College Physics 4.9 – The acceleration of a child in a wagon push by two children in opposite directions


Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second a force of 90.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 23.0 kg.
(a)What is the system of interest if the acceleration of the child in the wagon is to be calculated?
(b)Draw a free-body diagram, including all forces acting on the system.
(c)Calculate the acceleration.
(d)What would the acceleration be if friction were 15.0 N?


Solution:

Part A

If the acceleration of the child in the wagon is to be calculated, the system of interest is the wagon with the child in it.

Part B

Considering all the forces acting on the wagon and child with mass m, there will be 5 forces. The horizontal forces are the two forces coming from the two other children, denoted as Fr and Fl. We also have friction going against the direction of motion.

The vertical forces are the weight of the wagon and the child and the normal force. These forces are shown in the free-body diagram shown.

Considering all the forces acting on the wagon and child with mass m. there will be 5 forces. The horizontal forces are the two forces coming from the two other children, denoted as F_l: and F_r. We also have friction going against the direction of motion. The vertical forces are the weight of the wagon and the child and the normal force. These forces are shown in the free-body diagram shown.
Free-body diagram showing the 5 forces acting on the system.

Part C

From Newton’s second law of motion, we know that

\displaystyle \text{net F}=\text{ma}

Considering the horizontal direction to the right as positive, the forces involved are Fr, Fl and f. Therefore, we have

\displaystyle \text{F}_{\text{r}}-\text{F}_{\text{l}}-\text{f}=\text{ma}

Solving for acceleration, we have

\displaystyle \text{a}=\frac{\text{F}_\text{r}-\text{F}_\text{l}-\text{f}}{\text{m}}

\displaystyle \text{a}=\frac{90.0\:\text{N}-75.0\:\text{N}-12.0\:\text{N}}{23.0\:\text{kg}}

\displaystyle \text{a}=0.1304\:\text{m/s}^2       ◀

Part D

If the friction is 15 N, the acceleration would be

\displaystyle \text{a}=\frac{90.0\:\text{N}-75.0\:\text{N}-15.0\:\text{N}}{23.0\:\text{kg}}

\displaystyle \text{a}=0\:\text{m/s}^2       ◀

There would be no acceleration, meaning that the system is in equilibrium. The system is either not moving or it’s moving at a constant velocity.