Question

problem 9: there are two boxes in contact with each other, one with mass m1 = 13 kg and the other one with mass m2 = 17 kg. the boxes lie on a frictionless surface (see figure). you exert a force of f = 5 n horizontally on m1 (from the left hand side). what is the force on m2? next you go to the other side and exert the same force horizontally on m2 (from the right hand side). what is the force on m1?

Explanation:

Step1: Find the acceleration of the system

According to Newton's second - law $F = ma$. The total mass of the system $M = M_1+M_2=(13 + 17)\text{ kg}=30\text{ kg}$, and the applied force $F = 5\text{ N}$. The acceleration of the two - box system $a=\frac{F}{M_1 + M_2}$. Substituting the values, we get $a=\frac{5}{13 + 17}=\frac{5}{30}=\frac{1}{6}\text{ m/s}^2$.

Step2: Find the force on $M_2$ when $F$ is applied on $M_1$

The force on $M_2$ is what causes $M_2$ to accelerate with the system. Using Newton's second - law $F_2 = M_2a$. Substituting $M_2 = 17\text{ kg}$ and $a=\frac{1}{6}\text{ m/s}^2$, we get $F_2=17\times\frac{1}{6}=\frac{17}{6}\approx2.83\text{ N}$.

Step3: Find the force on $M_1$ when $F$ is applied on $M_2$

The force on $M_1$ is what causes $M_1$ to accelerate with the system. Using Newton's second - law $F_1 = M_1a$. Substituting $M_1 = 13\text{ kg}$ and $a=\frac{1}{6}\text{ m/s}^2$, we get $F_1=13\times\frac{1}{6}=\frac{13}{6}\approx2.17\text{ N}$.

Answer:

When the force is applied on $M_1$, the force on $M_2$ is $\frac{17}{6}\text{ N}\approx2.83\text{ N}$. When the force is applied on $M_2$, the force on $M_1$ is $\frac{13}{6}\text{ N}\approx2.17\text{ N}$.