1. Solving problems with Newton’s Laws
AP Physics
Section 4-7
Solving problems with Newton’s Laws

2. Free Body Diagrams
There are generally two types of force problems. One involves an object in equilibrium. This is when the net force equals zero. When all the forces are in equilibrium, the acceleration is zero. ∴
Fnet = ∑F = 0 N
a = 0
Keep in mind that this is not just for v = 0. It is also whenever v is constant! An object can be moving with constant velocity with no net force.
The second type of problem is when the object has a non-zero acceleration. In these problems:
Fnet = ∑F = ma
a ≠ 0

3. When solving force problems, you should always draw a diagram showing the forces on every object. This is a free-body diagram.
The easiest problems can be solved with only one free-body diagram, showing all the forces on one object. Never show the forces the object exerts on other objects unless the problem requires it.
There are several classic problems for which free-body diagrams are commonly drawn.
• An object being pulled in different directions.
• An object being pulled along a surface by a rope.
• Stacked blocks.
• An object sitting on, or sliding up or down, an incline.
• An Atwood machine (objects connected by a rope over a pulley).

4. 45° 26.1° 30° 100 N FN = 50 N pull FP = 100 N Ff = 34 N friction
An object being pulled along a surface by a rope.
150 N
Fg = 94 N
An object being pulled in different directions.
FN1
Stacked blocks.
Fg2
FN2
Ff21
friction m2 Ff FP
pull m1
Ff12
Fg2
Fg1
friction

5. a a See the text for example problems for some of these.
An object sitting on, or sliding down, an incline. FN a a Ff FT m1 FT
Fg1 m2 Fg
An Atwood machine.
See the text for example problems for some of these.
Fg2
HyperPhysics has excellent examples of 11 classic problems:

6. Using Newton’s third law in problem solving
Recall that Newton’s third law is about how two objects apply equal but opposite forces on each other.
Recall that Newton’s third law is about how two objects apply equal but opposite forces on each other.
Recall that Newton’s third law is about how two objects apply equal but opposite forces on each other.
FA on B = –FB on A
Examples:
Fbook on table = –Ftable on book
Fhammer on nail = –Fnail on hammer

7. Writing the force equations
FA on B = –FB on A
F1 on 2 = –F2 on 1
F1 = m1a1
F2 = m2a2
So,
m1a1 = –m2a2
The acceleration ratio will be the inverse of the mass ratio: m1 a2
a1 and a2 will point in opposite directions. = m2 a1

8. Steps for interaction pairs
1. Separate the system or systems from the external world.
2. Draw a free body diagram showing the forces on each object with a coordinate system for each.
3. Connect interaction pairs on the separate objects using dashed lines.
4. If there is acceleration, use Newton’s second law to relate net force to the acceleration of each object.
5. Use Newton’s third law to determine the magnitude of forces on other objects.
6. Solve problem and check units and signs.

9. Problems with a net force
A 50.0 kg bucket is being lifted by a rope. The rope will not break as long as the tension force in the rope is 525 N or less. The bucket starts from rest, and after being lifted 3.00 m, is moving 3.00 m/s. If acceleration is constant, will the rope break?

10. Problems with a net force
Free body diagrams
object 1: bucket
object 2: rope FP
force of rope on bucket FL
lifting force
a-r pair Fg FT
force of earth on bucket
force of bucket on rope

11. Net force equation, bucket:
Known values, bucket:
Unknown, rope:
m = 50.0 kg
FT = ? N
vi = 0.00 m/s
vf = 3.00 m/s
Δy = m
Net force equation, bucket:
Fnet = FP + Fg
Find magnitude of FP:
Solve for FP:
FP = ma – mg
FP = Fnet – Fg
FP = m(a – g)

12. ∴ Solve for acceleration using a kinematic equation: vf2 = vi2 + 2a∆y
vf2 – vi2 / 2∆y
a =
(3.00 m/s)2 – 02 / 2(+3.00 m)
a =
9.00 m2/s2 / m
a =
1.50 m/s2
g =
–9.80 m/s2
Solve for FP:
FP =
m(a – g)
FP =
50.0 kg[1.50 m/s2 – (–9.80 m/s2)]
FP =
50.0 kg[1.50 m/s m/s2]
Rope may break ∴
FP =
570 N
FT =
–570 N