Chapter 13.3, Problem 13.137P

A crash test is performed between an SUV A and a 2500-lb compact car B. The compact car is stationary before the impact and has its brakes applied. A transducer measures the force during the impact, and the force P varies as shown. Knowing that the coefficients of friction between the tires and road are ${\mu }_s=0.9$ and $\mu k=0.7$ , determine (a) the time at which the compact car will start moving. (b) the maximum speed of the car, (c) the time at which the car will come to a stop.

To determine

(a)

The time at which the compact car starts moving.

Answer to Problem 13.137P

Time at which the compact car starts moving $t=0.0075s$.

Explanation of Solution

Given information:

Weight of compact car B is equal to $2500lb$

$\begin{array}{l}{\mu }_s=0.9\\ {\mu }_k=0.7\end{array}$

“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:

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

Calculation:

Apply impulse momentum principle.

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

Substitute,

$m_B{v}_1+\int \left(P-F_f\right)dt=m_B{v}_2\to \left(1\right)$

Find Friction force,

$F_f={\mu }_{s}N=0.9\left(2500lb\right)=2250lb$

The force during impact P varies,

At $0\le t\le 0.1s$

$P=\frac{30000}{0.1}t=300000t$

Substitute in equation 1:

$\begin{array}{l}{m}_B{v}_1+\int \left(P-F_f\right)dt=m_B{v}_2\to \left(1\right)\\ 0+P-F_f=0\\ 300000t=2250lb\\ t=0.0075s\end{array}$

Conclusion:

The time when the compact car starts moving is calculated by putting the values of the weight of the car and friction force in the momentum principle equation.

To determine

(b)

The maximum speed of the car

Answer to Problem 13.137P

$v_{\max }=34.26ft/s$

Explanation of Solution

Given information:

Weight of compact car B is equal to $2500lb$

$\begin{array}{l}{\mu }_s=0.9\\ {\mu }_k=0.7\end{array}$

“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,

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

Calculation:

Find the friction factor when moving,

$F_f={\mu }_{s}N=0.7\left(2500lb\right)=1750lb$

At

$0.1\le t\le 0.2s$

$P=300000\left(0.2-t\right)$

But we know that,

$\begin{array}{l}P=300000\left(0.2-t\right)=1750lb\\ t=0.19417s\end{array}$

Apply impulse momentum principle,

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

Substitute,

$0+\underset{t_s}{\overset{0.1}{\int }}300000tdt+\underset{0.1}{\overset{t}{\int }}300000\left(0.2-t\right)dt-\underset{t_s}{\overset{t}{\int }}{\mu }_k{m}_{B}g=m_B{v}_2$

Therefore,

$v_2=34.26ft/s$

Conclusion:

The maximum speed is equal to $v_{\max }=34.26ft/s$

To determine

(c)

The time at which car comes to stop

Answer to Problem 13.137P

$t_{stop}=1.717s$

Explanation of Solution

Given information:

Weight of compact car B is equal to $2500lb$

$\begin{array}{l}{\mu }_s=0.9\\ {\mu }_k=0.7\end{array}$

“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,

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

Calculation:

To find the stopping time,

Apply impulse momentum principle,

$\begin{array}{l}{m}_B{v}_{\max }+\underset{t}{\overset{0.2}{\int }}300000\left(0.2-t\right)dt-\underset{t}{\overset{t_{stop}}{\int }}{\mu }_k{m}_{B}gdt=0\\ \left(2500lb\right)\left(34.26ft/s\right)+\underset{0.19417}{\overset{0.2}{\int }}300000\left(0.2-t\right)dt-\underset{0.19417}{\overset{t_{stop}}{\int }}\left(0.7\right)\left(2500lb\right)dt=0\end{array}$

Therefore,

$t_{stop}=1.717s$

Conclusion:

The car comes to stop at $t_{stop}=1.717s$