Chapter 4, Problem 57AP

Three objects are connected on a table as shown in Figure P4.73. 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 as shown, and the pulleys are frictionless. (a) Draw a diagram of the forces on each object. (b) Determine the acceleration of each object, including its direction. (c) Determine the tensions in the two cords. (d) If the tabletop were smooth, would the tensions increase, decrease, or remain the same? Explain.

Figure P4.73

To determine

The forces on the masses.

Answer to Problem 57AP

Tension, Normal force and gravity are the forces acting on the masses.

Explanation of Solution

Given Info: The masses are $m_1=4.00\text{kg}$, $m_2=1.00\text{kg}$ and $m_3=2.00\text{kg}$. The co-efficient of kinetic friction is 0.350.

The free body diagram is given below.

Conclusion:

Tension, Normal force and gravity are the forces acting on the masses.

To determine

The acceleration of the masses.

Answer to Problem 57AP

Solution: The acceleration of the masses is $2.31{\text{ms}}^{-2}$.

Explanation of Solution

Given Info: The masses are $m_1=4.00\text{kg}$, $m_2=1.00\text{kg}$ and $m_3=2.00\text{kg}$. The co-efficient of kinetic friction is 0.350.

The free body diagram is given below.

• $T_1$, $T_2$ and $T_3$ are the tension in the cords.
• $a_1$, $a_2$ and $a_3$ are the acceleration of the masses.
• $F_k$ is the kinetic friction force.

From the diagram,

$m_{1}g-T_1=m_{1}a$ (I)

$T_1-T_2-F_k=m_{2}a$ (II)

$T_2-m_{3}g=m_{3}a$ (III)

Substitute Equations (I) and (III) in (II) and re-arrange to get a.

$a=\frac{\left(m_1-m_3\right)g-F_k}{m_1+m_2+m_3}$ (IV)

The formula for the kinetic friction force is,

$F_k={\mu }_k{m}_{2}g$ (V)

Substitute Equation (V) in (IV) to get a.

$a=g\left(\frac{m_1-m_3-{\mu }_k{m}_2}{m_1+m_2+m_3}\right)$

Substitute 4.00 kg for $m_1$, 1.00 kg for $m_2$, 2.00 kg for $m_3$, 0.350 for ${\mu }_k$ and $9.8{\text{ms}}^{-2}$ for g to get a.

$\begin{array}{c}a=\left(9.8{\text{ms}}^{-2}\right)\left(\frac{\left(4.00\text{kg}\right)-\left(2.00\text{kg}\right)-\left(0.350\right)\left(1.00\text{kg}\right)}{\left(4.00\text{kg}\right)+\left(1.00\text{kg}\right)+\left(2.00\text{kg}\right)}\right)\\ =2.31{\text{ms}}^{-2}\end{array}$

Conclusion:

The acceleration of the masses is $2.31{\text{ms}}^{-2}$.

To determine

The tension on the two cords.

Answer to Problem 57AP

Solution:

The tension in the left cord is 29.96 N.

The tension in the right cord is 24.22 N.

Explanation of Solution

Given Info: The masses are $m_1=4.00\text{kg}$, $m_2=1.00\text{kg}$ and $m_3=2.00\text{kg}$. The co-efficient of kinetic friction is 0.350.

From Equation (I) of (b),

$T_1=m_1\left(g-a\right)$

Substitute 4.00 kg for $m_1$, $9.8{\text{ms}}^{-2}$ for g and $2.41{\text{ms}}^{-2}$ for a to get $T_1$.

$\begin{array}{c}{T}_1=\left(4.00\text{kg}\right)\left(9.8{\text{ms}}^{-2}-2.31{\text{ms}}^{-2}\right)\\ =29.96\text{}\text{N}\end{array}$

From Equation (III) of (b),

$T_2=m_3\left(g+a\right)$

Substitute 2.00 kg for $m_3$, $9.8{\text{ms}}^{-2}$ for g and $2.41{\text{ms}}^{-2}$ for a to get $T_1$.

$\begin{array}{c}{T}_2=\left(2.00\text{kg}\right)\left(9.8{\text{ms}}^{-2}+2.31{\text{ms}}^{-2}\right)\\ =24.22\text{}\text{N}\end{array}$

Conclusion:

The tension in the left cord is 29.96 N.

The tension in the right cord is 24.22 N.

To determine

The tension on the two cords.

Answer to Problem 57AP

The tension in the left cord decreases.

The tension in the right cord increases.

Explanation of Solution

Given Info: The masses are $m_1=4.00\text{kg}$, $m_2=1.00\text{kg}$ and $m_3=2.00\text{kg}$. The co-efficient of kinetic friction is 0.350.

From (b),

$a=g\left(\frac{m_1-m_3-F_k}{m_1+m_2+m_3}\right)$

$T_1=m_1\left(g-a\right)$

$T_2=m_3\left(g+a\right)$

For a smooth surface friction force is zero. Therefore, a increases. Hence, $T_1$ decreases and $T_2$ increases.

Conclusion:

The tension in the left cord decreases.

The tension in the right cord increases.