Equilibrium Quiz: Friction, Ladders, Noncentral Loads

Concept: Static friction is self-adjusting. Static friction can take on any value needed to prevent slipping, up to a maximum μ_s n. It only reaches μ_s n at the threshold of motion.

Concept: Normal force on level ground. On a flat surface with no other vertical forces, the normal force balances weight. Weight is mg = 20×9.8 ≈ 196 n, so n ≈ 196 n.

Concept: Static friction limit. The maximum static friction is f_s,max = μ_s n. The coefficient μ_s sets how large friction can become relative to the normal force.

Concept: Impending motion. Right before the box starts moving, static friction matches the applied force in magnitude. At that moment it is at its maximum value.

Concept: Smooth (frictionless) contact. A smooth wall provides only a normal reaction force, perpendicular to the wall. With no friction at the wall, there is no vertical friction force at that contact.

Concept: Center of mass of a uniform ladder. A uniform ladder’s weight acts at its midpoint. Half of 5.0 m is 2.5 m along the ladder from the bottom.

Concept: Vertical force balance. In equilibrium, total upward reaction forces must equal total downward loads. Since the only downward force is 90 n, the reactions must sum to 90 n.

Concept: Moments about a support. Taking moments about the left end removes r_l from the torque equation. Solving r_r(3.0) = 90(1.0) gives r_r = 90/3 = 30 n.

Concept: Using 'f_y' after moments. Once r_r is known, apply vertical equilibrium: r_l + r_r = 90 n. So r_l = 90 - 30 = 60 n.

Concept: Torque balance on a beam. Taking moments about the left end: r_r(4.0) = 400(0.50) = 200 n·m. So r_r = 200/4.0 = 50 n; since two choices are identical (50 n), either 50 n option matches.

Concept: Preventing slip with static friction. Since f_s,max = μ_s n, increasing μ_s or increasing n (when possible) raises the maximum available static friction. Reducing the required horizontal pull also makes it easier for static friction to 'keep up.'

Concept: Resolving tension components. The vertical components of the cable tensions must add up to the weight. If weight increases and angles stay the same, each cable must provide a larger vertical component, so tension increases.

Concept: Rotational equilibrium. Net torque must be zero in equilibrium. That means the total clockwise torque must balance the total counterclockwise torque.

Concept: Matching torques. For equal torques (perpendicular), fr must match. Since 150(0.20) = 30 n·m, the 50 n force needs distance d where 50d = 30, so d = 0.60 m.

Concept: Torque sign convention. Torque sign is determined by whether the force tends to rotate the object clockwise or counterclockwise about the pivot. That depends on geometry (where the force is applied) and direction, plus the sign convention you choose.

Concept: Net force with opposing forces. Subtract opposing forces to find the net. 40 n right minus 25 n left gives 15 n right, meaning there will be acceleration to the right.

Concept: Torque equilibrium about any point. If an object is truly in equilibrium, 'Στ=0' is true about any pivot point. Choosing different pivots changes the algebra convenience, not the final physical reactions/forces.

Concept: Equivalent point force for weight. In a uniform gravitational field, a distributed weight can be replaced by a single force acting at the center of mass. This gives the same net force and the same torque effect.