You now have every tool: free-body diagrams, net force, $F = ma$, the third law, connected bodies and friction. This lesson is about combining them. Whether a problem is "static" (nothing accelerates, forces balance) or "dynamic" (something accelerates), the same reliable recipe cracks it: draw the forces, choose a direction, sum them, and apply Newton's laws.
A lamp hangs motionless from a single vertical cable. A crane lifts a load that is speeding up as it rises.
Before reading: in which case does the cable tension equal the weight, and in which case is it greater? Write your prediction and one sentence of reasoning.
Draw, choose a direction, sum, then apply Newton's laws
Every force problem, a hanging sign or an accelerating lift, yields to the same four steps, so you never have to guess where to start.
Use this reliable recipe for any Newton's-law problem:
Deciding "static or dynamic" early tells you whether the forces should add to zero or to $ma$.
Solve any force problem with four steps: (1) free-body diagram, (2) choose a positive direction, (3) sum forces to find $F_{net}$, (4) apply $F_{net} = 0$ if static or $F_{net} = ma$ if accelerating. Decide static-or-dynamic first, it tells you whether the forces add to zero or to $ma$.
Pause, copy the highlighted recipe into your book before moving on.
Did you get this? True or false: an object moving at constant velocity has a net force of zero.
Putting weight, normal force, friction and applied force together
We just saw the four-step recipe. That raises a question: what does it look like on a full dynamic problem with several forces at once? This card answers it → work a box-with-friction problem end to end, combining $F = \mu N$ with $F_{net} = ma$.
A real dynamic problem just means several forces acting together, the recipe still handles it one direction at a time.
For a box on level ground being pushed, the vertical forces (weight and normal) balance, fixing $N = mg$. The horizontal direction is where the action is: the applied force drives forward, friction opposes, and the leftover is the net force. Then $F_{net} = ma$ gives the acceleration.
A $25\ \text{kg}$ crate is pulled across level ground by a horizontal rope with a tension of $90\ \text{N}$. The coefficient of kinetic friction is $\mu_k = 0.20$. Using $g = 9.8\ \text{m s}^{-2}$, find the crate's acceleration.
In a dynamic problem, balance the vertical forces to find $N = mg$, calculate friction $F_k = \mu_k N$, sum the horizontal forces to find $F_{net}$, then use $a = F_{net}/m$. Here $N = 245\ \text{N}$, $F_k = 49\ \text{N}$, $F_{net} = 41\ \text{N}$ and $a = 1.6\ \text{m s}^{-2}$.
Add the highlighted worked method to your notes before the check below.
Quick check: A $700\ \text{N}$ person stands in a lift accelerating downward. The scale reading (normal force) compared with $700\ \text{N}$ is…
Pattern A, Calculate then reason
A $60\ \text{kg}$ person stands on bathroom scales in a lift. Take $g = 9.8\ \text{m s}^{-2}$. In your book: (1) state the scale reading when the lift is at rest (static); (2) calculate the reading when the lift accelerates upward at $2.0\ \text{m s}^{-2}$ using $N = m(g + a)$; (3) explain in one sentence why the person feels heavier even though their mass is unchanged.
Two truths, one lie: which step in a force problem is the lie?
Dot point: Solve a variety of problems in static and dynamic real-world situations by applying Newton's laws of motion
Equilibrium means the forces add to zero in every direction
We just worked a dynamic problem with friction. That raises a question: what changes when nothing accelerates, a hanging sign or a parked car on a flat road? This card answers it → set $F_{net} = 0$ in each direction and solve.
When an object stays still or moves at steady speed, the forces in every direction must cancel exactly, that single condition solves the problem.
A static (equilibrium) object has zero acceleration, so the net force is zero in every direction. By Newton's first law, this is true whether the object is at rest or moving at constant velocity. For a hanging mass, the upward tension must equal the downward weight: $T = mg$. For an object held still on a level surface, the applied force must not exceed the maximum static friction. The recipe is identical to the dynamic case, you just set $F_{net} = 0$ instead of $ma$.
For example, a $5.0\ \text{kg}$ lamp hanging from one cable has $T = mg = 5.0 \times 9.8 = 49\ \text{N}$, because the cable tension exactly balances the weight.
In a static problem the object is in equilibrium: $F_{net} = 0$ in every direction, whether it is at rest or moving at constant velocity (Newton's first law). For a hanging mass, $T = mg$. Use the same diagram-first recipe, but set the force sum to zero instead of $ma$.
Pause, copy the highlighted reasoning into your book before the check below.
Last check: Decide if the statement is true or false.
