A uniform horizontal beam is pinned at its left end and rests against a smooth wall at its right end. A student places the torque pivot at the wall contact point. Which forces drop out of the torque equation?
AOnly gravity drops out because it acts at the center
BThe pin reaction at the left end drops out
CThe wall's normal force drops out because it acts right at the pivot and has zero moment arm
DAll forces drop out, making the torque equation useless at that pivot
Torque = force × perpendicular moment arm. When the pivot is placed at the point where the wall's normal force acts, that force has a zero moment arm and contributes zero torque — it drops out of the equation entirely. This leaves an equation involving only the pin reaction and gravity, making it straightforward to solve. Strategic pivot selection is the central technique in static equilibrium analysis.
Question 2 True / False
Two equal but opposite forces act at the two ends of a horizontal beam, one pushing up on the left and one pushing down on the right. The net force on the beam is zero, so the beam is in static equilibrium.
TTrue
FFalse
Answer: False
False. Static equilibrium requires BOTH conditions: ΣF = 0 AND Στ = 0. In this scenario, the net force is indeed zero, but the two forces form a couple that produces a net torque, causing the beam to rotate. Satisfying the force condition alone is insufficient for equilibrium of an extended object — the torque condition is a separate, independent requirement.
Question 3 True / False
The pivot point used in the torque equation for static equilibrium is expected to be located at an actual physical support or hinge.
TTrue
FFalse
Answer: False
False. The torque equilibrium condition Στ = 0 must hold about any point — this is a consequence of equilibrium itself. You are free to choose whichever pivot makes the algebra simplest. The standard strategy is to place the pivot at the location of an unknown force, which makes that force's torque contribution zero and removes it from the equation. There is no requirement that the pivot be a physical support.
Question 4 Multiple Choice
A ladder leans against a frictionless wall. To find the wall's normal force, a student writes Στ = 0 about the base of the ladder. Which forces appear in this torque equation?
AThe wall normal force and gravity only — base forces have zero moment arm at the base pivot
BAll four forces: gravity, wall normal, base normal, and base friction
COnly gravity, since it is the only downward force
DBase friction and gravity only — the wall force is too far away
With the pivot at the base, the two base forces (normal and friction) act right at the pivot — their moment arms are zero and they contribute zero torque. The torque equation then contains only the wall's normal force and gravity (each with nonzero moment arms from the base). This allows the wall force to be solved immediately, after which the force equations ΣFx = 0 and ΣFy = 0 yield the base forces.
Question 5 Short Answer
Explain why ΣF = 0 alone is not sufficient to guarantee static equilibrium of an extended object, and what additional condition is required.
Think about your answer, then reveal below.
Model answer: ΣF = 0 ensures the center of mass has no translational acceleration — the object won't slide or accelerate in any direction. But extended objects can also rotate. Forces can produce torques even when they balance: two equal and opposite forces applied at different points create zero net force but nonzero net torque, causing rotation. The additional condition is Στ = 0 about any point, ensuring no rotational acceleration. Both conditions together — zero net force and zero net torque — guarantee true static equilibrium.
A particle (point mass) only needs ΣF = 0. But rigid bodies have spatial extent, so forces applied at different points can cause rotation without causing translation. The torque equation is an independent constraint that the force equation cannot capture, which is why equilibrium problems with extended objects have three independent scalar equations (ΣFx, ΣFy, Στ) rather than two.