posted by Anonymous .
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 5.1 kN, and the circle’s radius is 10 m. At the top of the circle, what are the (a) magnitude FB and (b) direction (up or down) of the force on the car from the boom if the car’s speed is v= 4.0 m/s? What are the magnitude(c) FB and (d) the direction (up or down) of the force if the cars speed is v= 13m/s?
I have tried doing F=ma then A=v^2/R
but this answer is not correct. I know the direction just not the magnitude.
FB? is that the force on the boom?
FB=mg-mv^2/r If it is negative, the force is holding the car down. If it is positive, it is holding the car up.
A 1200 kg car is being driven up a 5:0 hill. The frictional force is directed opposite to
the motion of the car and has a magnitude of f = 524 N. A force ~F is applied to the
car by the road and propels the car forward. In addition to these two forces, two other
forces act on the car: its weight ~W and the normal force ~FN directed perpendicular
to the road surface. The length of the road up the hill is 290 m. What should be the
magnitude of ~F, so that the net work done by all the forces acting on the car is 150