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 μ_s = 0.9 and μ_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.

Fig. P13.137 and P13.138

To determine

Find the time (t) at which the compact car will start moving.

Answer to Problem 13.137P

The time (t) at which the compact car will start moving is $\underset{\_}{0.0075\sec }$.

Explanation of Solution

Given information:

The weight of the compact car (W) is $2500\text{lb}$.

The coefficient of static friction between tires and road $\left({\mu }_s\right)$ is 0.9.

The coefficient of friction between the tires and road $\left({\mu }_k\right)$ is 0.7.

The force P is $\text{30000lb}$.

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

Calculation:

Show the free body diagram of the compact car as in Figure (1).

Calculate the mass of the compact car (m) using the relation:

$\begin{array}{c}W=mg\\ m=\frac{W}{g}\end{array}$

Substitute $2500\text{lb}$ for W and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$\begin{array}{c}m=\frac{2500}{32.2}\\ =77.64\text{lb}\cdot \text{ft}/{\text{s}}^{\text{2}}\end{array}$

Calculate the normal force (N) acting on the compact car using the relation:

$N=mg$

Substitute $77.64\text{lb}\cdot \text{ft}/{\text{s}}^{\text{2}}$ for m and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$\begin{array}{c}N=77.64\text{lb}\cdot \text{ft}/{\text{s}}^{\text{2}}\left(32.2\right)\\ =2500.008\text{lb}\end{array}$

Calculate the force $\left(F_f\right)$ when the compact car starts moving using the relation:

$F_f={\mu }_{s}N$

Substitute 0.9 for ${\mu }_s$ and $2500.008\text{lb}$ for N.

$\begin{array}{c}{F}_f=0.9\times 2500.008\text{lb}\\ \text{=2250}\text{.0072lb}\end{array}$

The velocity at the point B $\left(v_{B,1}\right)$ is zero.

Calculate the initial momentum $\left(P_1\right)$ with which the compact car is moving using the relation:

$P_1=m{v}_{B,1}$

Substitute 0 for $v_{B,1}$.

$\begin{array}{c}{P}_1=m\left(0\right)\\ =0\end{array}$

Calculate the final momentum $\left(P_2\right)$ with which the compact car applied brake using the relation:

$P_2=m{v}_{B,2}$

After the brake applied the velocity $v_{B,2}$ will be zero.

Substitute 0 for $v_{B,2}$.

$\begin{array}{c}{P}_2=m\left(0\right)\\ =0\end{array}$

Calculate the force (P) by referring the graph using the relation:

For the interval of $0\le t\le 0.1$, the force (P) is $\frac{30000\text{lb}}{0.1}=300000\text{lb}$.

For the interval of $0.1\le t\le 0.2$, the force (P) is $\frac{30000\left(0.2-t\right)\text{lb}}{0.1}=300000\left(0.2-t\right)$.

The expression for the impulse force $\left(Im{p}_{1-2}\right)$ for the compact car as follows:

$Im{p}_{1-2}={\displaystyle \int \left(P-F_f\right)dt}$

Substitute $300000t$ for P and $\text{2250}\text{.0072lb}$ for $F_f$.

$Im{p}_{1-2}={\displaystyle \int \left(300000t-2250.0072\right)dt}$

Find the time (t) at which the compact car will start moving using the relation:

The expression for Impulse Momentum in the x direction as follows;

$m{v}_{B,1}+Im{p}_{1-2}=m{v}_{B,2}$

Substitute 0 for $m{v}_{B,1}$, ${\displaystyle \int \left(300000t-2250.0072\right)dt}$ for $Im{p}_{1-2}$ and 0 for $m{v}_{B,2}$.

$\begin{array}{l}0+{\displaystyle \int \left(300000t-2250.0072\right)dt}=0\\ 300000t=2250.0072\\ t=\frac{2250.0072}{300000}\\ t=0.0075\sec \end{array}$

Therefore, the time (t) at which the compact car will start moving is $\underset{\_}{0.0075\sec }$.

To determine

Find the maximum speed of the car $\left(v_{B,\max }\right)$.

Answer to Problem 13.137P

The maximum speed of the car $\left(v_{B,\max }\right)$ is $\underset{\_}{34.26\text{ft/s}}$.

Explanation of Solution

Given information:

The weight of the compact car (W) is $2500\text{lb}$.

The coefficient of static friction between tires and road $\left({\mu }_s\right)$ is 0.9.

The coefficient of friction between the tires and road $\left({\mu }_k\right)$ is 0.7.

The force P is $\text{30000lb}$.

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

Calculation:

Calculate the maximum time $\left(t_m\right)$ using the relation:

$P={\mu }_{k}N$

Substitute $300000\left(0.2-t\right)$ for P, $2500.008\text{lb}$ for N and 0.7 for ${\mu }_k$.

$\begin{array}{l}300000\left(0.2-t\right)=0.7\left(2500.008\right)\\ 0.2-t=\frac{1750.0056\text{lb}}{300000}\\ t=0.2-0.005833352\\ t=0.19417\sec \end{array}$

The expression for the final momentum $\left(P_2\right)$ with which the compact car applied brake as follows:

$P_2=m{v}_{B,\max }$

The expression for the impulse force $\left(Im{p}_{1-2}\right)$ for the compact car as follows:

$Im{p}_{1-2}={\displaystyle \underset{t_s}{\overset{0.1}{\int }}300000tdt}+{\displaystyle \underset{0.1}{\overset{t_m}{\int }}300000\left(0.2-t\right)dt}-{\displaystyle \underset{t_s}{\overset{t_m}{\int }}{\mu }_k{m}_{B}gdt}$

The expression for Impulse Momentum as follows:

$m{v}_{B,1}+Im{p}_{1-2}=m{v}_{B,\max }$

Substitute 0 for $m{v}_{B,1}$ and ${\displaystyle \underset{t_s}{\overset{0.1}{\int }}300000tdt}+{\displaystyle \underset{0.1}{\overset{t_m}{\int }}300000\left(0.2-t\right)dt}-{\displaystyle \underset{t_s}{\overset{t_m}{\int }}{\mu }_k{m}_{B}gdt}$ for $Im{p}_{1-2}$.

