Chapter 10, Problem 72P

Review. A block of mass m₁ = 2.00 kg and a block of mass m₂ = 6.00 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10.0 kg. The fixed, wedge-shaped ramp makes an angle of θ = 30.0° as shown in Figure P10.72. The coefficient of kinetic friction is 0.360 for both blocks. (a) Draw force diagrams of both blocks and of the pulley. Determine (b) the acceleration of the two blocks and (c) the tensions in the string on both sides of the pulley.

Figure P10.72

To determine

Sketch the force diagrams of both blocks and pulley.

Answer to Problem 72P

The free body diagram of the block of mass $m_1$ is shown below.

The free body diagram of the block of mass $m_2$ is shown below.

The free body diagram of the pulley is shown below.

Explanation of Solution

Consider the figure of given system

In physics and engineering free body diagrams are used for visualizing the forces, movements, and reaction forces acting on a body. The direction of the forces are also shown in free body diagram using straight arrows.

The free body diagram of the block of mass $m_1$ is shown below

The mass $m_1$ is resting on the plane, a normal force $n_1$ acting in upward direction, just opposite to normal force, the gravitational force due to the weight of the block $m_{1}g$ is acting downwards. ${\text{f}}_{k1}$ is the kinetic friction force, and $T_1$ is the tension of the string connected between mass $m_1$, and the pulley.

The free body diagram of the block of mass $m_2$ is shown below

The mass $m_1$ is resting on the inclined plane, a normal force $n_2$ acting in direction perpendicular to the surface of the block, the gravitational force due to the weight of the block $m_{2}g$ is acting downwards. ${\text{f}}_{k2}$ is the kinetic friction force, and $T_2$ is the tension of the string connected between mass $m_2$, and the pulley, both are in same direction.

The free body diagram of the pulley is shown below.

$Mg$ is the weight of the pulley, and $n_3$ is the normal force on pulley.

To determine

The acceleration of two blocks.

Answer to Problem 72P

The acceleration of two blocks is $\underset{\_}{0.309\text{m}/{\text{s}}^2}$.

Explanation of Solution

Consider the free body diagram of mass $m_1$.

Write the expression for the total force acting on mass $m_1$ in vertical direction.

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

Here, $m$ is the mass, and $a_y$ is the acceleration along $y$ direction.

The net force acting in vertical direction will be zero, since the normal force and $m_{1}g$ are opposite to each other.

Rewrite equation (I).

  $\begin{array}{c}{\displaystyle \sum F_y}=m{a}_y\\ n_1-m_{1}g=0\\ n_1=m_{1}g\end{array}$        (II)

Write the expression for kinetic friction force on $m_1$.

  ${\text{f}}_{k1}={\mu }_k{n}_1$        (III)

Here, ${\mu }_k$ is the coefficient of kinetic friction.

Write the expression for the force acting in $x$ direction.

  ${\displaystyle \sum F_x}=m{a}_x$        (IV)

Here, $a_x$ is the acceleration along $x$ direction.

Rewrite equation (IV) including the components of forces in horizontal direction.

  $-{\text{f}}_{k1}+T_1=m_{1}a$        (V)

Write the expression for the total torque.

  ${\displaystyle \sum \tau }=I\alpha$        (VI)

Here, $I$ is the moment of inertia, $\alpha$ is the angular acceleration.

Substitute, $-T_{1}R+T_{2}R$ for ${\displaystyle \sum \tau }$, $\frac{1}{2}M{R}^2$ for $I$, and $\frac{a}{R}$ for $\alpha$ in equation (VI).

  $\begin{array}{c}-T_{1}R+T_{2}R=\frac{1}{2}M{R}^2\left(\frac{a}{R}\right)\\ -T_1+T_2=\frac{1}{2}Ma\end{array}$        (VII)

Similarly the net force acting on $x$ direction in mass $m_2$ is also zero.

  $\begin{array}{c}{n}_2-m_{2}g\cos \theta =0\\ n_2=m_{2}g\cos \theta \end{array}$        (VIII)

Write the expression for the kinetic force of friction acting on mass $m_2$.

