Chapter 5, Problem 31P

Three objects are connected on a table as shown in Figure P5.31. 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 P5.31

To determine

To draw: The free body diagram of each object.

Answer to Problem 31P

Therefore, the free body diagrams of each object are shown in figure I.

Explanation of Solution

Given information: The mass of block $1$ is $4.00\text{kg}$, the mass of block $2$ is $1.00\text{kg}$, the mass of block $3$ is $2.00\text{kg}$ and the coefficient of friction between mass and table is $0.350$.

The free body diagram of each of the block is given blow.

Figure I

Conclusion:

Therefore, the free body diagrams of each object are shown in figure I.

To determine

The acceleration of each object including its direction.

Answer to Problem 31P

The acceleration of each object including its direction is $2.31\text{m}/{\text{s}}^2$ down for $m_1$, left for $m_2$ and up for $m_3$.

Explanation of Solution

Given information:

The mass of block $1$ is $4.00\text{kg}$, the mass of block $2$ is $1.00\text{kg}$, the mass of block $3$ is $2.00\text{kg}$ and the coefficient of kinetic friction between mass and table is $0.350$.

Consider $a$ is the positive magnitude of the acceleration $-a\widehat{\text{j}}$ of $m_1$, of the acceleration $-a\widehat{\text{i}}$ of $m_2$ and of the acceleration $+a\widehat{\text{j}}$ of $m_3$.

The force equation for block $1$ is,

$\begin{array}{c}{\displaystyle \sum F_{\text{y}}}=m{a}_{\text{y}}\\ T_{12}-m_{1}g=-m_{1}a\\ m_{1}a=-T_{12}+m_{1}g\end{array}$

• $T_{12}$ is the tension.
• $m_1$ is the mass of block $1$.
• $g$ is the acceleration due to gravity.

Substitute $4.00\text{kg}$ for $m_1$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in above equation.

$\left(4.00\text{kg}\right)a=-T_{12}+\left(4.00\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)$

$\left(4.00\text{kg}\right)a=-T_{12}+39.2\text{N}$ (I)

The force equation for block $2$ for horizontal direction is,

$\begin{array}{c}{\displaystyle \sum F_{\text{x}}}=m{a}_{\text{x}}\\ -T_{12}+{\mu }_{\text{k}}n+T_{23}=-m_{2}a\end{array}$

• ${\mu }_{\text{k}}$ is the coefficient of kinetic friction.
• $n$ is the normal reaction.
• $m_2$ is the mass of block $2$.
• $T_{23}$ is the tension in the string.

The force equation for vertical direction is,

$\begin{array}{c}{\displaystyle \sum F_{\text{y}}}=m{a}_{\text{y}}\\ n-m_{2}g=0\\ n=m_{2}g\end{array}$

• $g$ is the acceleration due to gravity.

Substitute $m_{2}g$ for $n$ in above equation.

$-T_{12}+{\mu }_{\text{k}}{m}_{2}g+T_{23}=-m_{2}a$

Substitute $1.00\text{kg}$ for $m_2$, $0.350$ for ${\mu }_{\text{k}}$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in above equation.

$\begin{array}{c}-T_{12}+\left(0.350\right)\left(1.00\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)+T_{23}=-\left(1.00\text{kg}\right)a\\ \left(1.00\text{kg}\right)a=T_{12}-\left(0.350\right)\left(9.80\text{N}\right)-T_{23}\end{array}$ (II)

The force equation for block $3$ is,

$\begin{array}{c}{\displaystyle \sum F_{\text{y}}}=m{a}_{\text{y}}\\ T_{23}-m_{3}g=m_{3}a\end{array}$

• $m_3$ is the mass of block $3$.

Substitute $2.00\text{kg}$ for $m_3$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in above equation.

$\begin{array}{c}{T}_{23}-\left(2.00\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)=\left(2.00\text{kg}\right)a\\ \left(2.00\text{kg}\right)a=T_{23}-\left(19.6\text{N}\right)\end{array}$ (III)

Now, add the above equation (I), (II) and (III) then,

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

Conclusion:

Therefore, the acceleration of each object including its direction is $2.31\text{m}/{\text{s}}^2$ down for $m_1$, left for $m_2$ and up for $m_3$.

To determine

The tension in the two cords.

Answer to Problem 31P

The tension in the first cord is $30.0\text{N}$ and the tension in the second cord is $24.2\text{N}$.

Explanation of Solution

Given information:

The mass of block $1$ is $4.00\text{kg}$, the mass of block $2$ is $1.00\text{kg}$, the mass of block $3$ is $2.00\text{kg}$ and the coefficient of kinetic friction between mass and table is $0.350$.

