Motor Torque required for 4 wheel drive electric car

[Sandesh_10]

Homework Statement

There is 14 kg of mass kept on a chassis (consider chassis weightless). Chassis is having 2 identical rear wheels and 2 Steel caster ball wheels at front. Consider the weight distribution to be equal on each of the wheel. All wheels are coupled to a DC motor-gearbox (1:1) each.The Rear Wheels are having 80mm diameter and made up of rubber and the chassis car is to be moved on a dry vinyl flooring. Adhesion coefficient,μ = 0.85.
Find the motor torque required if the car is to be moved at 1.2 m/s.

Homework Equations

Torque, τ = F * D/2
F : traction force required
D: diameter of wheel

Traction force, F= β * W
W : weight
β : adhesion factor

The Attempt at a Solution

F = 0.85 * 14*9.81 = 116.74 N
and thus
τ = 116.74 * 0.080/2 = 4.67 N.mm

Are these formulations and method correct?
Where and how does the velocity comes into picture?

 

[haruspex]

It seems to me that you have computed the maximum torque that can be applied without skidding. Since no rolling resistance or drag data have been provided, there is no lower limit to the torque required to keep the vehicle moving at a steady speed.

 

[barryj]

I think your calculations are correct but your units should be N-m not N.mm.
I (we) assume the velocity is constant and that the torque is merely to overcome the adhesion force.

 

[haruspex]

I think your calculations are correct but your units should be N-m not N.mm.
I (we) assume the velocity is constant and that the torque is merely to overcome the adhesion force.

That assumes 'adhesion coefficient' means that the wheels are tending to stick to the floor, resulting in a constant drag as the vehicle advances. The term was unfamiliar to me so I did a little research before posting (above). As far as I can make out, it is just another term for static friction. This led me to conclude, as I said, that what has been computed in the OP is the maximum torque that can be applied without skidding.