Difficulty: Intermediate | Prerequisites: 1D and 2D kinematics, vector components
Newton's laws connect forces to motion. Where kinematics describes how things move, dynamics explains why. The free-body diagram is the single most important tool in this unit: if you can draw one correctly, identifying every force and its direction, the rest is algebra. Most exam errors in this section trace back to a missing or misidentified force on the diagram, not to a calculation mistake.
Newton's second law (F_net = ma) is the central equation of mechanics. To use it, draw a free-body diagram showing every force on the object, resolve forces into components, then apply F_net = ma in each direction. For objects in equilibrium (constant velocity or at rest), F_net = 0 in every direction, which lets you solve for unknown forces.
Newton's first law (law of inertia)
An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted on by a net external force. In simple terms, things keep doing what they are doing unless something pushes or pulls them.
Newton's second law
F_net = ma. The net force on an object equals its mass times its acceleration. This is the workhorse equation of the course.
Newton's third law
For every force that object A exerts on object B, object B exerts an equal and opposite force on object A. Think of it as: forces always come in pairs, and the two forces in a pair act on different objects.
Free-body diagram (FBD)
A sketch showing a single object isolated from its surroundings, with arrows representing every force acting on that object. Each arrow should be labelled with the type of force, the object it acts on, and the object exerting it.
Normal force (N or F_N)
The contact force exerted by a surface, perpendicular to the surface. It adjusts to whatever value is needed to prevent the object from passing through the surface. It is not always equal to mg.
Weight (W or F_g)
The gravitational force on an object: W = mg, directed downward.
Friction (f)
A contact force parallel to the surface that opposes relative motion or the tendency of relative motion. Static friction prevents sliding; kinetic friction acts during sliding.
Equilibrium
A state where the net force on an object is zero. The object may be at rest or moving at constant velocity. In either case, a = 0.
A sled (mass 7.4 kg) carries a child (mass m). A horizontal force of 120 N produces acceleration 2.5 m/s^2 on a frictionless surface.
F_net = (m_sled + m_child) x a. So 120 = (7.4 + m_child)(2.5).
7.4 + m_child = 120 / 2.5 = 48. m_child = 48 - 7.4 = 40.6 kg, approximately 41 kg.
Answer: E) 41 kg.
A 23 kg suitcase is pulled at constant speed by a handle at 25 degrees above horizontal. Normal force = 180 N. Find the applied force F.
Constant speed means a = 0, so F_net = 0 in both x and y.
Vertical equilibrium: N + F sin(25) - mg = 0. So F sin(25) = mg - N = 23(9.80) - 180 = 225.4 - 180 = 45.4 N.
F = 45.4 / sin(25) = 45.4 / 0.4226 = 107.4 N, closest to 110 N.
Answer: C) 110 N.
Note: you could also check horizontal equilibrium (F cos(25) = friction force), but you do not need it here since the question only asks for F and the vertical equation alone gives it.
Two identical blocks (mass m each) are stacked: A on top of B. A hand pushes B to the right at constant speed. Block A does not move relative to B.
Constant speed means a = 0 for the system and for each block individually.
Block A (top block):
Weight: mg downward (gravitational force on A by Earth).
Normal force from B: N_AB upward (contact force on A by B).
Static friction from B: f_AB to the right (friction on A by B). This is what accelerates A to keep it moving with B. Since constant speed, this friction is actually zero. Both blocks move together at constant velocity, so A needs no net horizontal force.
Block B (bottom block):
Weight: mg downward (gravitational force on B by Earth).
Normal force from ground: N_ground upward (contact force on B by ground).
Normal force from A: N_BA downward (contact force on B by A, Newton's third law pair of N_AB).
Applied force: F_hand to the right (contact force on B by hand).
Kinetic friction from ground: f_ground to the left (friction on B by ground).
Static friction from A: f_BA to the left (friction on B by A, Newton's third law pair of f_AB). At constant speed this is zero.
For block A: N_AB = mg (vertical equilibrium of A).
For block B vertical equilibrium: N_ground = mg + N_BA = mg + mg = 2mg.
