Question

The total mass of the car, fuel, and driver at the start of the race was 840 kg.

54% of the weight is distributed over the driving wheels of the car.

The coefficient of friction between the tyres used and the track on this day was 1.70.

(a) Calculate the maximum accelerating force possible at the tyres, at the start, before the wheels spin.

(b)Calculate the maximum possible acceleration off the line.

Answer 1

Explanations & Calculations

• What happens when a vehicle moves forward is that its wheels coupled to the engine produce some force on the ground backward & in return the same is produced on it in the forward direction driving it.
• Friction helps this phenomenon & thus there is a limit that the wheels can rotate to produce the force as too much of it could let the driving wheels just spin & slip on the ground.
• The maximum possible force to be generated is the maximum friction available at the given moment which may depend on the nature of the tires & the ground.
• This force causes the vehicle to accelerate.
• Further calculations are based on this concept.

a)

• The weight which is distributed on the driving wheels is

$\qquad\qquad \begin{aligned} \small w&=\small 840kg \times 9.8ms^{-2} \times\frac{54}{100}=4445.28 \,N \end{aligned}$

• Then the normal force acting on them is

$\qquad\qquad \begin{aligned} \small R_d&=\small 4445.28\,N \end{aligned}$

• Therefore, the maximum possible friction at the driving wheels is

$\qquad\qquad \begin{aligned} \small F_d&=\small \mu\times R_d\\ &=\small 1.70\times4445.28\\ &=\small \bold{7556.98\,N} \end{aligned}$

b)

• As the vehicle moves, the friction generated at the front/driven wheels & the air resistance on the vehicle act as opposing forces to what is generated at the driver wheels resulting in a "net force:$\small 7556.98-f(t)$ ".
• Therefore, the maximum possible acceleration is experienced at the very beginning of the motion where the opposing forces are zero.
• Then the maximum acceleration would be

$\qquad\qquad \begin{aligned} \small a_{max}&=\frac{F_{max}}{M}\\ &=\small \frac{7556.98\,N}{840kg}\\ &=\small \bold{8.996\,ms^{-2}} \end{aligned}$