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The following is a problem from the examples section of my school textbook, so it is accompanied by a full solution. However, I do not fully understand it and need your guidance in clearing my doubts.

http://img151.imageshack.us/img151/7024/picture30rr.jpg [Broken]

"A cyclist is moving at speed v around a bend of radius r. The forces acting on the cyclist are N, the normal reaction due to the ground on the bicycle, mg, the force due to gravity, and f, the frictional side-thrust on the tyres of the bicycle.

For vertical equilibrium, N = mg --(1) [I agree with this.]

For circular motion, f = m(v^2)/r -- (2) [Yes, I understand this.]

Equations (1) and (2) taken together give f/N = (v^2)/rg. "

I do not understand the next section of the solution:

"The clockwise torque about the centre of gravity, G = fh cos(theta). The anticlockwise torque about the centre of gravity, G = Nh sin (theta).

For stability fh cos (theta) = Nh sin (theta), or f/N = tan (theta).

Substituting this into Equation (3), we have tan (theta) = (v^2)/rg."

I don't understand why it is possible to equate the net torque about the centre of gravity to be zero. As the cyclist is not in equilibrium, shouldn't both the net torque and the net force acting on the cyclist be more than zero? Also if we took the net torque about the point of contact of the bicycle with the ground, then it clearly isn't zero because the torque produced by mg isn't balanced out by any other force. :grumpy:

http://img151.imageshack.us/img151/7024/picture30rr.jpg [Broken]

"A cyclist is moving at speed v around a bend of radius r. The forces acting on the cyclist are N, the normal reaction due to the ground on the bicycle, mg, the force due to gravity, and f, the frictional side-thrust on the tyres of the bicycle.

For vertical equilibrium, N = mg --(1) [I agree with this.]

For circular motion, f = m(v^2)/r -- (2) [Yes, I understand this.]

Equations (1) and (2) taken together give f/N = (v^2)/rg. "

I do not understand the next section of the solution:

"The clockwise torque about the centre of gravity, G = fh cos(theta). The anticlockwise torque about the centre of gravity, G = Nh sin (theta).

For stability fh cos (theta) = Nh sin (theta), or f/N = tan (theta).

Substituting this into Equation (3), we have tan (theta) = (v^2)/rg."

I don't understand why it is possible to equate the net torque about the centre of gravity to be zero. As the cyclist is not in equilibrium, shouldn't both the net torque and the net force acting on the cyclist be more than zero? Also if we took the net torque about the point of contact of the bicycle with the ground, then it clearly isn't zero because the torque produced by mg isn't balanced out by any other force. :grumpy:

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