Question

A car is driving on a road that is tited at a 15° angle so that the car can turn around a curve. a.) The car feels a force that is perpendicular to the road, what component of this force will provide a centripetal force? b.) If the car also speeds up, with the wheel turned 20° towards the center, what fraction of this acceleration will be a torque?

Answer 1

N sin theta act as centripetal for theta N sin theta = m v^2/r and N cos theta = m g implies N = m g/cos theta centripetal force implies N sin theta therefore m g/cos theta times sin theta implies m g tan theta implies m g tan 15 degree F_c p = m g (0.27) (b) Torque = F_c p times r sin theta_2 = N si theta times r sin (20 degree) = N sin 15 degree times r sin 20 degree = m g sin 15 degree r sin 20 torque = m g r (0.089)