What is the chance of a light car safely rounding an unbanked curve on an icy road than for a heavy car?

Question

is it worse, the same of better? assume that other factors such as speed and tires are the same

please explain, i'm totally lost!

Answer 1

Consider the forces on the car if it rounds a curve with Radius R, at a speed v, and the friction coefficient is u

First, look at the minimum speed without sliding down the banked curve.
m*g*sin(th)-(m*cos(th)*v^2)/R<
(m*g*cos(th)+(m*sin(th)*v^2)/R)*u

note that mass divides out

Now the max speed without slippping up the slope
(m*cos(th)*v^2)/R-m*g*sin(th)<
(m*g*cos(th)+(m*sin(th)*v^2)/R)*u

again, mass divides out

So, the chances are the same.

j

Answer 2

Car M > car m will have friction force F = K Mg cos(theta) acting against the centrifugal/centripetal force C = MV^2/R so the sum of forces along the radius of turn (R) is net force = C - F = Ma = 0; where a = 0 along the radius of turn R and car M stays on the track around the curve. Theta is the embankment angle of the curve.

Similarly, car m has f = k mg cos(theta) and c = mv^2/R; so that net force = 0 = (c - f) and it, too, stays on track during the turn. So the issue is whether the net force = 0 when both cars (M > m) are going v = V the same speed when k = K the same coefficients of friction.

Let's look at F - C = 0, the specs for the big car when it is a balanced turn. Clearly F = K Mg cos(theta) = M V^2/R = C; so that K g cos(theta) = V^2/R where car M is not skidding. And, lo, the car's mass M cancels out; it has no bearing on skidding or not.

Similarly for car m, we would have k g cos(theta) = K g cos(theta) = V^2/R = v^2/R because k = K and v = V were given. That is, as the early answer said, weight mg or Mg makes no difference on which car skids and which does not. Both will be in a non-skid balanced turn for the same road conditions (theta and k) and same speeds (v and V).