Couple moments = free vectors?

Coupled moments are just two torques.

Suppose you have two masses, m, connected by a massless rod of length L. (I don't know about you engineers, but we physicists always keep a drawer full of frictionless pulleys, massless rods, and other apparatus of the sort.) Apply a force F at a distance r from mass 1, and the torque about mass 1 will be T=F*r*sin(angle). The torque about mass 2 will be T=-F*(L-r)*sin(angle). If r=L/2 the system will not rotate because there will be equal and opposite torques. It will simply translate.

Throw in two forces separated by a distance 2a with their center of separation at a distance r from mass 1. Then the torque about mass 1 would be

T = F*(r-a)*sin(angle) - F*(r+a)*sin(angle)

= -2 F a sin(angle)

r disappeared. You can maybe understand that better as a->0, because then you'll have two forces that exactly cancel each other out. Or you could think of it as one force acting as the fulcrum for another force.

The angular acceleration is not independent of r. But then the angular acceleration is proportional to the moment of inertia, which goes as r^2. If you change our dumbbell to a lollipop with a massless handle, the moment of inertia is simply I=mr^2. Angular acceleration goes as T=Ia, so we'd have

a = T/mr^2

The longer the handle is, the slower the acceleration, as you'd expect, while the torque is constant.

......... So how does a vehicular wheel work? You need to generate a net translational force at the axle. You have a frictional force,say f, at the road surface contact point. Assume that at a given instant, a single spoke is a massless rod, say of length r, connecting the axle and the frictional road force. There would be a torque moment, T, at the wheel axle, being supplied via the drive train.

Would the net translational force be (T-rf)/r? ... just summing moments on the spoke and dividing by r.

I'm not entirely sure what you're asking.

Torque is r cross f. Or, for a wheel on the road, we can dispense with the cross products and say T=rf. The drive train gives us some T, the wheel has an r, so the force on the road is f=T/r.

If you mean a case where the car (and the wheel) is accelerating, the moments of inertia of the wheels won't mean much compared with the mass of the car. But torques add (mind the signs).

T = I a

where T is the net torque, I the moment of inertia, and a the angular acceleration instead of using the customary Greek letter alpha. Or, for an accelerating wheel,

T_d + T_r = I a

where T_d is the torque from the drivetrain, T_r is the torque from the road, T_r=rf for a wheel radius r.

T_d - rf = I a

And solve for f or a or whatever you had in mind.

Nope. A free vector is a vector than has no origin. It only describes a magnitude and direction.

The force a car applies to the ground is a free vector. More specifically this example is a sliding vector.

Specifically, I had in mind the determination of the translational force on the chassis, required to accelerate the vehicle. An expression for the angular wheel acceleration is convertible to translational acceleration, and consequently, to the required translational force.

Initially, I had thought that it would be possible to sum moments about a 'spoke' or radius member connecting the axle and the road surface. However, I think that this is essentially the same thing as summing moments about the entire wheel.

Thanks for the insight. David Corliss