A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
Force on each front wheel = 3675 N (or 367.5 kg-wt) Force on each back wheel = 5145 N (or 514.5 kg-wt)
Explanation
This problem involves applying conditions for equilibrium of a rigid body. The car is in static equilibrium on level ground, so we apply both translational equilibrium (sum of vertical forces = 0) and rotational equilibrium (sum of moments = 0). Taking moments about the front axle eliminates the unknown front wheel reaction, making calculations simpler. The weight acts at the centre of gravity, and the back wheel reaction acts at the back axle.
Solution Steps
Step 1: Given data - Mass of car m = 1800 kg, Distance between axles = 1.8 m, Centre of gravity is 1.05 m behind front axle (so 0.75 m from back axle).
Step 2: Weight of car W = mg = = 17640 N. Let Rf = total reaction on front wheels and Rb = total reaction on back wheels.
Step 3: For translational equilibrium: Rf + Rb = W = 17640 N ... (i)
Step 4: For rotational equilibrium, taking moments about front axle: Rb × 1.8 = W × 1.05. Therefore, Rb = ()/1.8 = 10290 N.
Step 5: From equation (i): Rf = 17640 - 10290 = 7350 N.
Step 6: Force on each front wheel = 7350/2 = 3675 N. Force on each back wheel = 10290/2 = 5145 N.