Question

2. A car travels on a rough banked curved road of radius 200 m with velocity v where the road is inclined at an angle of 15° to the horizontal. If the coefficient of friction betweern the road and the car is 1/4, find the value of v if (a) the car is just about to slip up the road; (b) the car is just about to slip down the road.

Answer 1

When the car just about to slip up the road. R = 200 m mu = 1/4 f = mu N because N cos 15 degree = mg + F sin 15 degree N cos 15 degree - mu N sin 15 degree = mg - (1) Vertical-bar because F = mu N Also, N sin 15 degree + f cos 15 degree = mv^2/R N sin 15 degree + mu N cos 15 degree = mv^2/R - (2) from eq^n (1) & (2): sin 15 degree + mu cos 15 degree/cos 15 degree - mu sin 15 degree = v^2/Rg v = squareroot (200 times 9.81) times sin(15 degree) + 1/4 cos(15 degree)/cos(15 degree) - 1/4 sin(15 degree) v = 33.00 m/sec \r\n when the car is just about to slip down the rod. R = 200 m mu = 1/4 theta = 15 degree fBD of car, N cos theta + f sin theta = mg N cos theta + mu N sin theta = mg - (1) Also, N sin theta = mv^2/R + f cos theta N sin theta