Gr 11 · Dynamics · Deep Study
Understanding Forces
Forces are the cause of all motion changes. This guide builds from Newton's laws through the most common force scenarios — tension, friction, inclines, and the pendulum — with worked FBDs at each step.
6 sections
Free-body diagram examples
Checkpoint questions
Step-by-step solutions
1
Newton's Laws
Why things move — or don't
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Why do we need Newton's laws?
Newton's laws explain why things move (or don't). Without a net force, velocity never changes — this is why seatbelts exist. Your body keeps moving forward when the car stops; only the seatbelt (an external force) changes your velocity.
Newton's laws explain why things move (or don't). Without a net force, velocity never changes — this is why seatbelts exist. Your body keeps moving forward when the car stops; only the seatbelt (an external force) changes your velocity.
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The three laws at a glance:
1st: ΣF = 0 → constant velocity (or at rest)
2nd: ΣF = ma (net force = mass × acceleration)
3rd: Action-reaction pairs act on different objects
1st: ΣF = 0 → constant velocity (or at rest)
2nd: ΣF = ma (net force = mass × acceleration)
3rd: Action-reaction pairs act on different objects
Four forces act on the box: weight (down), normal (up), applied (right), kinetic friction (left).
Newton's 2nd Law
ΣF = ma | ΣFₓ = maₓ | ΣFₔ = maₔ
Units: 1 N = 1 kg·m/s²
★ Easy
Show solution
Finding acceleration from net force
5 kg box, F = 30 N applied, no friction. Find a.
1
Identify forces and draw FBD
Forces: Fapp = 30 N right, Fg = 49 N down, N = 49 N up. No friction.
2
Apply Newton's 2nd Law (x-axis)
ΣFₓ = maₓ
30 = 5 × a
a = 6 m/s²
30 = 5 × a
a = 6 m/s²
Answer: a = 6 m/s² to the right
★★ Medium
Show solution
Acceleration with friction
Same 5 kg box, F = 30 N, kinetic friction fk = 10 N. Find a.
1
Net force (x-axis)
ΣFₓ = Fapp − fk = 30 − 10 = 20 N
2
Apply ΣF = ma
20 = 5 × a
a = 4 m/s²
a = 4 m/s²
Answer: a = 4 m/s² (friction reduces acceleration from 6 to 4 m/s²)
Checkpoint 1
a) 8 kg box, F = 40 N applied, no friction. Find a.
b) 3 kg box on frictionless surface, a = 2.5 m/s². Find F.
c) You push a wall with 50 N. What force does the wall exert on you?
b) 3 kg box on frictionless surface, a = 2.5 m/s². Find F.
c) You push a wall with 50 N. What force does the wall exert on you?
a) ΣF = ma → 40 = 8a → a = 5 m/s²
b) F = ma = 3 × 2.5 = 7.5 N
c) By Newton's 3rd Law, the wall pushes back on you with 50 N in the opposite direction. That is why you feel resistance when you push.
2
Weight & Normal Force
Gravity vs the surface pushing back
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Why is N not always equal to mg?
Weight is the gravitational pull of Earth (always downward). Normal force is the surface pushing back (always perpendicular to the surface). On a flat surface they balance, but on a ramp the normal force only has to support the component of gravity perpendicular to the ramp — which is less than mg.
Weight is the gravitational pull of Earth (always downward). Normal force is the surface pushing back (always perpendicular to the surface). On a flat surface they balance, but on a ramp the normal force only has to support the component of gravity perpendicular to the ramp — which is less than mg.
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Key formulas:
Fg = mg (g = 9.8 m/s²)
N = mg on a flat horizontal surface
N = mg cosθ on an incline at angle θ
Fg = mg (g = 9.8 m/s²)
N = mg on a flat horizontal surface
N = mg cosθ on an incline at angle θ
Left: at rest, N balances mg exactly. Right: accelerating upward, the floor must push harder — N = m(g + a) > mg.
