Q.A mass of 6kg is suspended by a rope of length 2m from the ceiling. A force of 50N in the horizontal direction is applied at the mid-point P of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium? (Take g=10m s−2.) Neglect the mass of the rope.
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.
Point P is in equilibrium under three forces: the upper-segment tension, the horizontal 50N, and the pull of the lower segment (which equals the weight mg=60N). Resolving forces gives tanθ=6050, so θ=tan−1(5/6)≈39.8∘.
Setting up the geometry
The mass hangs vertically below P, so the lower segment (P to the mass) is vertical and its tension simply supports the weight. The horizontal force is applied at P, tilting the upper segment (ceiling to P) away from the vertical by the angle θ we want.
Step 1 — The hanging mass
The lower rope segment carries only the weight:
T2=mg=6×10=60N
Step 2 — Equilibrium at point P
Three forces meet at P: the upper tension T1 (at angle θ to the vertical), the horizontal force 50N, and the downward pull T2=60N of the lower segment.
Concept: Equilibrium of Concurrent Forces at Point P
Step 1: Analyse the lower segment of the rope (P to the mass)
It only carries the weight of the hanging mass, so its tension is
T2=mg=6×10=60N
Step 2: Identify the three forces meeting at P
The upper-segment tension T1 (at the unknown angle θ to the vertical), the horizontal applied force F=50N, and the downward pull T2=60N of the lower segment.
