Question

A 38.0-kg child swings in a swing supported by two chains, each 2.90 m long. The tension in each chain at the lowest point is 436 N.

(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

We can analyze the system as it follows in Figure 1., where x and y are the displacements on both axis, $\theta$ is the angle of displacement from the equilibrium point.

Then, since at the lowest point x = 0, y = 0, the forces acting on the child will be the sum of the tension forces acting on the swing and the weight mg of the child:

$\sum F_{net}=F_{net}=2T-mg \\ F_{net}=2(436\,N)-(38.0\,kg)(9.81\,m/s^2) \\ F_{net}=499.22\,N=m\frac{v^2}{r}; r = l \\ \therefore v= \sqrt{ \cfrac{lF_{net}}{m} } \\ \text{Now, we substitute and find v:} \\ v= \sqrt{ \cfrac{(2.9\,m)(499.22\,N)}{38\,kg} } \\ \implies v=6.172\frac{m}{s}$

(a) From there we were able to find that the speed at the lowest point was v = 6.172 m/s.

Then, the force exerted by the seat on the child at the lowest point can be found with the net force, defined as $\sum F_{ext}=F_{net}=F'-mg$, where F' = 2T is the force that we're looking for or the force that the seat is exerting on the child and is equal to twice the value for the tension T.

Thus (b) F' = (2)(436 N) = 872 N.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.