AP Physics 1 · Lesson 1

Systems, free-body diagrams and Newton’s laws

Know Know it

Start every force problem by choosing a system. Forces exerted by things outside it are external; forces between its parts are internal, come in equal and opposite pairs, and cancel for the system as a whole. The system’s center of mass, xcm=(m1x1+m2x2+… )/(m1+m2+… )x_{cm} = (m_1x_1 + m_2x_2 + \dots)/(m_1 + m_2 + \dots), accelerates as if the net external force acted on all the mass there.

A free-body diagram shows the object or system with an arrow for each external force, drawn from the object and labeled by what exerts it. Only real forces belong on it: each one needs a source you can name.

Newton’s laws. First: with zero net force, the velocity stays constant. Second: F⃗net=ma⃗\vec F_{\text{net}} = m\vec a, so the acceleration points along the net force. Third: forces come in pairs; if A pushes on B, then B pushes back on A just as hard, in the opposite direction.

Common forces
ForceExerted byDirection
Weight, Fg=mgF_g = mgEarth (or another planet or moon)Toward the planet’s center
Normal force, FNF_NA surface in contactPerpendicular to the surface, pushing out of it
Tension, FTF_TA rope, string or cableAlong the rope, pulling away from the object
Friction, ffA surface in contactAlong the surface, opposing sliding or attempted sliding
Spring force, Fs=kxF_s = kxA springBack toward the spring’s relaxed length

Calculator On the exam

On the exam you may use a four-function, scientific or graphing calculator on both sections, and the testing app also has built-in scientific and graphing calculators. You get an equation sheet with reference tables too, so spend your effort on choosing the right relationship, not memorizing it. In these lessons, use g = 9.8 m/s² unless a problem gives another value. Keep your calculator in degrees for force components.

Worked example Easy

A 1.5 kg weight plate is fixed at one end of a light 0.80 m bar, and a 4.5 kg weight plate is fixed at the other end. How far from the 1.5 kg plate is the center of mass of the bar and plates?

  1. 0.20 m
  2. 0.40 m
  3. 0.60 mAnswer
  4. 0.80 m

How to solve it

  1. Put the origin at the 1.5 kg plate, so it sits at x = 0 and the 4.5 kg plate at x = 0.80 m.
  2. xcm=(4.5)(0.80)/6.0=0.60x_{cm} = (4.5)(0.80)/6.0 = 0.60 m.
  3. Check: the center of mass is 0.20 m from the 4.5 kg plate and 0.60 m from the 1.5 kg plate, three times closer to the plate that has three times the mass.

Why each choice is right or wrong

  • A. Incorrect. 0.20 m is the distance from the heavier plate; the center of mass sits closer to the heavier plate, not the lighter one.
  • B. Incorrect. 0.40 m is the midpoint, which would be right only if the two plates had equal masses.
  • C. Correct. xcm=[(1.5)(0)+(4.5)(0.80)]/(1.5+4.5)=3.6/6.0=0.60x_{cm} = [(1.5)(0) + (4.5)(0.80)]/(1.5 + 4.5) = 3.6/6.0 = 0.60 m from the lighter plate.
  • D. Incorrect. 0.80 m is where the 4.5 kg plate sits; the 1.5 kg plate pulls the center of mass back toward the other end.

Trap Third-law partners never act on the same object

For a flowerpot resting on a level shelf, the normal force and the weight are equal in size, but they are not a third-law pair: both act on the flowerpot. The partner of Earth’s pull on the pot is the pot’s pull on Earth, and the partner of the shelf’s push on the pot is the pot’s push on the shelf.

Explain Explain it

Task verbs: calculate, justify

For connected objects that move together, first treat them as one system: only external forces appear, and a=Fnet/mtotala = F_{\text{net}}/m_{\text{total}}. Then choose a smaller system, one object, to find a force between the parts, such as the tension in a connecting rope.

To justify an answer with Newton’s laws, name the system, list the forces on it, and compare the net force with mama. If the velocity is constant, the forces must balance.

Worked example Medium

Two crates on a frictionless floor are joined by a light rope: a 12 kg crate in front and an 8.0 kg crate behind it. A worker pulls the front crate forward with a horizontal 50 N force. What is the tension in the rope between the crates?

  1. 20 NAnswer
  2. 25 N
  3. 30 N
  4. 50 N

How to solve it

  1. Whole system (both crates): the external horizontal force is 50 N on 20 kg, so a = 2.5 m/s².
  2. Rear crate alone: its only horizontal force is the rope tension.
  3. T=mreara=(8.0)(2.5)=20T = m_{\text{rear}}a = (8.0)(2.5) = 20 N.

Why each choice is right or wrong

  • A. Correct. Both crates together have a = 50 N / 20 kg = 2.5 m/s². The rope is the only horizontal force on the rear crate, so T=(8.0)(2.5)=20T = (8.0)(2.5) = 20 N.
  • B. Incorrect. 25 N assumes the 50 N splits evenly between the crates; the forces needed follow their masses.
  • C. Incorrect. 30 N is the net force on the front crate, (12)(2.5); the rope between the crates accelerates only the rear one.
  • D. Incorrect. 50 N is the worker’s pull; the rope needs to accelerate only the 8.0 kg rear crate at 2.5 m/s², which takes less force.

Calculate Your turn

A tugboat tows two barges in a line through calm water. The front barge has a mass of 3.0×1043.0 \times 10^4 kg and the rear barge 2.0×1042.0 \times 10^4 kg. The tug’s cable pulls the front barge, and a second cable joins the front barge to the rear one. The barges speed up at 0.050 m/s². Ignore water drag. Calculate the tension in each cable and explain why the two tensions differ.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Gives the tension in the tug’s cable as 2,500 N
  • Gives the tension in the cable between the barges as 1,000 N
  • Explains that the tug’s cable accelerates both barges while the other cable accelerates only the rear barge

Model answer

Treating both barges as one system, the tug’s cable is the only horizontal external force, so T1=(5.0×104)(0.050)=2,500T_1 = (5.0 \times 10^4)(0.050) = 2{,}500 N. The cable between the barges is the only horizontal force on the rear barge, so T2=(2.0×104)(0.050)=1,000T_2 = (2.0 \times 10^4)(0.050) = 1{,}000 N. The tug’s cable has to accelerate the whole 5.0×1045.0 \times 10^4 kg, while the cable between the barges accelerates only the rear barge’s 2.0×1042.0 \times 10^4 kg, so it needs less tension.

Apply Apply it

Practice these topics. AP Practice questions on exactly what this lesson covers, with an explanation for every answer.