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