Recall the equation (I).

$\left(4.00\text{kg}\right)a=-T_{12}+39.2\text{N}$

Substitute $2.31\text{m}/{\text{s}}^2$ for $a$ in above equation.

$\begin{array}{c}\left(4.00\text{kg}\right)\left(2.31\text{m}/{\text{s}}^2\right)=-T_{12}+39.2\text{N}\\ T_{12}=30.0\text{N}\end{array}$

Recall the equation (III).

$\left(2.00\text{kg}\right)a=T_{23}-\left(19.6\text{N}\right)$

Substitute $2.31\text{m}/{\text{s}}^2$ for $a$ in above equation.

$\begin{array}{c}\left(2.00\text{kg}\right)\left(2.31\text{m}/{\text{s}}^2\right)=T_{23}-\left(19.6\text{N}\right)\\ T_{23}=24.2\text{N}\end{array}$

Conclusion:

Therefore, the tension in the first cord is $30.0\text{N}$ and the tension in the second cord is $24.2\text{N}$.

To determine

To explain: The tension is increases, decreases or remains same if the table top is smooth.

Answer to Problem 31P

Therefore, the tension in first cord $T_{12}$ decreases and tension in the second cord $T_{23}$ increases.

Explanation of Solution

Given information:

The mass of block $1$ is $4.00\text{kg}$, the mass of block $2$ is $1.00\text{kg}$, the mass of block $3$ is $2.00\text{kg}$ and the coefficient of kinetic friction between mass and table is $0.350$.

If the tabletop is smooth, the friction disappears and the coefficient of the kinetic friction becomes zero.

Consider the following equation from the above calculations.

$\begin{array}{c}{m}_{1}a=-T_{12}+m_{1}g\\ -T_{12}+{\mu }_{\text{k}}{m}_{2}g+T_{23}=-m_{2}a\\ T_{23}-m_{3}g=m_{3}a\end{array}$

Then, the expression from the above three equation is,

$\begin{array}{c}{m}_{1}a+m_{2}a+m_{3}a=-T_{12}+m_{1}g+T_{12}+{\mu }_{\text{k}}{m}_{2}g-T_{23}+T_{23}-m_{3}g\\ a=\frac{\left(m_1+{\mu }_{\text{k}}{m}_2-m_3\right)g}{m_1+m_2+m_3}\end{array}$

Substitute $4.00\text{kg}$ for $m_1$, $1.00\text{kg}$ for $m_2$, $0$ for ${\mu }_{\text{k}}$, $2.00\text{kg}$ for $m_3$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in above expression.

$\begin{array}{c}a=\frac{\left(4.00\text{kg}+\left(0\right)\left(1.00\text{kg}\right)-2.00\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)}{4.00\text{kg}+1.00\text{kg}+2.00\text{kg}}\\ =\frac{\left(2.00\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)}{7.00\text{kg}}\\ =2.80\text{m}/{\text{s}}^2\end{array}$

Recall the equation (I).

$\left(4.00\text{kg}\right)a=-T_{12}+39.2\text{N}$

Substitute $2.80\text{m}/{\text{s}}^2$ for $a$ in above equation.

$\begin{array}{c}\left(4.00\text{kg}\right)\left(2.80\text{m}/{\text{s}}^2\right)=-T_{12}+39.2\text{N}\\ T_{12}=28.0\text{N}\end{array}$

Recall the equation (III).

$\left(2.00\text{kg}\right)a=T_{23}-\left(19.6\text{N}\right)$

Substitute $2.80\text{m}/{\text{s}}^2$ for $a$ in above equation.

$\begin{array}{c}\left(2.00\text{kg}\right)\left(2.80\text{m}/{\text{s}}^2\right)=T_{23}-\left(19.6\text{N}\right)\\ T_{23}=25.2\text{N}\end{array}$

Compare these tensions with the tensions in the cord due to $a=2.31\text{m}/{\text{s}}^2$.

The tensions in the cord for $a=2.31\text{m}/{\text{s}}^2$ is,

$\begin{array}{l}{T}_{12}=30.0\text{N}\\ T_{23}=24.2\text{N}\end{array}$

The tensions in the cord for ${\mu }_{\text{k}}=0$ is,

$\begin{array}{l}{T}_{12}=28.0\text{N}\\ T_{23}=25.2\text{N}\end{array}$

The tensions are changed when the tabletop is smooth. So the tension in first cord $T_{12}$ decreases and tension in the second cord $T_{23}$ increases.

Conclusion:

Therefore, the tension in first cord $T_{12}$ decreases and tension in the second cord $T_{23}$ increases.