Chapter 4, Problem 32P

To determine

a)

Free body diagram of each block.

Answer to Problem 32P

Solution:

Explanation of Solution

On each block, normal force by the surface acts in upward direction and force of gravity act in a downward direction. the contact force acts between the blocks in the equal and opposite direction

To determine

b)

The acceleration of the system.

Answer to Problem 32P

Solution:

$a=\frac{F}{m_A+m_B+m_C}$

Explanation of Solution

Consider the three blocks as one by combining the masses.

Given:

Mass of block A $=m_A$

Mass of block B $=m_B$

Mass of block C $=m_c$

Total Sum $=M$

The net force on the combination of block $=F$

Acceleration $=a$

Formula Used:

$F=ma$

Calculation:

The total mass of the blocks is given as

$M=m_A+m_B+m_C$

Acceleration is given as

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

To determine

c)

The net force on each block.

Answer to Problem 32P

Solution:

$F_A=\frac{m_{A}F}{m_A+m_B+m_C}$

$F_B=\frac{m_{B}F}{m_A+m_B+m_C}$

$F_C=\frac{m_{C}F}{m_A+m_B+m_C}$

Explanation of Solution

Given:

Mass of block A $=m_A$

Mass of block B $=m_B$

Mass of block C $=m_c$

Acceleration $=a=\frac{F}{m_A+m_B+m_C}$

The net force on block A $=F_A$

The net force on block B $=F_B$

The net force on block C $=F_C$

Formula Used:

The net force is given as

$F=ma$

Calculation:

The net force on block A is given as

$\begin{array}{l}{F}_A=m_{A}a\\ F_A=\frac{m_{A}F}{m_A+m_B+m_C}\end{array}$

The net force on block B is given as

$\begin{array}{l}{F}_B=m_{B}a\\ F_B=\frac{m_{B}F}{m_A+m_B+m_C}\end{array}$

The net force on block C is given as

$\begin{array}{l}{F}_C=m_{C}a\\ F_C=\frac{m_{C}F}{m_A+m_B+m_C}\end{array}$.

To determine

(d) To Determine:

The contact force on each block.

Answer to Problem 32P

Solution:

$F_{AB}=F_{BA}=\frac{(m_B+m_C)F}{m_A+m_B+m_C}$

$F_{BC}=F_{CB}=\frac{m_{C}F}{m_A+m_B+m_C}$

Explanation of Solution

The contact forces between the two blocks are equal in magnitude and opposite in direction as per Newton’s third law.

Hence

$\begin{array}{l}{F}_{AB}=F_{BA}\\ F_{BC}=F_{CB}\end{array}$

Given:

Mass of block A $=m_A$

Mass of block B $=m_B$

Mass of block C $=m_c$

Acceleration $=a=\frac{F}{m_A+m_B+m_C}$

The net force on block A $=F_A=\frac{m_{A}F}{m_A+m_B+m_C}$

The net force on block B $=F_B=\frac{m_{B}F}{m_A+m_B+m_C}$

The net force on block C $=F_C=\frac{m_{C}F}{m_A+m_B+m_C}$

Formula Used:

The net force is given as

$F=ma$

Calculation:

From the force diagram, force equation for block A along the horizontal direction is given as

The net force on block A is given as

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

Also

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

From the force diagram, force equation for block C along the horizontal direction is given as

$\begin{array}{l}{F}_{CB}=m_{C}a\\ F_{CB}=\frac{m_{C}F}{m_A+m_B+m_C}\end{array}$

Also

$\begin{array}{l}{F}_{BC}=F_{CB}\\ F_{BC}=\frac{m_{C}F}{m_A+m_B+m_C}\end{array}$

To determine

e)

Numerical answers to part b), c)and d)

Answer to Problem 32P

Solution:

$\begin{array}{l}a=3.2m/s^2\\ F_A=F_B=F_C=32N\\ F_{AB}=F_{BA}=64N\\ F_{BC}=F_{CB}=32N\end{array}$

Explanation of Solution

Given:

$F=96N$

Mass of block A $=m_A=10kg$

Mass of block B $=m_B=10kg$

Mass of block C $=m_c=10kg$

Acceleration $=a=\frac{F}{m_A+m_B+m_C}$

The net force on block A $=F_A$

The net force on block B $=F_B$

The net force on block C $=F_C$

Formula Used:

$a=\frac{F}{m_A+m_B+m_C}$

The net force is given as

$F=ma$

Calculation:

Using the equation

$a=\frac{F}{m_A+m_B+m_C}$

Inserting the values

$\begin{array}{l}a=\frac{96}{(10+10+10)}\\ a=3.2m/s^2\end{array}$

The net force on block A, B and C is given as

$\begin{array}{l}{F}_A=m_{A}a=10\times 3.2=32N\\ F_B=m_{B}a=10\times 3.2=32N\\ F_C=m_{C}a=10\times 3.2=32N\end{array}$

The contact force between A and B is given as

$\begin{array}{l}{F}_{AB}=F_{BA}=\frac{(m_B+m_C)F}{m_A+m_B+m_C}=\frac{(10+10)(96)}{10+10+10}=64N\\ F_{BC}=F_{CB}=\frac{m_{C}F}{m_A+m_B+m_C}=\frac{(10)(96)}{10+10+10}=32N\end{array}$