Question

Three blocks on a frictionless horizontal surface are in contact with each other, as shown in Fig. 4–54. A force$\overrightarrow{\mathbf{F}}$is applied to block A (mass m_A). (a) Draw a free-body diagram for each block. Determine (b) the acceleration of the system (in terms of m_A, m_B, and m_C), (c) the net force on each block, and (d) the force of contact that each block exerts on its neighbor. (e) If$m_A=m_B=m_C=10.0\mathrm{kg}$,and$F=96.0\mathrm{N}$,give numerical answers for (b), (c), and (d). Explain how your answers make sense intuitively.

FIGURE 4-54 Problem 34

Short answer

(a) The free-body diagram is shown as follows:

(b) The acceleration of the system is $\frac{F}{m_A+m_B+m_C}$.

(c) The expression for the net force on block A is $\frac{m_{A}F}{m_A+m_B+m_C}$. The expression for the net force on block B is $\frac{m_{B}F}{m_A+m_B+m_C}$. The expression for the net force on block C is $\frac{m_{C}F}{m_A+m_B+m_C}$.

(d) The net force acting on block A by block B, or the net force acting on block B by block A is $F\left(\frac{m_B+m_C}{m_A+m_B+m_C}\right)$. The net force acting on block B by block C, or the net force acting on block C by block B is $F\left(\frac{m_C}{m_A+m_B+m_C}\right)$.

(e) The acceleration of the system is $3.20\mathrm{m}{/\mathrm{s}}^2$. The net force on block A is $32.0\mathrm{N}$. The net force on block B is $32.0\mathrm{N}$. The net force on block C is $32.0\mathrm{N}$. The net force acting on block A by block B, or the net force acting on block B by block A is $64.0\mathrm{N}$. The net force acting on block B by block C, or the net force acting on block C by block B is $32.0\mathrm{N}$.

Step-by-step solution

Step 1. Newton’s second law

The value of the force applied to an object is altered linearly to the value of the object’s acceleration. Hence, as the value of the acceleration increases, the value of the applied force also increases.

Step 2. Given information

Given data:

The mass of block A is $m_A=10.0\mathrm{kg}$.

The mass of block B is $m_B=10.0\mathrm{kg}$.

The mass of block C is $m_C=10.0\mathrm{kg}$.

The force is $F=96.0\mathrm{N}$.

Step 3. Draw a free-body diagram

(a)

The free-body diagram for each block can be drawn as follows:

Here, $a$is the acceleration of the system, $F_{NA}$is the normal force on block A, $F_{NB}$is the normal force on block B, $F_{NC}$is the normal force on block C, $F_{AB}$is the force exerted on block A by block B, $F_{BA}$is the force exerted on block B by block A, $F_{BC}$is the force exerted on block B by block C, and $F_{CB}$is the force exerted on block C by block B.

Step 4. Calculate the expression for the acceleration of the system

(b)

The total mass of the system is given as

$M=m_A+m_B+m_C$.

The expression for the acceleration of the system can be written as

$\begin{array}{c}a=\frac{F}{M}\\ a=\frac{F}{m_A+m_B+m_C}\end{array}$

Thus, the acceleration of the system is $\frac{F}{m_A+m_B+m_C}$.

Step 5. Calculate the expression for the net force on each block

(c)

The expression for the net force on block A can be written as

$\begin{array}{c}{\left(F_A\right)}_{net}=m_{A}a\\ {\left(F_A\right)}_{net}=m_A\left[\frac{F}{m_A+m_B+m_C}\right]\\ {\left(F_A\right)}_{net}=\frac{m_{A}F}{m_A+m_B+m_C}\end{array}$

Thus, the expression for the net force on block A is $\frac{m_{A}F}{m_A+m_B+m_C}$.

The expression for the net force on block B can be written as

$\begin{array}{c}{\left(F_B\right)}_{net}=m_{B}a\\ {\left(F_B\right)}_{net}=m_B\left[\frac{F}{m_A+m_B+m_C}\right]\\ {\left(F_B\right)}_{net}=\frac{m_{B}F}{m_A+m_B+m_C}\end{array}$

Thus, the expression for the net force on block B is $\frac{m_{B}F}{m_A+m_B+m_C}$.

The expression for the net force on block C can be written as

$\begin{array}{c}{\left(F_C\right)}_{net}=m_{C}a\\ {\left(F_C\right)}_{net}=m_C\left[\frac{F}{m_A+m_B+m_C}\right]\\ {\left(F_C\right)}_{net}=\frac{m_{C}F}{m_A+m_B+m_C}\end{array}$

Thus, the expression for the net force on block C is $\frac{m_{C}F}{m_A+m_B+m_C}$.

Step 6. Calculate the expression for the force of contact that each block exerts on its neighbour

(d)

From the force diagram, the net force acting on block A is

$\begin{array}{c}F-F_{AB}=m_{A}a\\ F_{AB}=F-m_{A}a\\ F_{AB}=F-m_A\frac{F}{m_A+m_B+m_C}\\ F_{AB}=F\left(\frac{m_B+m_C}{m_A+m_B+m_C}\right)\end{array}$

Using Newton’s third law of motion, the magnitude of the force acting on block A by block B is equal to the force acting on block B by block A. Therefore,

$\begin{array}{l}{F}_{AB}=F_{BA}\\ F_{AB}=F\left(\frac{m_B+m_C}{m_A+m_B+m_C}\right)\end{array}$

Thus, the net force acting on block A by block B, or the net force acting on block B by block A is $F\left(\frac{m_B+m_C}{m_A+m_B+m_C}\right)$.