  ${\text{f}}_{k2}={\mu }_k{n}_2$        (IX)

Write the expression for the forces acting on $y$ direction.

  $-{\text{f}}_{k2}-T_2+m_{2}g\sin \theta =m_{2}a$        (X)

Add equation (V), (VII), and (X)

$\begin{array}{c}-{\text{f}}_{k1}+T_1+\left(-T_1+T_2\right)-{\text{f}}_{k2}-T_2+m_{2}g\sin \theta =m_{2}a+\frac{1}{2}Ma+m_{1}a\\ -{\text{f}}_{k1}-{\text{f}}_{k2}+m_{2}g\sin \theta =m_{2}a+\frac{1}{2}Ma+m_{1}a\end{array}$        (XI)

Substitute, equation (IX) and (III) in (XI), $m_{1}g$ for $n_1$, $m_{2}g\cos \theta$ for $n_2$ to obtain an expression for acceleration.

  $\begin{array}{c}-\left({\mu }_k{m}_{1}g\right)-\left({\mu }_k{m}_{2}g\cos \theta \right)+m_{2}g\sin \theta =\left(m_2+\frac{1}{2}M+m_1\right)a\\ a=\frac{m_2\left(\sin \theta -{\mu }_k\cos \theta \right)-{\mu }_k{m}_1}{m_2+\frac{1}{2}M+m_1}g\end{array}$        (XII)

Conclusion:

Substitute, $6.00\text{kg}$ for $m_2$, $30.0°$ for $\theta$, $0.360$ for ${\mu }_k$, $2.00\text{kg}$ for $m_1$, $9.80\text{m}/{\text{s}}^2$ for $g$, $10.0\text{kg}$ for $M$ in equation (XII).

  $\begin{array}{c}a=\frac{6.00\text{kg}\left(\sin 30.0°-0.360\times \cos 30.0°\right)-0.360\times 2.00\text{kg}}{6.00\text{kg}+\frac{1}{2}\left(10.0\text{kg}\right)+2.00\text{kg}}\left(9.80\text{m}/{\text{s}}^2\right)\\ =0.309\text{m}/{\text{s}}^2\end{array}$

Therefore, the acceleration of two blocks is $\underset{\_}{0.309\text{m}/{\text{s}}^2}$.

To determine

The tension in the string on both sides of the pulley.

Answer to Problem 72P

The tension in the string on both sides of the pulley is $\underset{\_}{T_1=7.67\text{N}}$, and $\underset{\_}{T_2=9.22\text{N}}$.

Explanation of Solution

Use equation (V) and (VII) to obtain the answer.

Conclusion:

Substitute, $2.00\text{kg}$ for $m_1$, $0.309\text{m}/{\text{s}}^2$ for $a$, and $7.06\text{N}$ for ${\text{f}}_{k1}$ in equation (V) to find $T_1$.

  $\begin{array}{c}-\left(7.06\text{N}\right)+T_1=\left(2.00\text{kg}\right)\left(0.309\text{m}/{\text{s}}^2\right)\\ T_1=\left(2.00\text{kg}\right)\left(0.309\text{m}/{\text{s}}^2\right)+7.06\text{N}\\ =7.67\text{N}\end{array}$

Substitute, $10.0\text{kg}$ for $M$, $0.309\text{m}/{\text{s}}^2$ for $a$, $7.67\text{N}$ for $T_1$ in equation (VII) to find $T_2$.

  $\begin{array}{c}-7.67\text{N}+T_2=\frac{1}{2}\left(10.0\text{kg}\right)\left(0.309\text{m}/{\text{s}}^2\right)\\ T_2=7.67\text{N}+\frac{1}{2}\left(10.0\text{kg}\right)\left(0.309\text{m}/{\text{s}}^2\right)\\ =9.22\text{N}\end{array}$

Therefore, the tension in the string on both sides of the pulley is $\underset{\_}{T_1=7.67\text{N}}$, and $\underset{\_}{T_2=9.22\text{N}}$.