Answer to Question #151665 in Classical Mechanics for Cassie Curtis

Question

The object of curling is for a player to slide a stone on a sheet of ice toward the center of concentric circles. The coefficient of friction between the stone and the ice can be as little as .01. During the delivery you are only allowed to touch the 40 lb stone for 1216mm until you reach the "hog line". If the center of the circle is 6400mm from the hogline how much force needs to be applied to the curling stone?

Answer 1

Answer

Using friction force $F=\mu mg$

This implies

For two different cases

$\frac{F_2}{F_1}=\frac{\mu_2 m_2g}{\mu_1m_1g}$

Therefore using data as above question

$F_2=\frac{F_1 \mu_2 m_2g}{\mu_1m_1g}$

Because

$m\propto \frac{1}{\lambda}$

So $\frac{m_2}{m_1}=\frac{\lambda_1}{ \lambda_2 }=\frac{1216}{6400}=0.2$

All other are same as previous so

$F_2=40\times0.2=8lb$