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Statics studies objects that remain at rest or move with constant velocity because all forces and torques balance. This cheat sheet helps students translate physical situations into free-body diagrams and equilibrium equations. It is especially useful for beams, ladders, cables, trusses, and objects on inclined surfaces.

Clear diagrams and sign conventions prevent most statics errors before algebra begins.

The core conditions for equilibrium are that the net force and net torque are zero. Forces are resolved into components using coordinate axes chosen for convenience, often along and perpendicular to a surface. Torques are calculated from a lever arm and force, with the choice of pivot used to simplify unknowns.

Friction, tension, normal forces, and support reactions are modeled as external forces acting on an isolated body.

Key Facts

  • Translational equilibrium requires Fx=0\sum F_x = 0, Fy=0\sum F_y = 0, and Fz=0\sum F_z = 0 when using three-dimensional coordinates.
  • Rotational equilibrium requires τ=0\sum \tau = 0 about any chosen pivot or axis for a rigid body in static equilibrium.
  • Torque magnitude is τ=rFsinθ\tau = rF\sin\theta, where rr is the distance from the pivot to the force application point and θ\theta is the angle between r\vec{r} and F\vec{F}.
  • A force produces no torque about a pivot if its line of action passes through the pivot, so τ=0\tau = 0 for that force.
  • The weight of an object is W=mgW = mg and acts vertically downward through the center of mass.
  • Static friction satisfies 0fsμsN0 \le f_s \le \mu_s N, while maximum static friction is fs,max=μsNf_{s,\max} = \mu_s N.
  • Kinetic friction has magnitude fk=μkNf_k = \mu_k N and acts opposite relative sliding motion.
  • For an incline at angle θ\theta, weight components are mgsinθmg\sin\theta parallel to the plane and mgcosθmg\cos\theta perpendicular to the plane.

Vocabulary

Static equilibrium
A condition in which an object has zero linear acceleration and zero angular acceleration, so F=0\sum \vec{F} = \vec{0} and τ=0\sum \vec{\tau} = \vec{0}.
Free-body diagram
A diagram that isolates one object and shows every external force acting on it with correct directions and points of application.
Torque
The rotational effect of a force about a point or axis, calculated by τ=r×F\vec{\tau} = \vec{r} \times \vec{F}.
Line of action
The straight line extending through a force vector, used to determine the perpendicular lever arm for torque.
Normal force
A contact force perpendicular to a surface that prevents objects from passing through one another.
Coefficient of static friction
The dimensionless constant μs\mu_s that sets the maximum static friction force by fs,max=μsNf_{s,\max} = \mu_s N.

Common Mistakes to Avoid

  • Including forces from the wrong object in the free-body diagram is incorrect because a free-body diagram must show only forces acting on the isolated body, not forces it exerts on others.
  • Assuming the normal force always equals mgmg is wrong because NN depends on acceleration, incline angle, and other vertical or perpendicular forces.
  • Using τ=rF\tau = rF for every force is wrong because the correct magnitude is τ=rFsinθ\tau = rF\sin\theta or equivalently τ=Fd\tau = Fd_\perp.
  • Choosing a pivot and then including torque from forces that pass through it is wrong because those forces have zero lever arm and produce τ=0\tau = 0.
  • Setting static friction equal to μsN\mu_s N automatically is wrong because static friction adjusts as needed up to the limit fs,max=μsNf_{s,\max} = \mu_s N.

Practice Questions

  1. 1 A 12kg12\,\text{kg} box rests on a horizontal floor. What are the magnitudes of its weight and normal force if no other vertical forces act on it?
  2. 2 A uniform 4.0m4.0\,\text{m} beam weighing 200N200\,\text{N} is supported at its left end and by a cable at the right end. If a 300N300\,\text{N} load hangs 3.0m3.0\,\text{m} from the left end, what upward force must the cable provide for equilibrium?
  3. 3 A 25kg25\,\text{kg} crate rests on a 3030^\circ incline with coefficient of static friction μs=0.60\mu_s = 0.60. Determine whether the crate can remain at rest without slipping.
  4. 4 Why can the torque equilibrium equation τ=0\sum \tau = 0 be written about any pivot point for a rigid body in static equilibrium?