From the force diagram, the net force acting on block B is

$\begin{array}{c}{F}_{BC}=F_{BA}-m_{B}a\\ F_{BC}=F\left(\frac{m_B+m_C}{m_A+m_B+m_C}\right)-m_B\left(\frac{F}{m_A+m_B+m_C}\right)\\ F_{BC}=m_C\left(\frac{F}{m_A+m_B+m_C}\right)\\ F_{BC}=F\left(\frac{m_C}{m_A+m_B+m_C}\right)\end{array}$

Using Newton’s third law of motion, the magnitude of the force acting on block B by block C is equal to the force acting on block C by block B. Therefore,

$\begin{array}{l}{F}_{BC}=F_{CB}\\ F_{BC}=F\left(\frac{m_C}{m_A+m_B+m_C}\right)\end{array}$

Thus, the net force acting on block B by block C, or the net force acting on block C by block B is $F\left(\frac{m_C}{m_A+m_B+m_C}\right)$.

Step 7. Calculate the numerical answers for (b), (c), and (d)

(e)

The acceleration of the system can be calculated as

$\begin{array}{l}a=\frac{F}{m_A+m_B+m_C}\\ a=\frac{\left(96.0\mathrm{N}\right)}{\left(10.0\mathrm{kg}\right)+\left(10.0\mathrm{kg}\right)+\left(10.0\mathrm{kg}\right)}\\ a=3.20\mathrm{m}{/\mathrm{s}}^2\end{array}$

Thus, the acceleration of the system is $3.20\mathrm{m}{/\mathrm{s}}^2$.

The net force on block A can be calculated as

localid="1654881097209" $\begin{array}{l}{\left(F_A\right)}_{net}=\frac{m_{A}F}{m_A+m_B+m_C}\\ {\left(F_A\right)}_{net}=\frac{\left(10.0\mathrm{kg}\right)\left(96.0\mathrm{N}\right)}{\left(10.0\mathrm{kg}\right)+\left(10.0\mathrm{kg}\right)+\left(10.0\mathrm{kg}\right)}\\ {\left(F_A\right)}_{net}=32.0\mathrm{N}\end{array}$

Thus, the force on block A is $32.0\mathrm{N}$.

The net force on block B can be calculated as

localid="1654881059016" $\begin{array}{l}{\left(F_B\right)}_{net}=\frac{m_{B}F}{m_A+m_B+m_C}\\ {\left(F_B\right)}_{net}=\frac{\left(10.0\mathrm{kg}\right)\left(96.0\mathrm{N}\right)}{\left(10.0\mathrm{kg}\right)+\left(10.0\mathrm{kg}\right)+\left(10.0\mathrm{kg}\right)}\\ {\left(F_B\right)}_{net}=32.0\mathrm{N}\end{array}$

Thus, the force on block B is $32.0\mathrm{N}$.

The net force on block C can be calculated as

localid="1654881033440" $\begin{array}{l}{\left(F_C\right)}_{net}=\frac{m_{C}F}{m_A+m_B+m_C}\\ {\left(F_C\right)}_{net}=\frac{\left(10.0\mathrm{kg}\right)\left(96.0\mathrm{N}\right)}{\left(10.0\mathrm{kg}\right)+\left(10.0\mathrm{kg}\right)+\left(10.0\mathrm{kg}\right)}\\ {\left(F_C\right)}_{net}=32.0\mathrm{N}\end{array}$

Thus, the force on block C is $32.0\mathrm{N}$.

The net force acting on block A by block B, or the net force acting on block B by block A can be calculated as

$\begin{array}{c}{F}_{AB}=F_{BA}\\ F_{AB}=F\left(\frac{m_B+m_C}{m_A+m_B+m_C}\right)\\ F_{AB}=\left(96.0\mathrm{N}\right)\left[\frac{\left(10.0\mathrm{kg}\right)+\left(10.0\mathrm{kg}\right)}{\left(10.0\mathrm{kg}\right)+\left(10.0\mathrm{kg}\right)+\left(10.0\mathrm{kg}\right)}\right]\\ F_{AB}=F_{BA}=64.0\mathrm{N}\end{array}$

Thus, the net force acting on block A by block B, or the net force acting on block B by block A is $64.0\mathrm{N}$.

The net force acting on block B by block C, or the net force acting on block C by block B can be calculated as

$\begin{array}{l}{F}_{BC}=F_{CB}\\ F_{BC}=F\left(\frac{m_C}{m_A+m_B+m_C}\right)\\ F_{BC}=\left(96.0\mathrm{N}\right)\left[\frac{\left(10.0\mathrm{kg}\right)}{\left(10.0\mathrm{kg}\right)+\left(10.0\mathrm{kg}\right)+\left(10.0\mathrm{kg}\right)}\right]\\ F_{BC}=F_{CB}=32.0\mathrm{N}\end{array}$

Thus, the net force acting on block B by block C, or the net force acting on block C by block B is $32.0\mathrm{N}$.