Weight
Fg = mg
Normal (flat)
N = mg
Normal (incline)
N = mg cosθ
★ Easy
Show solution
Normal force on flat ground
70 kg person standing on a flat floor. Find Fg and N.
1
Calculate weight
Fg = mg = 70 × 9.8 = 686 N (downward)
2
Apply ΣFₔ = 0 (person not accelerating vertically)
N − Fg = 0
N = Fg = 686 N
N = Fg = 686 N
Answer: Fg = 686 N downward, N = 686 N upward
★★ Medium
Show solution
Apparent weight in elevator
Same 70 kg person in an elevator accelerating upward at 2 m/s². Find N.
1
Draw FBD — net force is upward (same direction as acceleration)
N acts upward, Fg = 686 N acts downward. Net force must be upward (a = 2 m/s² up).
2
Apply ΣFₔ = maₔ
N − mg = ma
N = m(g + a) = 70(9.8 + 2) = 70 × 11.8 = 826 N
N = m(g + a) = 70(9.8 + 2) = 70 × 11.8 = 826 N
Answer: N = 826 N (person feels heavier than usual)
Checkpoint 2
a) 12 kg box resting on a flat surface. Find N.
b) 60 kg person in elevator accelerating downward at 3 m/s². Find apparent weight.
c) Why does N decrease when an elevator accelerates downward?
b) 60 kg person in elevator accelerating downward at 3 m/s². Find apparent weight.
c) Why does N decrease when an elevator accelerates downward?
a) N = mg = 12 × 9.8 = 117.6 N
b) N = m(g − a) = 60(9.8 − 3) = 60 × 6.8 = 408 N
c) When accelerating downward, the net force is downward: mg − N = ma. So N = m(g − a) < mg. The floor needs to push less because gravity is already partially responsible for the downward acceleration.
3
Tension
The pull force in ropes and strings
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Why is tension the same throughout a massless rope?
Tension is the pull force in a rope or string. For a massless rope, tension is the same at every point — because if it weren't, there would be a net force on a section with zero mass, which would produce infinite acceleration. It always pulls — never pushes. A rope cannot push; it just goes slack.
Tension is the pull force in a rope or string. For a massless rope, tension is the same at every point — because if it weren't, there would be a net force on a section with zero mass, which would produce infinite acceleration. It always pulls — never pushes. A rope cannot push; it just goes slack.
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Key facts:
Massless rope → T constant throughout
Atwood machine: a = (m₂ − m₁)g / (m₁ + m₂)
Rising mass: T = m₁(g + a) Falling mass: T = m₂(g − a)
Massless rope → T constant throughout
Atwood machine: a = (m₂ − m₁)g / (m₁ + m₂)
Rising mass: T = m₁(g + a) Falling mass: T = m₂(g − a)
m₂ > m₁, so m₂ falls and m₁ rises. The rope transmits the same tension T to both masses throughout.
★ Easy
Show solution
Tension in a stationary hanging rope
4 kg box hanging from a rope, stationary. Find T.
1
FBD: T upward, Fg downward. Object is stationary → a = 0.
ΣFₔ = 0
T − mg = 0
T = mg = 4 × 9.8 = 39.2 N
T − mg = 0
T = mg = 4 × 9.8 = 39.2 N
Answer: T = 39.2 N
★★ Medium
Show solution
Atwood machine
Atwood machine: m₁ = 3 kg, m₂ = 5 kg. Find acceleration and tension.
