Chapter 5, Problem 14P

Three objects are connected on a table as shown in Figure P5.14. The coefficient of kinetic friction between the block of mass m₂ and the table is 0.350. The objects have masses of m₁ = 4.00 kg, m₂ = 1.00 kg, and m₃ = 2.00 kg, and the pulleys are frictionless. (a) Draw a free-body diagram of each object. (b) Determine the acceleration of each object, including its direction. (c) Determine the tensions in the two cords. What If? (d) If the tabletop were smooth, would the tensions increase, decrease, or remain the same? Explain.

Figure P5.14

To determine

Draw free body diagram of the each object.

Answer to Problem 14P

The free body diagram of the mass of block $m_1$ is shown in Figure 1.

The free body diagram of the mass of block $m_2$ is shown in Figure 2.

The free body diagram of the mass of block $m_3$ is shown in Figure 3.

Explanation of Solution

The free body diagram is the graphical illustration used to visualize the movements and forces applied on a body.

Let $f$ be the force due to the friction, $n$ be the normal force, $mg$ is the force due to gravitation, and $T$ be the tension on the string.

The free body diagram of the mass of block $m_1$ is shown in Figure 1.

The free body diagram of the mass of block $m_2$ is shown in Figure 2.

The free body diagram of the mass of block $m_3$ is shown in Figure 3.

Conclusion:

Therefore, the free body diagram of the mass of block $m_1$ is shown in Figure 1, the free body diagram of the mass of block $m_2$ is shown in Figure 2, and the free body diagram of the mass of block $m_3$ is shown in Figure 3.

To determine

The magnitude and direction of acceleration of each object.

Answer to Problem 14P

The magnitude of acceleration of each object is $\underset{\_}{2.31\text{m}/{\text{s}}^2}$, and the direction is $\underset{\_}{\text{down for }{m}_1,\text{left for }{m}_2,\text{and up for }{m}_3}$.

Explanation of Solution

Apply Newton’s second law for in the $y$ direction of Figure 1.

    ${\displaystyle \sum F_y=}{m}_1{a}_y$        (I)

Here, $F_y$ is the net force in the $y$ direction, $m$ is the mass, and $a_y$ is the acceleration in the $y$ direction.

From the Figure 1, write the expression for net force in the $y$ direction, and equate with the equation (I).

    $T_{\text{12}}-m_{1}g=m_1{a}_y$        (II)

Here, $T_{12}$ is the tension in the left rope.

Substitute, $-a$ for $a_y$ in the equation (II).

    $\begin{array}{c}{T}_{\text{12}}-m_{1}g=-m_{1}a\\ -T_{\text{12}}+m_{1}g=m_{1}a\end{array}$        (III)

From the Figure 2, write the expression for net force in the $x$ direction, and equate with the Newton’s second law.

    $\begin{array}{c}-T_{12}+{\mu }_{\text{k}}n+T_{23}=-m_{2}a\\ T_{12}-{\mu }_{\text{k}}n-T_{23}=m_{2}a\end{array}$        (IV)

Since, $a_y=-a$.

Here, $T_{23}$ is the tension in the right rope.

From the Figure 2, write the expression for net force in the $y$ direction, and equate with the equation (I).

    $n-m_{2}g=0$        (V)

From the Figure 3, write the expression for net force in the $y$ direction, and equate with the equation (I).

    $T_{23}-m_{3}g=m_{3}a$        (VI)

Add equation (III), (IV), and (VI).

    $\begin{array}{c}-T_{\text{12}}+m_{1}g+T_{12}-{\mu }_{\text{k}}n-T_{23}+T_{23}-m_{3}g=m_{1}a+m_{2}a+m_{3}a\\ m_{1}g-{\mu }_{\text{k}}n-m_{3}g=a\left(m_1+m_2+m_3\right)\end{array}$        (VII)

Write the expression for normal force for mass $m_2$.

    $n=m_{2}g$

Use, $m_{2}g$ for $n$ in the equation (VII).

    $m_{1}g-{\mu }_{\text{k}}{m}_{2}g-m_{3}g=a\left(m_1+m_2+m_3\right)$        (VIII)

Conclusion:

Substitute, $4.00\text{kg}$ for $m_1$, $1.00\text{kg}$ for $m_2$, $2.00\text{kg}$ for $m_3$, $9.80\text{m}/{\text{s}}^2$ for $g$, and $0.350$ for ${\mu }_{\text{k}}$ in the equation (VIII), to find $a$.

    $\begin{array}{c}\left(4.00\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)-0.350\left(1.00\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)-\left(2.00\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)=a\left(4.00\text{kg}+1.00\text{kg}+2.00\text{kg}\right)\\ 39.2\text{N}-3.43\text{N}-19.6\text{N}=\left(7.00\text{kg}\right)a\\ a=\frac{\left(7.00\text{kg}\right)a}{39.2\text{N}-3.43\text{N}-19.6\text{N}}\\ =2.31\text{m}/{\text{s}}^2\end{array}$

Therefore, the magnitude of acceleration of each object is $\underset{\_}{2.31\text{m}/{\text{s}}^2}$, and the direction is $\underset{\_}{\text{down for }{m}_1,\text{left for }{m}_2,\text{and up for }{m}_3}$.

To determine

The tension in the two cords.

Answer to Problem 14P

Tension in the first code is $\underset{\_}{30.0\text{N}}$, and the tension in the second string is $\underset{\_}{24.2\text{N}}$.

Explanation of Solution

Conclusion:

Substitute, $2.31\text{m}/{\text{s}}^2$ for $a$, $4.00\text{kg}$ for $m_1$, and $9.80\text{m}/{\text{s}}^2$ for $g$ in the equation (III), and solve for $T_{12}$.

    $\begin{array}{c}-T_{\text{12}}+\left(4.00\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)=\left(4.00\text{kg}\right)\left(2.31\text{m}/{\text{s}}^2\right)\\ T_{\text{12}}=30.0N\end{array}$

Substitute, $2.31\text{m}/{\text{s}}^2$ for $a$, $2.00\text{kg}$ for $m_3$, and $9.80\text{m}/{\text{s}}^2$ for $g$ in the equation (VI), and solve for $T_{23}$.

    $\begin{array}{c}{T}_{23}-\left(2.00\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)=\left(2.00\text{kg}\right)\left(2.31\text{m}/{\text{s}}^2\right)\\ T_{23}=24.2\text{N}\end{array}$

Therefore, tension in the first code is $\underset{\_}{30.0\text{N}}$, and the tension in the second string is $\underset{\_}{24.2\text{N}}$.

To determine

Whether the tension increase, decrease, or remain the same if the tabletop were smooth.

Answer to Problem 14P

The tension in the left rope will decreases, and the tension in the right rope will increases.

Explanation of Solution

If the table top is smooth, the friction on the top will disappear. So the acceleration become larger.

From the equation (III), $T_{12}=m_{1}g-m_{1}a$, and from the equation (VI), $T_{23}=m_{3}g+m_{3}a$. According to these equations, for large acceleration, tension in the left rope decreases, and for the right rope tension increases.

Conclusion:

Therefore, the tension in the left rope will decreases, and the tension in the right rope will increases.