$\delta = \dfrac{PL}{AE}$

Where:

δ = π (1500.5 - 1500) = 0.5π mm

P = T

L = 1500π mm

A = 10(80) = 800 mm^{2}

E = 200 000 MPa

Thus,

$0.5\pi = \dfrac{T( 1500 \pi )}{800(200\,000)}$

$T = 53\,333.33 \, \text{N}$

$F = 2T$

$p(1500)(80) = 2(53\,333.33)$

$p = 0.8889 \, \text{MPa}$ → internal pressure

Total normal force, N:

N = p × contact area between tire and wheel

N = 0.8889 × π(1500.5)(80)

N = 335 214.92 N

Friction resistance, f:

f = μN = 0.30(335 214.92)

f = 100 564.48 N = 100.56 kN

Torque = f × ½(diameter of wheel)

Torque = 100.56 × 0.75025

Torque = **75.44 kN · m**

## Comments

## Mr. Verterra,

Mr. Verterra,

I think this problem needs more illustration, detailed illustration. I cannot seem to distinguish between a tire and a wheel. Also, why does the value of the length has a pie symbol next to it. Kindly explain. Thank you Sir.

Benjamin