Chapter 10, Problem 56A

Harry Hotrod rounds a corner in his sports car at 50 km/h. Fortunately, a force of friction holds him on the road. If he rounds the corner at twice the speed, how much greater must the force of friction be to prevent him from skidding off the road? (You can solve this by using a simple proportion.)

To determine

The force of friction to prevent a person H from skidding off the road if the speed is changed.

Answer to Problem 56A

The force of friction to prevent a person H from skidding off the road if the speed is doubled is four times the initial frictional force.

Explanation of Solution

Given:

Rotational speed is $50\text{km}/\text{h}$

Speed is doubled.

Formula used:

The expression for the frictional force is,

  $f=\frac{m{v}^2}{r}$

Here, $m$ is the mass, $v$ is the linear speed, and $r$ is the radius of curvature.

Calculation:

Consider the mass and the radius of curvature is constant.The frictional force is directly proportional to the speed.

  $f\alpha v^2$

The force of friction to prevent person H from skidding off the road if the speed is doubled is,

  $\begin{array}{l}\frac{{\left(f_{\text{friction}}\right)}_{\text{i}}}{{\left(f_{\text{friction}}\right)}_{\text{f}}}=\frac{\frac{m{v}^2}{r}}{\frac{m{\left(2v\right)}^2}{r}}\\ \frac{{\left(f_{\text{friction}}\right)}_{\text{i}}}{{\left(f_{\text{friction}}\right)}_{\text{f}}}=\frac{1}{4}\\ {\left(f_{\text{friction}}\right)}_{\text{f}}=4{\left(f_{\text{friction}}\right)}_{\text{i}}\end{array}$

Conclusion:

Thus, the force of friction to prevent a person H from skidding off the road if the speed is doubled is four times the initial frictional force.