Chapter 13.3, Problem 13.148P

To determine

(a)

Impulse exerted on the rivet and the energy absorbed by the rivet if anvil has infinite mass.

Answer to Problem 13.148P

When anvil has infinite mass, the energy absorbed by the rivet is $9.316ft.lb$.

The impulse exerted on the rivet is $0.9316lb.s$.

Explanation of Solution

Given information:

Weight of hammer is $1.5lb$.

The velocity of the rivet is $20ft/s$.

“A force acting on a particle during a very short time interval but large enough to produce a definite change in momentum is called an impulsive force.”

Impulse momentum principle for impulsive motion is defined as

$m{v}_1+\sum F_{avg}\Delta t=m{v}_2$

The total linear momentum of two particles is conserved. Therefore

$m_A{v}_A+m_B{v}_B=m_A{v}_A{}^1+m_B{v}_B{}^1$

Calculation:

Mass $m_H$ of hammer

$m_H=\frac{1.5lb}{32.2ft/s^2}=0.04658lb.s^2/ft$

Kinetic energy $T_1$ before the impact

Assume $v_H$ as the velocity of the hammer.

$T_1=\frac{1}{2}{m}_H{v}_H{}^2=\frac{1}{2}\left(0.04658lb.s^2/ft\right){\left(20ft/s\right)}^2=9.316ft.lb$

Assume $v_2$ as the common velocity of hammer and anvil just after the impact and $m_A$ as the mass of anvil.

Apply conservation of linear momentum.

$\downarrow m_H{v}_H=\left(m_H+m_A\right)v_2$

Rearrange

$v_2=\frac{m_H{v}_H}{m_H+m_A}\to \left(1\right)$

Kinetic energy $T_2$ after the impact

$\begin{array}{l}{T}_2=\frac{1}{2}\left(m_H+m_A\right)v_2{}^2\\ =\frac{1}{2}\left(m_H+m_A\right){\left(\frac{m_H{v}_H}{m_H+m_A}\right)}^2\\ =\frac{1}{2}\frac{m_H{}^2{v}_H{}^2}{m_H+m_A}\\ T_2=\frac{m_H}{m_H+m_A}{T}_1\to \left(2\right)\end{array}$

Find the impulse $F\Delta t$ exerted on the hammer.

$\begin{array}{l}m{v}_1+\sum F_{avg}\Delta t=m{v}_2\\ \downarrow m_H{v}_H-F\Delta t=m_H{v}_2\end{array}$

Rearrange:

$F\Delta t=m_H\left(v_H-v_2\right)\to \left(3\right)$

If anvil has infinite mass

According to equation 1:

$v_2=0$

According to equation 2:

$T_2=0$

Energy absorbed by the rivet.

$T_1-T_2=9.316ft.lb$

Impulse exerted on the rivet.

$F\Delta t=m_H{v}_H=\left(0.04658lb.s^2/ft\right)\left(20ft/s\right)=0.9316lb.s$

Conclusion:

When anvil has infinite mass, the energy absorbed by the rivet is $9.316ft.lb$.

The impulse exerted on the rivet is $0.9316lb.s$.

To determine

(b)

Impulse exerted on the rivet and the energy absorbed by the rivet if anvil has a weight of $9lb.$

Answer to Problem 13.148P

When anvil has a weight of $9lb$, the energy absorbed by the rivet is $7.9853ft.lb$.

The impulse exerted on the rivet is $0.7985lb.s$.

Explanation of Solution

Given information:

Weight of hammer is $1.5lb$.

The velocity of the rivet is $20ft/s$.

“A force acting on a particle during a very short time interval but large enough to produce a definite change in momentum is called an impulsive force.”

Impulse momentum principle for impulsive motion is defined as

$m{v}_1+\sum F_{avg}\Delta t=m{v}_2$

The total linear momentum of two particles is conserved. Therefore

$m_A{v}_A+m_B{v}_B=m_A{v}_A{}^1+m_B{v}_B{}^1$

Calculation:

According to sub part a

$v_2=\frac{m_H{v}_H}{m_H+m_A}\to \left(1\right)$

$T_2=\frac{m_H}{m_H+m_A}{T}_1\to \left(2\right)$

$F\Delta t=m_H\left(v_H-v_2\right)\to \left(3\right)$

Mass $m_A$ of anvil

$m_A=\frac{9lb}{32.2ft/s^2}=0.2795lb.s^2/ft$

The common velocity $v_2$ of anvil and hammer just after the impact.

$v_2=\frac{\left(0.04658lb.s^2/ft\right)\left(20ft/s\right)}{0.32608lb.s^2/ft}=2.857ft/s$

The kinetic energy $T_2$ just after the impact.

$T_2=\frac{\left(0.04658lb.s^2/ft\right)}{\left(0.32608lb.s^2/ft\right)}\left(9.316ft.lb\right)=1.3307ft.lb$

Energy absorbed by the rivet.

$T_1-T_2=9.316ft.lb-1.3307ft.lb=7.9853ft.lb$

The impulse $F\Delta t$ exerted on the rivet.

$F\Delta t=\left(0.04658lb.s^2/ft\right)\left(20ft/s-2.857ft/s\right)=0.7985lb.s$

Conclusion:

When anvil has a weight of $9lb$, the energy absorbed by the rivet is $7.9853ft.lb$.

The impulse exerted on the rivet is $0.7985lb.s$.