Chapter 13.3, Problem 13.148P

(a)

To determine

Find the impulsive force $F\Delta t$ and the energy absorbed $\left(\Delta T\right)$ by the rivet which has infinite mass.

Answer to Problem 13.148P

The impulsive force $F\Delta t$ and the energy absorbed $\left(\Delta T\right)$ by the rivet are $\underset{\_}{0.932\text{lb}\cdot \text{s}}$ and $\underset{\_}{9.32\text{ft}\cdot \text{lb}}$ respectively.

Explanation of Solution

Given information:

The weight of the hammer (W) is $1.5\text{lb}$.

The velocity of the hammer $\left(v_H\right)$ is $20\text{ft/s}$.

The acceleration due to gravity (g) is $32.2\text{ft}/{\text{s}}^{\text{2}}$.

Calculation:

Show the free body momentum diagram of the hammer head and anvil as Figure (1).

Use the principle of conservation of momentum to the impact of hammer head and anvil, to obtain the final velocity of anvil and hammer after the impact.

The expression for the principle of conservation of momentum as follows;

$m_H{v}_H+m_A{v}_A=m_H{v}_2+m_A{v}_2$

Initially the anvil is at rest, so the velocity will be zero.

Substitute 0 for $v_A$.

$\begin{array}{l}{m}_H{v}_H+0=\left(m_H+m_A\right)v_2\\ v_2=\frac{m_H{v}_H}{m_H+m_A}\end{array}$ (1)

Show the free body impulse-momentum diagram of the hammer head as in Figure (2).

Use the principle of the impulse momentum to the hammer head to find the impulse exerted $\left(F\Delta t\right)$ on the rivet.

The expression for the principle of the impulse momentum as follows;

$\begin{array}{l}{m}_H{v}_H-F\Delta t=m_H{v}_2\\ F\Delta t=m_H\left(v_H-v_2\right)\end{array}$ (2)

Calculate the mass of the hammer $\left(m_H\right)$ using the formula:

$m_H=\frac{W_H}{g}$

Substitute $1.5\text{lb}$ for $W_H$, and 32.2 $\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{m}_H=\frac{1.5\text{lb}}{32.2\text{ft}/{\text{s}}^2}\\ =0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}\end{array}$

The expression for the kinetic energy of the hammer before impact $\left(T_1\right)$ as follows;

$T_1=\frac{1}{2}{m}_H{v}_H^2$

Substitute $0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_H$, and $20\text{ft}/\text{s}$ for $v_H$.

$\begin{array}{c}{T}_1=\frac{1}{2}{m}_H{v}_H^2\\ =\frac{1}{2}\left(0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}\right){\left(20\text{ft}/\text{s}\right)}^2\\ =\left(0.02329\right)\left(400\right)\\ =9.316\text{ft}\text{.lb}\end{array}$

Calculate the final kinetic energy of the hammer and anvil system after the impact $\left(T_2\right)$ using the formula:

$T_2=\frac{1}{2}\left(m_H+m_A\right)v_2^2$

Substitute $\frac{m_H{v}_H}{m_H+m_A}$ for $v_2$.

$\begin{array}{c}{T}_2=\frac{1}{2}\left(m_H+m_A\right){\left(\frac{m_H{v}_H}{m_H+m_A}\right)}^2\\ =\frac{1}{2}\frac{m_H^2{v}_H^2}{m_H+m_A}\\ =\frac{1}{2}\left(m_H{v}_H^2\right)\left(\frac{m_H}{m_H+m_A}\right)\end{array}$

Substitute $T_1$ for $\frac{1}{2}{m}_H{v}_H^2$.

$T_2=\left(\frac{m_H}{m_H+m_A}\right)T_1$ (3)

Calculate the mass of the anvil $\left(m_A\right)$ using the formula:

$m_A=\frac{W_A}{g}$

Here, $W_A$ is the weight of the anvil.

Substitute $\infty$ for $W_A$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{m}_A=\frac{\infty }{32.2\text{ft}/{\text{s}}^2}\\ =\infty \end{array}$

Consider the equation (1).

Substitute $0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_H$, $20\text{ft}/\text{s}$ for $v_H$, and $\infty$ for $m_A$.

$\begin{array}{c}{v}_2=\frac{m_H{v}_H}{m_H+m_A}\\ =\frac{\left(0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}\right)\left(20\text{ft}/\text{s}\right)}{0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}+\infty }\\ =0\end{array}$

Consider the equation (2).

Substitute $0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_H$, $20\text{ft}/\text{s}$ for $v_H$, and 0 for $v_2$.

$\begin{array}{c}F\Delta t=m_H\left(v_H-v_2\right)\\ =\left(0.04658\text{lb}\cdot {\text{s}}^2/\text{ft}\right)\left(20\text{ft}/\text{s}-0\right)\\ =0.932\text{lb}\cdot \text{s}\end{array}$

Calculate the energy absorbed by the rivet $\left(\Delta T\right)$ using the relation:

Consider the equation (3).