$\begin{array}{c}0+{\displaystyle \underset{t_s}{\overset{0.1}{\int }}300000tdt}+{\displaystyle \underset{0.1}{\overset{t_m}{\int }}300000\left(0.2-t\right)dt}-{\displaystyle \underset{t_s}{\overset{t_m}{\int }}{\mu }_{k}mgdt}=m{v}_{B,\max }\\ {\left[\frac{300000t^2}{2}\right]}_{t_s}^{0.1}+{\left[60000t-300000\frac{t^2}{2}\right]}_{0.1}^{t_m}-{\left[{\mu }_{k}g\right]}_{t_s}^{t_m}=m{v}_{B,\max }\end{array}$

Calculate the maximum velocity $\left(v_{B,\max }\right)$:

${\left[\frac{300000t^2}{2}\right]}_{t_s}^{0.1}+{\left[60000t-300000\frac{t^2}{2}\right]}_{0.1}^{t_m}-{\left[{\mu }_{k}g\right]}_{t_s}^{t_m}=m{v}_{B,\max }$

Substitute $0.0075\sec$ for $t_s$, $0.19417\sec$ for $t_m$, $77.64\text{lb}\cdot \text{ft}/{\text{s}}^{\text{2}}$ for m, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g, and 0.7 for ${\mu }_k$.

$\begin{array}{l}{\left[\frac{300000t^2}{2}\right]}_{0.0075}^{0.1}+{\left[60000t-300000\frac{t^2}{2}\right]}_{0.1}^{0.19417}-\left[0.7\left(9.81\right)\right]=77.64v_{B,\max }\\ \left\{\begin{array}{l}\left[\frac{300000}{2}\left({0.1}^2-{0.0075}^2\right)\right]\\ +\left[60000\left(0.19417-0.1\right)-300000\frac{\left({0.19417}^2-{0.1}^2\right)}{2}\right]\\ -\left(0.7\times 9.81\right)\end{array}\right\}=77.64v_{B,\max }\\ v_{B,\max }=34.26\text{ft/s}\end{array}$

Therefore, the maximum speed of the car $\left(v_{B,\max }\right)$ is $\underset{\_}{34.26\text{ft/s}}$.

To determine

Find the time at which the car $\left(t_{stop}\right)$ will come to a stop.

Answer to Problem 13.137P

The time at which the car $\left(t_{stop}\right)$ will come to a stop is $\underset{\_}{1.717\sec }$.

Explanation of Solution

Given information:

The weight of the compact car (W) is $2500\text{lb}$.

The coefficient of static friction between tires and road $\left({\mu }_s\right)$ is 0.9.

The coefficient of friction between the tires and road $\left({\mu }_k\right)$ is 0.7.

The force P is $\text{30000lb}$.

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

Calculation:

The expression for the initial momentum $\left(P_1\right)$ as follows:

$P_1=m{v}_{B,\max }$

The final momentum $\left(P_2\right)$ is zero.

The expression for the impulse force $\left(Im{p}_{1-2}\right)$ for the compact car as follows:

$Im{p}_{1-2}={\displaystyle \underset{t_m}{\overset{0.2}{\int }}300000\left(0.2-t\right)dt}-{\displaystyle \underset{t_m}{\overset{t_{stop}}{\int }}{\mu }_k{m}_{B}gdt}$

The expression for Impulse Momentum as follows:

$m{v}_{B,\max }+Im{p}_{1-2}=m{v}_{B,2}$

Substitute ${\displaystyle \underset{t_m}{\overset{0.2}{\int }}300000\left(0.2-t\right)dt}-{\displaystyle \underset{t_m}{\overset{t_{stop}}{\int }}{\mu }_k{m}_{B}gdt}$ for $Im{p}_{1-2}$.

$m{v}_{B,\max }+{\displaystyle \underset{t_m}{\overset{0.2}{\int }}300000\left(0.2-t\right)dt}-{\displaystyle \underset{t_m}{\overset{t_{stop}}{\int }}{\mu }_k{m}_{B}gdt}=0$

Substitute $34.26\text{ft/s}$ for $v_{B,\max }$, $0.19417\sec$ for $t_m$, $77.64\text{lb}\cdot \text{ft}/{\text{s}}^{\text{2}}$ for m, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g and 0.7 for ${\mu }_k$.

$\begin{array}{l}\left\{\begin{array}{l}\left(77.64\times 34.26\right)+{\left[60000t-300000\frac{t^2}{2}\right]}_{0.19417}^{0.2}\\ -{\left[0.7\times 9.81\times 77.64\times t\right]}_{0.19417}^{t_m}\end{array}\right\}=0\\ \left\{\begin{array}{l}\left(77.64\times 34.26\right)+\left[60000t-300000\frac{\left({0.2}^2-{0.19417}^2\right)}{2}\right]\\ -\left[0.7\times 9.81\times 77.64\times \left(t_m-0.19417\right)\right]\end{array}\right\}=0\\ t_m=1.717\sec \end{array}$

Therefore, the time at which the car $\left(t_{stop}\right)$ will come to a stop is $\underset{\_}{1.717\sec }$.