Chapter 8, Problem 110A

To determine

Friction force exerted between skates and ice.

Answer to Problem 110A

  $F_{friction}=972\text{ N}$

Explanation of Solution

Given:

Mass of skater, $m=60\text{ kg}$

Velocity of skater, $v=18\text{ m/s}$

Radius of the curve, $r=20\text{ m}$

Formula used:

  $F_c=\frac{m{v}^2}{r}$

Here, $F_c$ is centripetal force, m is mass, v is velocity and risthe radius.

Calculation:

To negotiate the curve, frictional force must be equal to centripetal force:

Therefore, $F_c=F_{friction}$

Now, substitute the value of m, v and r and solve.

  $\begin{array}{c}{F}_{friction}=\frac{60\times {18}^2}{20}\\ =972\text{ N}\end{array}$

Conclusion:

Therefore, friction force required to negotiate on the curve is 972 N.