Chapter 3, Problem 79SP

In Fig. 3-28, how large a force F is needed to give the blocks an acceleration of 3.0

m/s2 if the coefficient of kinetic friction between blocks and table is 0.20? How large

a force does the 1.50-kg block then exert on the 2.0-kg block?

To determine

The maximum force $F$ that is needed to accelerate the blocks by $3{\text{ m/s}}^2$, and the then reaction force between the blocks $2\text{ kg}$ and $1.5\text{ kg}$ while the coefficient of friction between blocks and table is $0.20$ using the Fig. $3.28$.

Answer to Problem 79SP

Solution:

$22\text{ N}$ and $15\text{ N}$

Explanation of Solution

Given data:

Arrangement of the blocks of mass $1.5\text{ kg}$, $2\text{ kg}$, and $1\text{ kg}$ isprovided in the Fig. 3.28.

Coefficient of friction between blocks and table is $0.20$.

The needed acceleration of the blocksis $3{\text{ m/s}}^2$.

Formula used:

Write the expression for the first condition of the force’s equilibrium:

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ {\displaystyle \sum F_y}=0\end{array}$

Here, ${\displaystyle \sum F_x}$ is the net force in x- direction or horizontal direction and ${\displaystyle \sum F_y}$ is the net force in y-direction or vertical direction.

From the Newton’ s second law of the motion, the expression of the force is

${\displaystyle \sum F_x}=m{a}_x$

Here, ${\displaystyle \sum F_x}$ is the netforce in the x direction, $a$ is the acceleration of the blocks in x-direction, and $m$ is the mass of the block.

Write the expression for the friction force:

$F_f=\mu F_N$

Here, $F_f$ is the friction force and $F_N$ is the normal force.

Sign convention: Consider the direction of the upward forces is positive and the direction of the downward forces is negative. Also, consider the leftward force as negative and rightward forces are positive.

Explanation:

Draw the free body diagram of a $1\text{ kg}$ block:

In $\text{1 kg}$ block, $W_3$ is the weight of the block $\text{1 kg}$, $F_{N3}$ is the normal force that act on the block $3$, $F_{f3}$ is the friction force that oppose the motion of the block $3$, and $a$ is acceleration.

Recall the expression for the first condition of the force’s equilibrium:

${\displaystyle \sum F_y}=0$

From the free body diagram:

$F_{N3}-m_{3}g=0$

Substitute $1\text{ kg}$ for $m_3$ and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}{F}_{N3}-\left(1\text{ kg}\right)\left(9.81{\text{ m/s}}^2\right)=0\\ F_{N3}=9.81\text{ N}\end{array}$

Recall the expression for the friction force:

$F_{f3}=\mu F_{N3}$

Substitute $9.81\text{ N}$ for $F_{N3}$ and $0.20$ for $\mu$

$\begin{array}{c}{F}_{f3}=\left(0.20\right)\left(9.81\text{ N}\right)\\ =1.962\text{ N}\end{array}$

Recall the expression for Newton’s second law of the motion:

${\displaystyle \sum F_x}=ma$

From free body diagram, the force equation is

$N_2-F_{f3}=m_{3}a$

Substitute $1\text{ kg}$ for $m_3$, $1.962\text{ N}$ for $F_{f3}$, and ${\text{3 m/s}}^2$ for $g$

$\begin{array}{c}{N}_2-1.962\text{ N}=\left(1\text{ kg}\right)\left({\text{3 m/s}}^2\right)\\ N_2=4.9\text{ N}\end{array}$

Draw the free body diagram of the all the blocks:

In $\text{2 kg}$ block, $W_2$ is the weight of the $\text{2 kg}$ block, $F_{N2}$ is the normal force that act on the block, $N_2$ is the contact force between the $\text{2 kg}$ block and $\text{1 kg}$ block, $F_{f2}$ is the friction force that oppose the motion of the block, and $a$ is acceleration.

Now, consider the $\text{2 kg}$ block and recall the expression for the first condition of the force’s equilibrium:

${\displaystyle \sum F_y}=0$

From free body diagram:

$F_{N2}-m_{2}g=0$

Substitute $\text{2 kg}$ for $m_2$ and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}{F}_{N2}-\left(2\text{ kg}\right)\left(9.81{\text{ kg/m}}^2\right)=0\\ F_{N3}=19.62\text{ N}\end{array}$

Recall the expression of the friction force:

$F_{f2}=\mu F_{N2}$

Substitute $\text{19}\text{.62 N}$ for $F_{N2}$ and $0.20$ for $\mu$

$\begin{array}{c}{F}_{f2}=\left(0.20\right)\left(19.62\text{ N}\right)\\ =3.924\text{ N}\end{array}$

Recall the expression for Newton’s second law of the motion:

${\displaystyle \sum F_x}=ma$

From free body diagram:

$N_1-N_2-F_{f2}=m_{2}a$

Substitute $\text{2 kg}$ for $m_3$, $3.924\text{ N}$ for $F_{f3}$, $4.9\text{ N}$ for $N_2$, and ${\text{3 m/s}}^2$ for $a$

$\begin{array}{c}{N}_1-4.9\text{ N}-3.924\text{ N}=\left(\text{2 kg}\right)\left({\text{3 m/s}}^2\right)\\ N_1=14.8\text{ N}\\ \approx 15\text{ N}\end{array}$

Draw the free body diagram of the all the blocks:

In the above diagram, $1.5\text{ kg}$ mass of the block, $W_1$ is the weight of the block, $F_{N1}$ is the normal force that act on the block, $N_1$ is the contact force between the $1.5\text{ kg}$ block and $\text{2 kg}$ block, $F_{f1}$ is the friction force that oppose the motion of the block, and $a$ is acceleration.

Now, consider the $\text{1}\text{.5 kg}$ block $1$ andrecall the expression for the first condition of the force’s equilibrium:

${\displaystyle \sum F_y}=0$

From free body diagram:

$F_{N1}-m_{1}g=0$

Substitute $\text{1 kg}$ for $m_1$ and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}{F}_{N1}-\left(1.5\text{ kg}\right)\left(9.81{\text{ m/s}}^2\right)=0\\ F_{N1}=14.71\text{ N}\end{array}$

Recall the expression for the friction force:

$F_{f1}=\mu F_{N1}$

Substitute $14.71\text{ N}$ for $F_{N1}$ and $0.20$ for $\mu$

$\begin{array}{c}{F}_{f1}=\left(0.20\right)\left(14.71\text{ N}\right)\\ =2.9\text{ N}\end{array}$

Recall the expression for Newton’s second law of the motion:

${\displaystyle \sum F_x}=ma$

From free body diagram:

$F-N_1-F_{f1}=m_{1}a$

Substitute $\text{1}\text{.5 kg}$ for $m_1$, $\text{2}\text{.9 N}$ for $F_{f1}$, $14.8\text{ N}$ for $N_1$, and ${\text{3 m/s}}^2$ for $a$

$\begin{array}{c}F-14.8\text{ N}-2.9\text{ N}=\left(\text{1}\text{.5 kg}\right)\left({\text{3 m/s}}^2\right)\\ F=22.2\text{ N}\\ \simeq 22\text{ N}\end{array}$

Conclusion:

Therefore, the required force $F$ will be $22\text{ N}$ to accelerate the blocks and the reaction force between the $1.5\text{ kg}$ block and $2\text{ kg}$ block is $15\text{ N}$.