Chapter 17.1, Problem 17.14P

The double pulley shown has a mass of 15 kg and a centroidal radius of gyration of 160 mm. Cylinder A and block B are attached to cords that are wrapped on the pulleys as shown. The coefficient of kinetic friction between block B and the surface is 0.2. Knowing that the system is at rest in the position shown when a constant force P = 200 N is applied to cylinder A, determine (a) the velocity of cylinder A as it strikes the ground, (b) the total distance that block B moves before coming to rest.

Fig. P17.14

To determine

Find the velocity of cylinder A as it strikes the ground.

Answer to Problem 17.14P

The velocity of the cylinder A when it strikes the ground is $\underset{\_}{4.79\text{m/s}}$.

Explanation of Solution

Given information:

The mass of the cylinder A is $m_A=5\text{kg}$.

The mass of the block B is $m_B=15\text{kg}$.

The radius of the outer pulley is $r_A=250\text{mm}$.

The radius of the inner pulley is $r_B=150\text{mm}$.

The centroidal radius of gyration is $k_C=160\text{mm}$.

The coefficient of friction between the surface and the block B is ${\mu }_k=0.20$.

The constant force applied at cylinder A is $P=200\text{N}$.

Calculation:

Consider the acceleration due to gravity is $g=9.81{\text{m/s}}^2$.

Consider the radius of the outer pulley as $r_A$.

Consider the radius of the inner pulley as $r_B$.

Find the velocity in the outer pulley $\left(v_A\right)$ using the relation.

$\begin{array}{l}{v}_A=r_A{\omega }_C\\ {\omega }_C=\frac{v_A}{r_A}\end{array}$ (1)

Here, the angular velocity of the pulley is ${\omega }_C$.

Find the velocity in the inner pulley $\left(v_B\right)$ using the relation.

$v_B=r_B{\omega }_C$

Substitute $\frac{v_A}{r_A}$ for ${\omega }_C$.

$v_B=r_B\frac{v_A}{r_A}$ (2)

Find the distance of the outer pulley $\left(s_A\right)$ is in contact using the relation.

$\begin{array}{l}{s}_A=r_A{\theta }_C\\ {\theta }_C=\frac{s_A}{r_A}\end{array}$

Here, the number of revolutions in the pulley C is ${\theta }_C$.

Find the distance of the inner pulley $\left(s_B\right)$ is in contact using the relation.

$s_B=r_B{\theta }_C$

Substitute $\frac{s_A}{r_A}$ for ${\theta }_C$.

$s_B=r_B\frac{s_A}{r_A}$ (3)

The initial total kinetic energy at rest is zero.

$T_1=0$

Find the mass moment of inertia in the pulley C $\left({\overline{I}}_C\right)$ using the equation.

${\overline{I}}_C=m_C{k}_C^2$

Here, the mass in the pulley C is $m_C$.

Substitute 15 kg for $m_C$ and 160 mm for $k_C$.

$\begin{array}{c}{\overline{I}}_C=15\times {\left(160\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^2\\ =0.384\text{kg}\cdot {\text{m}}^2\end{array}$

Find the total kinetic energy $\left(T_2\right)$ at the final stage using the relation.

$T_2=\frac{1}{2}{m}_A{v}_A^2+\frac{1}{2}{m}_B{v}_B^2+\frac{1}{2}{\overline{I}}_C{\omega }_C^2$

Substitute 5 kg for $m_A$, 15 kg for $m_B$, $r_B\frac{v_A}{r_A}$ for $v_B$, $0.384\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_C$, and $\frac{v_A}{r_A}$ for ${\omega }_C$.

$T_2=\frac{1}{2}\times 5\times v_A^2+\frac{1}{2}\times 15\times {\left(r_B\frac{v_A}{r_A}\right)}^2+\frac{1}{2}\times 0.384\times {\left(\frac{v_A}{r_A}\right)}^2$

Substitute 250 mm for $r_A$ and 150 mm for $r_B$.

$\begin{array}{c}{T}_2=\frac{1}{2}\times 5\times v_A^2+\frac{1}{2}\times 15\times {\left(\frac{150}{250}{v}_A\right)}^2+\frac{1}{2}\times 0.384\times {\left(\frac{v_A}{250\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}}\right)}^2\\ =2.5v_A^2+2.7v_A^2+3.072v_A^2\\ =8.272v_A^2\end{array}$

When $s_A=1\text{m}$;

Substitute 150 mm for $r_B$, 1 m for $s_A$ and 250 mm for $r_A$ in Equation (3).

$\begin{array}{c}{s}_B=150\times \frac{1}{250}\\ =0.6\text{m}\end{array}$

Show free-body diagram the block B as in Figure 1.

Resolve the vertical component of forces as follows;

$\begin{array}{c}{\displaystyle \sum F_y=0}\\ N_B-m_{B}g\cos 30°=0\\ N_B=15\times 9.81\times \cos 30°\\ =127.44\text{N}\end{array}$

Find the frictional force $F_f$ using the relation.

