Q.Block A, of weight 100N, rests on a frictionless inclined plane whose slope makes an angle of 30∘ with the horizontal. A flexible cord tied to A runs up along the slope, passes over a frictionless pulley at the top of the incline, and then hangs vertically, supporting a block B of weight W. Find the weight W for which the system is in equilibrium.
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.
On a frictionless incline the string tension only has to balance A's weight component along the slope. The hanging block's weight equals that tension. …
Because the incline is frictionless, the cord tension must exactly balance the component of A's weight down the slope, 100sin30∘=50N. The hanging block's weight equals this tension, so W=50N.
Concept
In equilibrium the net force on each block is zero. The single light cord over a frictionless pulley has the same tension T throughout.
With no friction, the cord tension must exactly balance the component of A's weight along the slope; the same tension supports the hanging block B (single cord, frictionless pulley).
Step 1: Draw the forces on block A along the incline
Along the slope: tension T (up the slope) balances the component of A's weight down the slope, WAsinθ.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
KEAM 2026Set eng-2026-04204 marksMCQ
Q.An oil drop of mass m carrying a charge q is descending under its own weight. If it is made to remain stationary by applying an electric field of intensity E, then the value of q is (g = acceleration due to gravity)
(A) mgE
(B) Emg
(C) mEg
(D) Egm
(E) Eg2m
›Reveal solutionSolution
Equilibrium of the charged drop requires qE=mg, giving q=mg/E.
Balance of forces. For the drop to remain stationary the upward electric force must equal the downward weight: …
Q.A traffic light of mass 103 kg is suspended by two cables making 30∘ with the vertical. The tension in each cable is:
(A) 10 N
(B) 9.8 N
(C) 98 N
(D) 19.6 N
(E) 20 N
›Reveal solutionSolution
Each cable makes 30∘ with the vertical; the two vertical components support the weight: 2Tcos30∘=mg, giving T=98 N.
The traffic light of mass m=103 kg hangs in equilibrium from two symmetric cables, each at 30∘ to the vertical. The horizontal components cancel; the vertical components add to balance the weight:
Q.A garden roller of weight 100 kg is pulled with a force of 300 N acting at an angle of 30∘ with the ground. The effective pulling weight of the roller in (kg wt) is (g=10 ms−2)
(A) 850
(B) 725
(C) 800
(D) 820
(E) 700
›Reveal solutionSolution
The vertical component of the pull lifts part of the load: effective weight =W−Fsinθ=1000−150=850 N.
The true weight of the roller is
W=mg=100×10=1000N.
The applied force F=300 N acts at 30∘ above the ground, so its upward vertical component is