Chapter 12.1, Problem 12.47P

The roller-coaster track shown is contained in a vertical plane. The portion of track between A and B is straight and horizontal, while the portions to the left of A and to the right of B have radii of curvature as indicated. A car is traveling at a speed of 72 when the brakes are suddenly applied; causing the wheels of the car to slide on the track $\left({\mu }_k=0.20\right)$ . Determine the initial deceleration of the car if the brakes are applied as the car (a) has almost reached A, (b) is traveling between A and B, (c) has just passed B.

  

To determine

(a)

The initial deceleration of the car if the brakes are applied as the car reaches at A at given condition.

Answer to Problem 12.47P

The required value of the deceleration is $4.63m/s^2.$

Explanation of Solution

Given information:

$\begin{array}{l}{\mu }_k=0.20\\ v=72km/h=20m/s\\ g=9.81m/s^2\\ \rho =30m\end{array}$

Formula used:

$F=ma$

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

Calculation:

To solve this problem, let us find the sum of normal forces.

$\Rightarrow \sum F_n=m{a}_n$.

$\Rightarrow N-mg=m\frac{v^2}{\rho }$.

$\Rightarrow N=m\left(g+\frac{v^2}{\rho }\right)$.

We know that, $F={\mu }_{k}N$.

Put the value of N in the above equation.

$\Rightarrow {\mu }_{k}m\left(g+\frac{v^2}{\rho }\right)$.

Again, $\sum F_t=m{a}_t$.

$\Rightarrow F=m{a}_t$.

$\Rightarrow a_t=\frac{F}{m}$.

$\Rightarrow a_t={\mu }_k\left(g+\frac{v^2}{\rho }\right)$.

Put the values in the above equation.

$\Rightarrow a_t=0.20\left[9.81+\frac{{\left(20\right)}^2}{30}\right]$.

$\Rightarrow a_t=4.63m/s^2$

To determine

(b)

The initial deceleration of the car if the brakes are applied as the car is travelling between A and B at given condition.

Answer to Problem 12.47P

The required value of the deceleration is $1.962m/s^2.$

Explanation of Solution

Given information:

$\begin{array}{l}{\mu }_k=0.20\\ v=72km/h=20m/s\\ g=9.81m/s^2\\ \rho =30m\end{array}$

Formula used:

$F=ma$

Calculation:

Now, $a_n=0$.

The sum of the normal forces would be.

$\Rightarrow \sum F_n=m{a}_n=0$.

$\Rightarrow N-mg=0$.

$\Rightarrow N=mg$.

$\Rightarrow F={\mu }_{k}N$.

$\Rightarrow {\mu }_{k}mg$.

Again, $\sum F_t=m{a}_t$.

$\Rightarrow F=m{a}_t$.

$\Rightarrow a_t=\frac{F}{m}$.

$\Rightarrow {\mu }_{k}g$.

$\Rightarrow 0.20\left(9.81\right)$.

$\Rightarrow a_t=1.962m/s^2$

To determine

(c)

The initial deceleration of the car if the brakes are applied as the car has just passed B at given condition.

Answer to Problem 12.47P

The required value of the deceleration is $0.1842m/s^2.$

Explanation of Solution

Given information:

$\begin{array}{l}{\mu }_k=0.20\\ v=72km/h=20m/s\\ g=9.81m/s^2\\ \rho =30m\end{array}$

Formula used:

$F=ma$

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

Calculation:

Now we will find the sum of downward normal forces.

$\Rightarrow \sum F_n=m{a}_n$.

$\Rightarrow mg-N=\frac{m{v}^2}{\rho }$.

$\Rightarrow N=m\left(g-\frac{v^2}{\rho }\right)$.

$\Rightarrow F={\mu }_{k}N$.

$\Rightarrow {\mu }_{k}m\left(g-\frac{v^2}{\rho }\right)$.

Again, $\sum F_t=m{a}_t$.

$\Rightarrow F=m{a}_t$.

$\Rightarrow a_t=\frac{F}{m}$.

$\Rightarrow {\mu }_k\left(g-\frac{v^2}{\rho }\right)$.

$\Rightarrow 0.20\left[9.81-\frac{{20}^2}{45}\right]$.

$\Rightarrow a_t=0.1842m/s^2$