Question

A 35.0-kg child swings in a swing supported by two chains, each 2.92 m long. The tension in each chain at the lowest point is 342N.

(a) Find the child's speed at the lowest point.
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)

Answer 1

part (A),

sum the forces in the normal direction F(n), so we have:

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*(342) - (35)*(9.81) = (35)*(v^2 / 2.92)
v = 5.33 m/s {ANSWER}


Part B)

Since the tension in the chain at the lowest point is 342 N, we know that since there are two chains, the force exerted by the seat is simply:

F = 2*(342)
F = 684 N {ANSWER}

Or if you need proof just sum the forces on the child since you already know the velocity, calling N the force exerted by the seat:

N - m*g = m*(v^2 / r)
N - (35)*(9.81) = (35)*(5.33^2 / 2.92)
N = 683.86 N=684 N {ANSWER}