$F_f={\mu }_k{N}_B$

Substitute 0.20 for ${\mu }_k$ and 127.44 N for $N_B$.

$\begin{array}{c}{F}_f=0.20\times 127.44\\ =25.487\text{N}\end{array}$

Apply the principle of work and energy for the cylinder A, the block B and the double pulley C as follows;

$U_{1\to 2}=P{s}_A+m_{A}g{s}_A-F_f{s}_B-m_{B}g{s}_B\sin 30°$

Substitute 200 N for P, 1 m for $s_A$, 5 kg for $m_A$, $9.81{\text{m/s}}^2$ for g, 25.487 N for $F_f$, 0.6 m for $s_B$, and 15 kg for $m_B$.

$\begin{array}{c}{U}_{1\to 2}=\left(200\times 1\right)+\left(5\times 9.81\times 1\right)-\left(25.487\times 0.60\right)-\left(15\times 9.81\times 0.60\times \sin 30°\right)\\ =200+49.05-15.2922-44.145\\ =189.6128\text{J}\end{array}$

Write the equation of work and energy for the system using the equation.

$T_1+U_{1\to 2}=T_2$

Substitute 0 for $T_1$, 189.6128 J for $U_{1\to 2}$, and $8.272v_A^2$ for $T_2$

$\begin{array}{l}0+189.6128=8.272v_A^2\\ v_A=4.7877\text{m/s}\end{array}$

Therefore, the velocity of the cylinder A when it strikes the ground is $\underset{\_}{4.79\text{m/s}}$.

To determine

Find the total distance the block B moves before coming to rest.

Answer to Problem 17.14P

The total distance travelled by the block B before coming to rest is $\underset{\_}{1.936\text{m}}$.

Explanation of Solution

Given information:

The mass of the cylinder A is $m_A=5\text{kg}$.

The mass of the block B is $m_B=15\text{kg}$.

The radius of the outer pulley is $r_A=250\text{mm}$.

The radius of the inner pulley is $r_B=150\text{mm}$.

The centroidal radius of gyration is $k_C=160\text{mm}$.

The coefficient of friction between the surface and the block B is ${\mu }_k=0.20$.

The constant force applied at cylinder A is $P=200\text{N}$.

Calculation:

Refer part (a) for $v_A$ calculation.

Substitute $4.7877\text{m/s}$ for $v_A$ and 250 mm for $r_A$ in Equation (1).

$\begin{array}{c}{\omega }_C=\frac{4.7877}{250\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}}\\ =19.1508\text{rad/s}\end{array}$

Substitute 150 mm for $r_B$, $4.7877\text{m/s}$ for $v_A$, and 250 mm for $r_A$ in Equation (2).

$\begin{array}{c}{v}_B=150\times \frac{4.7877}{250}\\ =2.87262\text{m/s}\end{array}$

Find the total kinetic energy $\left(T_3\right)$ when the cylinder A strikes the ground.

$T_3=\frac{1}{2}{m}_B{v}_B^2+\frac{1}{2}{\overline{I}}_C{\omega }_C^2$

Substitute 5 kg for $m_A$, $2.87262\text{m/s}$ for $v_B$, 15 kg for $m_B$, $0.384\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_C$, and $19.1508\text{rad/s}$ for ${\omega }_C$.

$\begin{array}{c}{T}_3=\frac{1}{2}\times 15\times {2.87262}^2+\frac{1}{2}\times 0.384\times {19.1508}^2\\ =61.89+70.4166\\ =132.3066\text{J}\end{array}$

At the final position, the system comes at rest.

The kinetic energy at rest is zero.

$T_4=0$

Apply the principle of work and energy for the block B as follows;

$U_{3\to 4}=-F_f{s^{\prime }}_B-m_{B}g{s^{\prime }}_B\sin 30°$

Here, the additional distance travelled by the block is ${s^{\prime }}_B$.

Substitute $9.81{\text{m/s}}^2$ for g, 25.487 N for $F_f$, and 15 kg for $m_B$.

$\begin{array}{c}{U}_{3\to 4}=-25.487\times {s^{\prime }}_B-15\times 9.81\times {s^{\prime }}_B\sin 30°\\ =-99.062{s^{\prime }}_B\end{array}$

Write the equation of work and energy for the system using the equation.

$T_3+U_{3\to 4}=T_4$

Substitute 132.3066 J for $T_3$, $-99.062{s^{\prime }}_B$ for $U_{3\to 4}$, and 0 for $T_4$

$\begin{array}{l}132.3066-99.062{s^{\prime }}_B=0\\ {s^{\prime }}_B=1.3356\text{m}\end{array}$

Find the total distance $\left(d_B\right)$ travelled by the block B before coming to rest using the relation.

$d_B=s_B+{s^{\prime }}_B$

Substitute 0.6 m for $s_B$ and 1.3356 m for ${s^{\prime }}_B$.

$\begin{array}{c}{d}_B=0.6+1.3356\\ =1.9356\text{m}\end{array}$

Therefore, the total distance travelled by the block B before coming to rest is $\underset{\_}{1.936\text{m}}$.