Ranking from largest to smallest: N_ground (= 2mg) > N_AB = N_BA = W_A = W_B (all equal to mg).
N_AB and N_BA are a Newton's third law pair, so they have the same magnitude.
F_{\text{net}} = m aW = m gVertical equilibrium with an angled pull (angle theta above horizontal):
N + F \sin\theta = m g \quad \Rightarrow \quad F = \frac{m g - N}{\sin\theta}Horizontal equilibrium with friction:
F \cos\theta = fFree-body diagrams are used by every engineer who analyses structures, machines, or vehicles. Structural engineers use them to determine the forces in bridge trusses. Automotive engineers use them to calculate braking forces and tyre loads. Even physiotherapists use force diagrams to understand the loads on joints during exercise.
Students often assume the normal force equals mg. It does not when there is a vertical component from another force (like an angled pull or push), or when the object is on an incline, or when it is accelerating vertically.
Students forget Newton's third law pairs act on different objects. The normal force on A from B and the normal force on B from A are equal in magnitude, but they appear on different free-body diagrams.
"Constant speed" means a = 0, which means F_net = 0. Students sometimes apply F = ma with a nonzero acceleration for constant-speed problems.
When drawing FBDs, students sometimes include forces that act on other objects. Each diagram should show only the forces acting on that one object.
Free-body diagram questions appear on every exam. Practise labelling each force with its type (gravity, normal, friction, applied, tension) and the two objects involved.
Ranking force magnitudes is a common exam format. Use equilibrium conditions (F_net = 0 in each direction) to deduce relationships between forces.
Problems with angled applied forces (like the suitcase) are very common. Remember that the applied force has both a horizontal and a vertical component, and the vertical component changes the normal force.
True or false: The normal force on an object always equals its weight. (False. It depends on other vertical forces and whether the object is accelerating vertically.)
A 50 kg box is pushed across a frictionless floor with 200 N. What is its acceleration? (a = F/m = 200/50 = 4.0 m/s^2)
Fill in the blank: Newton's third law says forces come in ______, acting on ______ objects. (pairs, different)
True or false: If an object moves at constant velocity, the net force on it is zero. (True.)
You pull a suitcase with force F at angle theta above horizontal. The vertical component of F is ______. (F sin(theta))
Q: You pull a 7.4 kg sled carrying your sister across a frictionless surface with 120 N horizontally. The acceleration is 2.5 m/s^2. What is your sister's mass?
A: Total mass = F/a = 120/2.5 = 48 kg. Sister's mass = 48 - 7.4 = 40.6 kg, approximately 41 kg.
Q: A 23 kg suitcase is pulled at constant speed by a force F at 25 degrees above horizontal. The normal force is 180 N. Find F.
A: Vertical equilibrium: N + F sin(25) = mg. So F = (mg - N) / sin(25) = (225.4 - 180) / 0.4226 = 45.4 / 0.4226 = 107 N, approximately 110 N.
Q: Two identical blocks of mass m are stacked (A on B). A hand pushes B horizontally at constant speed on a surface with friction. Draw the FBD for block B and list all forces.
A: Block B has: weight mg downward, normal force from ground N_ground upward, normal force from A downward (N_BA = mg), applied force F_hand to the right, kinetic friction from ground f_ground to the left, and static friction from A (f_BA) to the left (which equals zero at constant speed since A has no net horizontal force).
Q: For the stacked blocks above, rank all vertical forces by magnitude.
A: N_ground = 2mg (largest), then W_A = W_B = N_AB = N_BA = mg (all equal). N_ground must support both blocks.
Newton's laws are the foundation for everything that follows in mechanics: work and energy, momentum, circular motion, and rotation. The free-body diagram technique transfers directly to all of these. In particular, the normal force you calculate here will reappear in friction problems (f = mu N), and the force-component approach is identical to what you used in projectile motion for velocity components.
Newton's laws, Newton's first law, Newton's second law, Newton's third law, F = ma, free-body diagram, FBD, normal force, weight, friction, static friction, kinetic friction, equilibrium, constant speed, net force, angled force, stacked blocks, contact force, action-reaction pair, PHYSICS 114, UW general physics