Chapter 5, Problem 11Q

To determine

1. The point at which the normal force acting on the car is the largest.
2. The point at which the normal force acting on the car is thesmallest.
3. The point at which the driver feels heaviest.
4. The point at which the driver feels lightest.
5. The fastest speed at which the car can go without losing the contact with the road at point A.

Answer to Problem 11Q

Solution:

1. Normal force action on a car is the largest at point C.
2. Normal force action on a car is the smallest at point A.

Weight felt by the driver is same as the normal force, so,

3. Driver feels heaviest at point C
4. Driver feelslightest at point A
5. The car can go with the road at point A without losing the contact if its speed must be less than or equal to $\sqrt{\text{gr}}$ .

Explanation of Solution

Given:

Both valley and hill have a radius of curvature R.

Formula Used:

Newton’s second law

  ${\displaystyle \sum \text{F=ma}}$

  ${\displaystyle \sum \text{F}}$ : Summation of all the forces acting

m : a mass of the body

a : acceleration

Centripetal acceleration

  $a_n=\frac{v^2}{r},$

  $\begin{array}{l}v\text{:}\text{speed,}\\ r\text{: radius}\end{array}$

  $a_n$ : Centripetal acceleration

Calculation:

The circular path toward the center of the curve.

Newton’s law at point A

  $\begin{array}{l}mg-N=m{a}_n\\ mg-N=m\left(\frac{v^2}{r}\right)\\ N=m\left(g-\frac{v^2}{r}\right)\cdots \cdots \cdots (1)\end{array}$

Newton’s law at point B

  $\begin{array}{l}mg-N=0\\ N=mg\cdots \cdots \cdots \cdots (2)\end{array}$

Newton’s at point C

  $\begin{array}{l}N-mg=m{a}_n\\ N-mg=m\frac{v^2}{r}\\ N=m\left(\frac{v^2}{r}+g\right)\cdots \cdots \cdots \cdots (3)\end{array}$

From equation (1), (2) and (3), it becomes clear that

1. Normal force has the largest value at point C.
2. Normal force has the smallest value at point A

Weight felt by driver is apparent weight and apparent weight is normal reaction exerted by the road.

3. Driver feels heaviest at C because normal reaction is the greatest value at point C
4. Driver feels lightest at point A because normal reaction is the least value at point A
5. Putting the value $\text{N=0}$, in reaction (1)

  $\begin{array}{l}N=m\left(g-\frac{v^2}{r}\right)\\ v=\sqrt{gr}\end{array}$

The maximum speed of car so that it goes without losing contact at point A is $\sqrt{rg.}$

Conclusion:

Normal force and weight felt by the driver is maximum value at point C (lowest point) ofthe valley.

Normal force and weight felt by the driver is minimum value at point A (top point) of the hill

A car goes without losing contact with the road at A should have a maximum speed $\sqrt{rg.}$