Substitute $0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_H$, $9.316\text{ft}\text{.lb}$ for $T_1$, and $\infty$ for $m_A$.

$\begin{array}{c}{T}_2=\left(\frac{0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}}{0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}+\infty }\right)9.316\text{ft}\text{.lb}\\ =0\end{array}$

Calculate the energy absorbed by the rivet under each blow $\left(\Delta T\right)$ when the anvil has an infinite mass support using the relation:

$\Delta T=T_1-T_2$

Substitute $9.316\text{ft}\text{.lb}$ for $T_1$ and 0 for $T_2$.

$\begin{array}{c}\Delta T=9.316\text{ft}\text{.lb}-0\\ =9.32\text{ft}\text{.lb}\end{array}$

Therefore, the impulsive force $F\Delta t$ and the energy absorbed $\left(\Delta T\right)$ by the rivet are $\underset{\_}{0.932\text{lb}\cdot \text{s}}$ and $\underset{\_}{9.32\text{ft}\cdot \text{lb}}$ respectively.

(b)

To determine

Find the impulsive force $\left(F\Delta t\right)$ and the energy absorbed $\left(\Delta T\right)$ by the rivet which has weight of $9\text{lb}$.

Answer to Problem 13.148P

The impulsive force $\left(F\Delta t\right)$ and the energy absorbed $\left(\Delta T\right)$ by the rivet has weight of $9\text{lb}$ are $\underset{\_}{0.799\text{lb}\cdot \text{s}}$ and $\underset{\_}{7.99\text{ft}\cdot \text{lb}}$ respectively.

Explanation of Solution

Given information:

The weight of the hammer (W) is $1.5\text{lb}$.

The velocity of the hammer $\left(v_H\right)$ is $20\text{ft/s}$.

The acceleration due to gravity (g) is $32.2\text{ft}/{\text{s}}^{\text{2}}$.

Calculation:

Calculate the impulse exerted by the rivet $\left(F\Delta t\right)$ and the energy absorbed by the rivet under each blow $\left(\Delta T\right)$ when the anvil has a support weight of $9\text{lb}$ using the relation:

Calculate the mass of the anvil $\left(m_A\right)$ using the relation:

$m_A=\frac{W_A}{g}$

Substitute $9\text{lb}$ for $W_A$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{m}_A=\frac{9\text{lb}}{32.2\text{ft}/{\text{s}}^2}\\ =0.2795\text{lb}{\text{.s}}^2/\text{ft}\end{array}$

Consider the equation (1).

Substitute $0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_H$, $20\text{ft}/\text{s}$ for $v_H$, and $0.2795\text{lb}{\text{.s}}^2/\text{ft}$ for $m_A$ in the Equation (1).

$\begin{array}{c}{v}_2=\frac{m_H{v}_H}{m_H+m_A}\\ =\frac{\left(0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}\right)\left(20\text{ft}/\text{s}\right)}{0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}+0.2795\text{lb}{\text{.s}}^2/\text{ft}}\\ =\frac{0.9316}{0.32608}\\ =2.857\text{ft}/\text{s}\end{array}$

Consider the equation (2).

Substitute $0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_H$, $20\text{ft}/\text{s}$ for $v_H$, and $2.857\text{ft}/\text{s}$ for $v_2$ in the Equation (2).

$\begin{array}{c}F\Delta t=m_H\left(v_H-v_2\right)\\ =\left(0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}\right)\left(20\text{ft}/\text{s}-2.857\text{ft}/\text{s}\right)\\ =\left(0.04658\right)\left(17.143\right)\\ =0.799\text{lb}\text{.s}\end{array}$

Consider the equation (3).

Substitute $0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_H$, $9.316\text{ft}\text{.lb}$ for $T_1$, and $0.2795\text{lb}{\text{.s}}^2/\text{ft}$ for $m_A$ in the Equation (3).

$\begin{array}{c}{T}_2=\left(\frac{0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}}{0.04658\text{lb}\text{.}{\text{s}}^2/\text{ft}+0.2795\text{lb}{\text{.s}}^2/\text{ft}}\right)9.316\text{ft}\text{.lb}\\ =\left(\frac{0.04658}{0.32608}\right)9.316\\ =1.331\text{ft}\text{.lb}\end{array}$

Calculate the energy absorbed by the rivet under each blow $\left(\Delta T\right)$ when the anvil has a support weight of $9\text{lb}$ using the relation:

$\Delta T=T_1-T_2$

Substitute $1.331\text{ft}\text{.lb}$ for $T_1$ and 0 for $T_2$.

$\begin{array}{c}\Delta T=9.316\text{ft}\cdot \text{lb}-1.331\text{ft}\cdot \text{lb}\\ =7.99\text{ft}\cdot \text{lb}\end{array}$