Chapter 10, Problem 26A

A 70-kg person sits on the edge of a horizontal rotating platform 2 m from the center of the platform and has a tangential speed of 3 m/s. Calculate the force of friction that keeps the person in place.

To determine

Frictional force.

Answer to Problem 26A

  $f=315\text{N}$

Explanation of Solution

Given:

Mass of the person, $m=70\text{kg}$

Tangential velocity, $v=3\text{m/s}$

Distance of person from the center of the platform, $r=2\text{m}$

Formula Used:

The equation for centripetal force is,

  $F_c=\frac{m{v}^2}{r}$

Where, $m$ is mass, $v$ is tangential velocity and $r$ is radius of circular path.

Calculation:

When the person rotates with horizontal platform in the circular path, the friction force that keeps the person in place is equal to the centripetal force. So,

  $\begin{array}{c}f=F_c\\ =\frac{m{v}^2}{r}\\ =\frac{70\times 3^2}{2}\\ =315\text{N}\end{array}$

Conclusion:

Therefore, the frictional force is $f=315\text{N}$