What if I pick the wrong pivot point?

There is no wrong pivot point for a system in static equilibrium. The net torque is zero about any point. However, picking a point where unknown forces act simplifies the algebra by zeroing out their torques.

How do I handle a force applied at an angle?

Break the force into components parallel and perpendicular to the moment arm. Only the perpendicular component generates torque. Alternatively, use $ au = rF\sin heta$, where $ heta$ is the angle between the position vector and the force vector.

Does the weight of the object itself matter?

Yes, unless the problem specifies a 'massless' or 'light' object. For a uniform object, treat the entire weight as a single downward force acting exactly at its geometric center (center of mass).

How to solve a torque and equilibrium problem

A rigid body is in static equilibrium when it experiences zero net force and zero net torque. Apply this method when a non-rotating, non-accelerating object is subjected to multiple forces at different locations.

The setup

Define the rigid body of interest. Draw a detailed free-body diagram (FBD) showing all external forces applied to the body and their exact points of application. Establish a standard Cartesian coordinate system. Choose a rotation axis (pivot point) to calculate torques. Selecting a pivot point where an unknown force acts will eliminate that force from the torque equation.

The steps

Draw the FBD and place the pivot point.
Resolve all forces into $x$ and $y$ components.
Apply Newton's First Law for translation: $\sum F_x = 0$ and $\sum F_y = 0$ .
Apply Newton's First Law for rotation: $\sum au = 0$ . Use the convention that counterclockwise (CCW) torques are positive and clockwise (CW) torques are negative. Calculate each torque using $au = r F \sin heta$ or $au = F_{\perp} r$ .
Solve the resulting system of linear equations for the unknown variables.

Checking the result

Select a completely different pivot point on the rigid body. Recalculate $\sum au$ using the forces you just found. If the sum is not exactly zero, there is an algebraic or sign error in the initial calculation. Verify that all force answers are in Newtons (N) and torques in Newton-meters (N\cdot m).

Common errors

A frequent error is assigning incorrect signs to torques; strict adherence to the CCW positive/CW negative rule is required. Another standard mistake is omitting the weight of the beam itself, which must be modeled as a force vector acting downward at the object's center of mass.

Worked example

A uniform horizontal beam of length $L = 4.0 ext{ m}$ and mass $M = 20.0 ext{ kg}$ is supported by two pillars at its ends. A point mass $m = 50.0 ext{ kg}$ sits on the beam at a distance $x = 1.0 ext{ m}$ from the left pillar. Find the normal forces exerted by the left pillar ( $F_L$ ) and right pillar ( $F_R$ ). Use $g = 9.8 ext{ m/s}^2$ .

Let the left end of the beam be the origin ( $x=0$ ) and choose it as the pivot point.

Identify the forces and their positions:

$F_L$ acts upward at $x = 0$ .
Weight of mass $m$ , $W_m = mg = (50.0)(9.8) = 490 ext{ N}$ , acts downward at $x = 1.0 ext{ m}$ .
Weight of beam, $W_b = Mg = (20.0)(9.8) = 196 ext{ N}$ , acts downward at the center of mass $x = 2.0 ext{ m}$ .
$F_R$ acts upward at $x = 4.0 ext{ m}$ .

Apply $\sum au = 0$ about the left end: $au_{FL} + au_{Wm} + au_{Wb} + au_{FR} = 0$ $(0)(F_L) - (1.0)(490) - (2.0)(196) + (4.0)(F_R) = 0$ $0 - 490 - 392 + 4.0 F_R = 0$ $4.0 F_R = 882$ $F_R = 220.5 ext{ N}$

Apply $\sum F_y = 0$ to find $F_L$ : $F_L + F_R - W_m - W_b = 0$ $F_L + 220.5 - 490 - 196 = 0$ $F_L - 465.5 = 0$ $F_L = 465.5 ext{ N}$

The left pillar exerts $465.5 ext{ N}$ and the right pillar exerts $220.5 ext{ N}$ .

FAQ

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References: University Physics (Young and Freedman) Chapter 11 · OpenStax University Physics Volume 1, Chapter 12