1
Find acceleration
a = (m₂ − m₁)g / (m₁ + m₂)
a = (5 − 3)(9.8) / (3 + 5)
a = 19.6 / 8 = 2.45 m/s²
a = (5 − 3)(9.8) / (3 + 5)
a = 19.6 / 8 = 2.45 m/s²
2
Find tension (using rising mass m₁)
T = m₁(g + a) = 3(9.8 + 2.45) = 3 × 12.25 = 36.75 N
3
Verify with falling mass m₂
T = m₂(g − a) = 5(9.8 − 2.45) = 5 × 7.35 = 36.75 N ✓
Answer: a = 2.45 m/s², T = 36.75 N
Checkpoint 3
a) 6 kg box hanging in equilibrium. Find T.
b) Atwood machine: m₁ = 2 kg, m₂ = 6 kg. Find a.
c) If a rope has mass, is T the same throughout? Why?
b) Atwood machine: m₁ = 2 kg, m₂ = 6 kg. Find a.
c) If a rope has mass, is T the same throughout? Why?
a) T = mg = 6 × 9.8 = 58.8 N
b) a = (6 − 2)(9.8) / (2 + 6) = 39.2 / 8 = 4.9 m/s²
c) No. A rope with mass has weight, so the tension must be greater at the top (supporting more rope below it) than at the bottom. The assumption of a massless rope simplifies problems by making T uniform throughout.
4
Friction
Static vs kinetic — and why they're different
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Why is static friction larger than kinetic friction?
Friction is caused by microscopic surface irregularities that interlock. When an object is at rest, those irregularities have time to mesh together more deeply — so more force is needed to break them apart (static). Once sliding starts, there's less time for them to mesh, so kinetic friction is lower. This is why it's harder to start sliding a heavy box than to keep it sliding.
Friction is caused by microscopic surface irregularities that interlock. When an object is at rest, those irregularities have time to mesh together more deeply — so more force is needed to break them apart (static). Once sliding starts, there's less time for them to mesh, so kinetic friction is lower. This is why it's harder to start sliding a heavy box than to keep it sliding.
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Key formulas:
Static: fs ≤ μsN (object not moving)
Kinetic: fk = μkN (object sliding)
μs > μk always fk opposes direction of motion
Static: fs ≤ μsN (object not moving)
Kinetic: fk = μkN (object sliding)
μs > μk always fk opposes direction of motion
Left: static friction exactly matches the applied force (up to its maximum). Right: once sliding begins, kinetic friction is constant and lower than the static maximum. The graph shows the sharp drop at breakaway.
Kinetic friction (most used in problems)
fk = μk N
μk is dimensionless (no units) — just a ratio
★ Easy
Show solution
Calculating kinetic friction force
10 kg box sliding on floor, μk = 0.3. Find fk.
1
Find normal force (flat surface)
N = mg = 10 × 9.8 = 98 N
2
Apply kinetic friction formula
fk = μkN = 0.3 × 98 = 29.4 N
Answer: fk = 29.4 N (opposing the direction of motion)
★★ Medium
Show solution
Does the box slide? Find acceleration.
15 kg box, Fapp = 80 N, μs = 0.45, μk = 0.35. Does it slide? If yes, find a.
1
Find N and maximum static friction
N = mg = 15 × 9.8 = 147 N
fs,max = μsN = 0.45 × 147 = 66.15 N
fs,max = μsN = 0.45 × 147 = 66.15 N
2
Compare Fapp to fs,max
Fapp = 80 N > 66.15 N = fs,max → the box slides.
3
Switch to kinetic friction, find acceleration
fk = μkN = 0.35 × 147 = 51.45 N
ΣF = 80 − 51.45 = 28.55 N
a = 28.55 / 15 = 1.9 m/s²
ΣF = 80 − 51.45 = 28.55 N
a = 28.55 / 15 = 1.9 m/s²
Answer: Yes, it slides. a = 1.9 m/s²
Checkpoint 4
a) μk = 0.4, m = 8 kg, flat surface, pushed at constant velocity. Find applied force.
b) μs = 0.5, m = 5 kg on flat surface. Find minimum force to start sliding.
c) Why does increasing speed NOT increase kinetic friction?
b) μs = 0.5, m = 5 kg on flat surface. Find minimum force to start sliding.
c) Why does increasing speed NOT increase kinetic friction?
a) Constant velocity → a = 0 → ΣF = 0. So Fapp = fk = μkmg = 0.4 × 8 × 9.8 = 31.4 N
b) Fmin = fs,max = μsN = 0.5 × 5 × 9.8 = 24.5 N
c) In the standard model, fk = μkN. Since N doesn't change with speed (on a flat surface) and μk is a constant, fk stays constant regardless of speed. (At very high speeds, aerodynamic effects would matter — but not in this model.)
