A car of mass 1300 kg is moving at a speed of 28 m/s on a 100 m radius curve....?

Question

Find:
(a)
the centripetal force
(b)
the coefficient of friction if there is no banking to the curve
(c)
the banking angle if there is no friction

Answer 1

F = m(v^2/r)
F = 1300 * ((28^2)/100 )
F = 1 300 * ((28^2) / 100) = 10,192N

Answer 2

to make confident that the vehicle to holiday in uniform around action, it could have a centripetal acceleration, an acceleration in path of the centre element the vehicle is traveling approximately. This acceleration is given through v^2/r, the region v is the fee of the vehicle and r is radius of the circle. you may artwork this out to be 4.32 m/s^2. the vehicle is moved through friction unfavorable to the floor. even because of the fact the wheel applies a tension on the floor, through Newton's 0.33 regulation, that perfect comparable tension is utilized from the floor decrease back on the wheel. If friction is extreme sufficient, then the wheel is caught on the floor, and the stress is spent shifting the vehicle forwards. whether, if friction isn't extreme sufficient, the wheel does no longer stick unfavorable to the floor, starts off spinning, and the vehicle gained't pass contained interior the path you p.c.. it to. an comparable applies to turning the vehicle. If the frictional tension isn't great sufficient, the vehicle will skid out. to make confident that the vehicle to be waiting to preserve the turn, the frictional tension must be as great because of the fact the centripetal tension. The centripetal tension is given through centripetal acceleration extra beneficial through mass. 51.3 kg*4.32 m/s^2 = 221.6 N, and for this reason the fee of the frictional tension must be 221.6 N to save the vehicle from skidding out of the curve.

Answer 3

http://en.wikipedia.org/wiki/Coefficient_of_friction