An Accelerating Equipment Cart

A physics teacher pushes a cart of demonstration equipment to a classroom, as in Figure 4.11. Her mass is 65.0 kg, the cart’s mass is 12.0 kg, and the equipment’s mass is 7.0 kg. To push the cart forward, the teacher’s foot applies a force of 150 N in the opposite direction (backward) on the floor. Calculate the acceleration produced by the teacher. The force of friction, which opposes the motion, is 24.0 N.

A teacher is standing and facing to the right. She is pushing a cart with various items on it including books and boxes. The teacher and cart are labeled, "System." A force vector labeled F teacher shows the force of the teacher's push on the cart and points to the right. A force vector of equal length labeled F cart show the force exerted by the cart and points to the left. A friction force vector at the bottom area of the cart points to the left. A force vector labeled F floor shows the force of the floor on the teacher's feet and points to the right. A force vector of equal length labeled F foot show the force exerted by teacher foot as she walks and points to the left. A free-body diagram of the system illustrates the friction force pointing to the left, the weight force pointing downward, the Newton force pointing upward, and the F floor force pointing to the right.

Figure 4.11

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
Because they accelerate together, we define the system to be the teacher, the cart, and the equipment. The teacher pushes backward with a force $F_{foot}$ of 150 N. According to Newton’s third law, the floor exerts a forward force $F_{floor}$ of 150 N on the system. Because all motion is horizontal, we can assume that no net force acts in the vertical direction, and the problem becomes one dimensional. As noted in the figure, the friction f opposes the motion and therefore acts opposite the direction of $F_{floor} .$
We should not include the forces $F_{teacher}$ , $F_{cart}$ , or $F_{foot}$ because these are exerted by the system, not on the system. We find the net external force by adding together the external forces acting on the system (see the free-body diagram in the figure) and then use Newton’s second law to find the acceleration.
Solution
Newton’s second law is

$a = \frac{F_{net}}{m} .$
4.21

The net external force on the system is the sum of the external forces: the force of the floor acting on the teacher, cart, and equipment (in the horizontal direction) and the force of friction. Because friction acts in the opposite direction, we assign it a negative value. Thus, for the net force, we obtain

$F_{net} = F_{floor} - f = 150 N - 24 .0 N = 126 N.$
4.22

The mass of the system is the sum of the mass of the teacher, cart, and equipment.

$m = (65 .0 + 12 .0 + 7 .0) kg = 84 kg$
4.23

Insert these values of net F and m into Newton’s second law to obtain the acceleration of the system.

$\begin{array}{l} a = \frac{F_{net}}{m} \\ a = \frac{126 N}{84 kg} = 1 {.5 m/s}^{2} \end{array}$
4.24

$F_{1} < F_{2}$
4.25

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Strategy

Solution