5
Inclined Planes
Resolving gravity into components along a ramp
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Why do we resolve gravity on an incline?
On a ramp, gravity pulls straight down but the surface is tilted. We can't use N = mg anymore. Instead, we split mg into two components: one parallel to the surface (which causes the object to slide) and one perpendicular (which N must balance). This is the key insight for all incline problems.
On a ramp, gravity pulls straight down but the surface is tilted. We can't use N = mg anymore. Instead, we split mg into two components: one parallel to the surface (which causes the object to slide) and one perpendicular (which N must balance). This is the key insight for all incline problems.
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The two components (incline angle θ):
N = mg cosθ (perpendicular — balanced by surface)
F‖ = mg sinθ (parallel — pulls object down the slope)
a = g(sinθ − μk cosθ) (if sliding with friction)
N = mg cosθ (perpendicular — balanced by surface)
F‖ = mg sinθ (parallel — pulls object down the slope)
a = g(sinθ − μk cosθ) (if sliding with friction)
Gravity (mg) is split into two components: mg sinθ pulls the block down the slope; mg cosθ is balanced by N. Kinetic friction fₖ opposes motion (acts up the slope when block slides down).
★ Easy
Show solution
Frictionless incline
10 kg box on a 30° frictionless ramp. Find N, F‖, and a.
1
Resolve gravity into components
N = mg cos 30° = 10 × 9.8 × 0.866 = 84.9 N
F‖ = mg sin 30° = 10 × 9.8 × 0.5 = 49 N
F‖ = mg sin 30° = 10 × 9.8 × 0.5 = 49 N
2
Find acceleration (no friction)
ΣF‖ = ma
49 = 10 × a
a = 4.9 m/s² (down the slope)
49 = 10 × a
a = 4.9 m/s² (down the slope)
Answer: N = 84.9 N, F‖ = 49 N, a = 4.9 m/s² down the slope
★★ Medium
Show solution
Incline with friction
Same 10 kg box on 30° ramp, but now μk = 0.2. Find a.
1
Calculate kinetic friction
N = 84.9 N (from above)
fk = μkN = 0.2 × 84.9 = 17 N (up the slope)
fk = μkN = 0.2 × 84.9 = 17 N (up the slope)
2
Net force along slope
ΣF‖ = mg sinθ − fk = 49 − 17 = 32 N (down)
3
Find acceleration
a = 32 / 10 = 3.2 m/s²
Answer: a = 3.2 m/s² down the slope (friction reduced it from 4.9 m/s²)
Checkpoint 5
a) 5 kg box on a 45° frictionless ramp. Find a.
b) 8 kg box on a 20° ramp, μk = 0.25. Find a.
c) At what angle does sinθ = μk cosθ? What does that mean physically?
b) 8 kg box on a 20° ramp, μk = 0.25. Find a.
c) At what angle does sinθ = μk cosθ? What does that mean physically?
a) a = g sin 45° = 9.8 × 0.707 = 6.93 m/s²
b) N = 8 × 9.8 × cos 20° = 73.7 N; fk = 0.25 × 73.7 = 18.4 N; F‖ = 8 × 9.8 × sin 20° = 26.8 N; Net = 26.8 − 18.4 = 8.4 N; a = 8.4/8 = 1.05 m/s²
c) sinθ = μk cosθ → tanθ = μk. Physically this is the critical angle where the gravitational pull down the slope exactly equals the kinetic friction. At this angle a = g(sinθ − μkcosθ) = 0 — the object slides at constant velocity.