For a mass hanging at rest from a single vertical cable, the cable tension equals the weight of the mass.
Fill the gaps. A static problem has an acceleration of , so the forces . A dynamic problem has a non-zero net force, which is found with $F_{net} = $ .
Dot point: Solve a variety of problems in static and dynamic real-world situations by applying Newton's laws of motion
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 6(5 marks) 1. A $70\ \text{kg}$ person stands on scales inside a lift. Using $g = 9.8\ \text{m s}^{-2}$: (a) state the scale reading when the lift is stationary; (b) calculate the reading when the lift accelerates upward at $1.5\ \text{m s}^{-2}$; (c) explain, with reference to the net force, why the reading differs from the person's true weight.
1 mark: stationary reading $= mg = 686\ \text{N}$ · 1 mark: states $N - mg = ma$ for upward acceleration · 1 mark: $N = m(g + a) = 70 \times 11.3$ · 1 mark: $N \approx 791\ \text{N}$ · 1 mark: explains the floor must supply extra force to produce the upward net force
ApplyBand 6(4 marks) 2. A $15\ \text{kg}$ suitcase is pulled along level ground by a handle at $25^\circ$ above the horizontal with a force of $60\ \text{N}$. The coefficient of kinetic friction is $0.20$ and $g = 9.8\ \text{m s}^{-2}$. Determine the acceleration.
1 mark: resolves the pull into components · 1 mark: normal force reduced by the vertical component, $N = 121.6\ \text{N}$ · 1 mark: friction $= 24.3\ \text{N}$ · 1 mark: $a = 2.0\ \text{m s}^{-2}$
Dot point: Solve a variety of problems in static and dynamic real-world situations by applying Newton's laws of motion
AnalyseBand 6(3 marks) 3. Outline a general procedure for solving any force problem, and explain how you decide whether it is a static or a dynamic problem.
1 mark: isolate the object and draw a free-body diagram · 1 mark: choose axes and resolve, then apply $F_{net} = ma$ per direction · 1 mark: static if the acceleration is zero (at rest or constant velocity), dynamic otherwise
Dot point: Solve a variety of problems in static and dynamic real-world situations by applying Newton's laws of motion
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (5 marks): (a) When the lift is stationary the person is in equilibrium, so the scale reads the true weight: $N = mg = 70 \times 9.8 = 686\ \text{N}$. (b) When the lift accelerates upward, the net upward force is $N - mg = ma$, so $N = m(g + a) = 70 \times (9.8 + 1.5) = 70 \times 11.3 \approx 791\ \text{N}$. (c) The reading is larger than the true weight because, to accelerate the person upward, the floor must push up with more force than gravity pulls down; this extra normal force is what produces the upward net force $ma$, and it is what the scales register.
Q2 (4 marks): Resolve the pull: horizontal $F_x = 60\cos 25^\circ = 54.4\ \text{N}$, vertical $F_y = 60\sin 25^\circ = 25.4\ \text{N}$ upward. The vertical component lifts part of the suitcase’s weight, so the normal force is reduced: $$N = mg - F_y = 15(9.8) - 25.4 = 147 - 25.4 = 121.6\ \text{N}$$ The friction is therefore $$f = \mu N = 0.20 \times 121.6 = 24.3\ \text{N}$$ Horizontally, $$F_{net} = F_x - f = 54.4 - 24.3 = 30.1\ \text{N}$$ so $$a = \frac{30.1}{15} = 2.0\ \text{m s}^{-2}$$ The key step is that the angled pull reduces $N$, so assuming $N = mg$ would overstate the friction and understate the acceleration.
Q3 (3 marks): First isolate the object of interest and draw a free-body diagram showing every force acting on it. Then choose a set of axes, usually one along the direction of motion, define a positive direction and resolve every force into components. Apply Newton’s second law separately in each direction. Finally solve the resulting equations for the unknown and check the sign and units of the answer. Whether the problem is static or dynamic is decided by the acceleration, not by whether the object is moving: if it is at rest OR moving at constant velocity the acceleration is zero, so $F_{net} = 0$ and the forces balance. If the speed or direction is changing, the acceleration is non-zero and $F_{net} = ma$ must be used.
You can now solve both static and dynamic problems with one reliable recipe. Next, you take this onto a slope, starting with the normal force and free-body diagram on an inclined plane.
↩ Back to focus-area hubAt the start you predicted when the cable tension equals the weight. The answer: the tension equals the weight only when the load is static (or at constant velocity), and is greater when the load accelerates upward, because $T - mg = ma$. If you thought tension always equals weight, revisit Cards 1 and 3.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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