Q.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.
Imagine a book lying flat on a table. It doesn't move. A lamp hanging from the ceiling — still. A bridge holding up cars — motionless. What do all these have in common? They are in static equilibrium.
The word "static" means unchanging or stationary. "Equilibrium" comes from Latin aequilibrium — "equal balance." Put them together: a state where an object is completely at rest, with no tendency to start moving or rotating.
But here's the key insight: being at rest doesn't mean nothing is happening. Forces are still acting on that book — gravity pulls it down, the table pushes it up. The lamp feels tension from the ceiling pulling up and gravity pulling down. These forces are cancelling each other out perfectly.
Note
Static equilibrium is dynamic balance — forces are present, but their net effect is zero. The object "chooses" to stay still because all pushes and pulls are perfectly matched.
The Two Conditions for Static Equilibrium
For an object to be truly static (not moving or rotating), two separate things must be true simultaneously.
Condition 1: No Net Force (Translational Equilibrium)
The sum of all forces acting on the object must be zero. In vector form:
∑F=0
This means:
All upward forces equal all downward forces
All leftward forces equal all rightward forces
All forward forces equal all backward forces
If you break it into components (the standard exam approach):
∑Fx=0,∑Fy=0,∑Fz=0
Why this alone isn't enough: Imagine pushing a door at its handle — it rotates open. The forces might balance (you push, the hinges push back), but the door still moves. That's why we need the second condition.
Condition 2: No Net Torque (Rotational Equilibrium)
The sum of all torques (twisting effects) about any point must be zero:
∑τ=0
Torque depends on three things: the force applied, the distance from the pivot point, and the angle at which you push. For a force F applied at distance r from the pivot, at angle θ:
τ=rFsinθ
Watch out
A common mistake: thinking torque only matters if the object is actually rotating. Torque can be present even when nothing moves — it's just balanced by other torques. A seesaw with two kids of equal weight at equal distances is a perfect example.
Putting It All Together
For an object to be in static equilibrium:
∑F=0and∑τ=0
Both conditions must hold simultaneously. If either fails, the object will either accelerate (move in a straight line) or start rotating (or both).
A Simple Example: The Book on the Table
Consider a 2 kg book on a horizontal table. Gravity pulls down with force Fg=mg=2×9.8=19.6 N.
Torque balance about the front axle gives the back-axle load; force balance then gives the front-axle load; dividing by two wheels per axle gives the per-wheel forces.
W=mg=1800×9.8=17640 N. Torque about the front axle: Fb×1.8=W×1.05⇒Fb=10290 N. Force balance: Ff=W−Fb=7350 N. Per wheel: front =Ff/2=3675 N, back …
Taking torques about the front axle, the back axle carries 10290 N and the front axle carries 7350 N in total; dividing by two wheels per axle gives 3675 N on each front wheel and 5145 N on each back wheel.
Setting up
The car (mass 1800 kg) is in static equilibrium, so both the net force and net torque on it are zero. Its weight is
W=mg=1800×9.8=17640 N
acting at the centre of gravity, which is 1.05 m behind the front axle (and so 0.75 m in front of the back axle, since the wheelbase is 1.8 m). Let Ff be the total upward force from the ground on both front wheels, and Fb the total force on both back wheels.
Torque balance about the front axle
Choosing the front axle as the pivot eliminates Ff from the equation (its moment arm is zero). The weight, 1.05 m behind the front axle, creates a torque balanced by Fb acting at the full wheelbase distance:
Concept: Torque Balance for a Rigid Body on Two Supports
Treat the car as a rigid body in equilibrium; the ground pushes up on the front and back wheels, and the weight acts at the centre of gravity.
Step 1: Set up
W=mg=1800×9.8=17640 N. Wheelbase =1.8 m; centre of gravity is 1.05 m behind the front axle (so 0.75 m ahead of the back axle). Let Ff, Fb be the total forces at the front and back axles.