6
Pendulum Forces
Circular motion, tension, and gravity
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Why is tension greatest at the bottom of the swing?
A pendulum bob moves in a circular arc. At every point, tension pulls toward the pivot and gravity pulls down. The net centripetal force must point toward the centre of the circle. At the bottom, both tension (up) and the centripetal requirement combine — T must support the weight AND provide centripetal force, so T > mg. At the side of the swing, only the component of T perpendicular to gravity contributes to centripetal force.
A pendulum bob moves in a circular arc. At every point, tension pulls toward the pivot and gravity pulls down. The net centripetal force must point toward the centre of the circle. At the bottom, both tension (up) and the centripetal requirement combine — T must support the weight AND provide centripetal force, so T > mg. At the side of the swing, only the component of T perpendicular to gravity contributes to centripetal force.
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Key formulas:
At angle θ (general): T = mg cosθ + mv²/r
At rest / slow swing: T ≈ mg cosθ
At bottom (θ = 0): T = mg + mv²/r
At angle θ (general): T = mg cosθ + mv²/r
At rest / slow swing: T ≈ mg cosθ
At bottom (θ = 0): T = mg + mv²/r
Main: pendulum at angle θ — T acts along the string toward the pivot, mg acts straight down, and the net centripetal force points toward the arc center. Inset: at the bottom, T must exceed mg to provide the upward centripetal force.
General pendulum tension
T − mg cosθ = mv²/r → T = mg cosθ + mv²/r
Centripetal direction is always toward the pivot (the centre of the circle)
★ Easy
Show solution
Tension at rest at an angle
0.5 kg pendulum bob at rest at 30° from vertical. Find T.
1
At rest, v = 0, so mv²/r = 0
T = mg cosθ
T = 0.5 × 9.8 × cos 30°
T = 4.9 × 0.866 = 4.24 N
T = 0.5 × 9.8 × cos 30°
T = 4.9 × 0.866 = 4.24 N
Answer: T = 4.24 N
★★ Medium
Show solution
Tension at the bottom of the swing
0.5 kg bob at bottom of swing, v = 2 m/s, string length r = 0.8 m. Find T.
1
At bottom, θ = 0, cos 0° = 1, net centripetal force = T − mg
T − mg = mv²/r
2
Solve for T
T = mg + mv²/r
T = (0.5)(9.8) + (0.5)(2)²/(0.8)
T = 4.9 + (0.5)(4)/0.8
T = 4.9 + 2.5 = 7.4 N
T = (0.5)(9.8) + (0.5)(2)²/(0.8)
T = 4.9 + (0.5)(4)/0.8
T = 4.9 + 2.5 = 7.4 N
Answer: T = 7.4 N (greater than mg = 4.9 N, as expected)
Checkpoint 6
a) 0.3 kg pendulum bob at rest at 45° from vertical. Find T.
b) 0.4 kg bob at the bottom of swing, v = 3 m/s, r = 1.2 m. Find T.
c) Why is tension always greater at the bottom than at the side of the swing?
b) 0.4 kg bob at the bottom of swing, v = 3 m/s, r = 1.2 m. Find T.
c) Why is tension always greater at the bottom than at the side of the swing?
a) T = mg cos 45° = 0.3 × 9.8 × 0.707 = 2.08 N
b) T = mg + mv²/r = (0.4)(9.8) + (0.4)(9)/(1.2) = 3.92 + 3.0 = 6.92 N
c) At the bottom, the centripetal acceleration points straight up (toward the pivot). Tension must support the full weight (mg) AND provide the centripetal force (mv²/r). At the side, the centripetal direction is horizontal, so gravity does not directly oppose tension — less tension is needed to supply the centripetal force while letting gravity act freely downward.