AP Physics 1 · Free-response skill

From a loaded plank to an equation and a graph

The move. Turn a described equilibrium setup into torque and force equations, write a support force as a function of a load’s position, and read that function as a graph.

How to do it

  1. List every force on the plank with where it acts. A uniform plank’s weight acts at its middle.
  2. Take torques about a point where an unknown force acts; that force drops out. Then use the sum of vertical forces for the other unknown.
  3. When a load can sit anywhere, keep its position as a letter. The torque equation then turns into a formula you can read like a line: what is fixed is the intercept, and what multiplies the letter is the slope.
  4. An object resting on supports is about to tip when one support’s force falls to zero. Set that force to zero and solve for the position.

Practice prompt Try it

A uniform plank 3.0 m long weighs 200 N and lies level on two supports. Support P is under the plank’s left end, and support Q is 2.0 m from the left end, so 1.0 m of the plank sticks out past Q. A 300 N toolbox sits on the plank. Measure every position x from the plank’s left end, in meters.

Answer each part. Show your reasoning.

  1. The toolbox is at x = 0.50 m. Find the upward force that support Q exerts on the plank.
  2. With the toolbox still at x = 0.50 m, find the upward force that support P exerts.
  3. Now let the toolbox sit at any position x on the plank, as long as the plank stays level. Write the force from support Q, in newtons, as a formula in x.
  4. Describe the shape of a graph of the force from support Q (vertical axis) against the toolbox’s position x (horizontal axis).
  5. The toolbox is slid toward the overhanging right end. At what position x is the plank just about to tip?

Write each part, then check yourself against the checklist. Nothing you type here is saved or graded.

Check yourself and see a model answer

Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 5 points

Model answer, annotated

(a) Take torques about P, so P’s force drops out: Q(2.0)=(200)(1.5)+(300)(0.50)=300+150=450Q(2.0) = (200)(1.5) + (300)(0.50) = 300 + 150 = 450 N·m, so Q=225Q = 225 N.

  • Point 1 Balances torques about P, with the plank’s weight at its middle, 1.5 m from P.

(b) The upward forces balance the 500 N of weight: P+225=200+300P + 225 = 200 + 300, so P=275P = 275 N.

  • Point 2 Uses the sum of vertical forces with the part a result.

(c) About P with the toolbox at x: Q(2.0)=(200)(1.5)+300xQ(2.0) = (200)(1.5) + 300x, so Q=150+150xQ = 150 + 150x newtons. At x = 0.50 m this gives 225 N, matching part a.

  • Point 3 Keeps the position as a letter in the torque equation.

(d) The formula is linear in x, so the graph is a straight line that rises as x grows, with slope 150 N per meter and a vertical intercept of 150 N.

  • Point 4 Reads the formula as a line with a positive slope.

(e) The plank is about to tip about Q when P’s force reaches zero, so Q carries all 500 N: 150+150x=500150 + 150x = 500, giving x=7/3≈2.33x = 7/3 \approx 2.33 m, one-third of a meter past Q. (Check about Q: 300(1/3)=200(0.50)=100300(1/3) = 200(0.50) = 100 N·m.)

  • Point 5 Sets P’s force to zero to find the tipping point.