Forces at angles, static equilibrium and friction AP Physics 1 · Lesson 3 of 7
Most real forces are not horizontal. This lesson shows you how to resolve a force into components, how an angled pull changes the normal force and friction, why pulling beats pushing, and how to solve the cable-and-knot equilibrium problems that appear on every AP Physics 1 test.
On this page
In this lesson
This lesson covers forces at angles without friction, static equilibrium, and forces at angles with friction:
- Resolving forces into components
- Forces at angles (no friction)
- Static equilibrium
- Forces at angles with friction
All numbers use g = 9.8 m/s², as on the AP Physics 1 equation sheet. Every formula used here is on that sheet: AP Physics 1 equation sheet.
Resolving forces into components
When a force acts at an angle, split it into two perpendicular components and apply Newton's second law separately along each axis. Pick axes so that as many forces as possible lie along them: on level ground, horizontal and vertical; on a slope (Lesson 4), along and perpendicular to the slope.
| Force at angle θ measured from… | Horizontal component | Vertical component |
|---|---|---|
| the horizontal | F cosθ | F sinθ |
| the vertical | F sinθ | F cosθ |
Key idea
Always check the angle is measured from the axis you think it is. “Cos goes with the side next to the angle” works whatever the orientation. Quick check: as θ → 0, the component along the reference line should become the whole force.
Worked example: components
A rope pulls on a crate with 50 N at 37° above the horizontal. Find the components.
- Horizontal: 50 cos 37° = 50 × 0.80 = 40 N.
- Vertical: 50 sin 37° = 50 × 0.60 = 30 N upward.
- Check: √(40² + 30²) = 50 N.
Forces at angles, without friction
Pull a sled with a rope at angle θ above the horizontal on frictionless ice. Only the horizontal component F cosθ accelerates the sled. The vertical component F sinθ does not make the sled move up (unless it is bigger than the weight); it reduces how hard the sled presses on the ice.
x: F cosθ = ma y: FN + F sinθ − mg = 0 ⇒ FN = mg − F sinθ
Worked example: sled on ice
A child pulls a 15 kg sled across frictionless ice with a rope at 30° above the horizontal. The tension is 60 N. Find the acceleration and the normal force.
- x: a = F cosθ/m = 60 cos 30°/15 = 51.96/15 = 3.46 m/s².
- y: FN = mg − F sinθ = 15(9.8) − 60(0.5) = 147 − 30 = 117 N, less than the sled's weight.
Common trap
FN = mg is a special case, not a rule. Any force with a vertical component changes the normal force. Always find FN from the vertical forces.
Static equilibrium: cables and knots
An object is in translational equilibrium when its acceleration is zero: it is at rest, or moving at constant velocity. Then the net force is zero in every direction:
ΣFx = 0 ΣFy = 0
The classic problem is a weight hung from a knot held by two or three cables. Draw the free-body diagram of the knot, where all the forces meet, then write the two equations.
Method for a knot in equilibrium
- Free-body diagram of the knot: each cable pulls along its own direction, away from the knot.
- Resolve each tension into horizontal and vertical components (watch where each angle is measured from).
- ΣFx = 0: the horizontal parts of the two tensions are equal and opposite.
- ΣFy = 0: the vertical parts add up to the weight.
- Solve the two equations for the two unknown tensions.
Worked example 1: symmetric cables
A 20 kg sign hangs from two cables, each at 25° above the horizontal. Find the tension in each cable.
- By symmetry, T1 = T2 = T, so the horizontal parts cancel.
- Vertical: 2T sin 25° = mg = 196 N.
- T = 196 / (2 × 0.423) = 232 N, more than the sign's weight.
Worked example 2: cables at 30° and 60°
A 10 kg sign hangs as in the diagram: cable 1 at 30° and cable 2 at 60° to the ceiling. Find both tensions.
- x: T1 cos 30° = T2 cos 60°, so T2 = T1(0.866/0.5) = 1.732 T1.
- y: T1 sin 30° + T2 sin 60° = 98, so T1(0.5 + 1.732 × 0.866) = 2.0 T1 = 98.
- T1 = 49 N and T2 = 85 N. The steeper cable carries more of the load.
Key idea
The flatter the cables, the bigger the tension. Each cable's vertical part must still hold up the weight, and a nearly horizontal cable has only a tiny vertical part, so the tension becomes huge. That is why a washing line can never be pulled perfectly straight with clothes on it.
In AP Physics 1, equilibrium also needs zero net torque so that nothing starts to rotate. That half of the story, with beams and ladders, comes in Unit 5 (torque guide). Here every force acts at one point, so only the force balance matters.
Forces at angles, with friction
Add friction and the angle matters twice: the horizontal component drives the motion, and the vertical component changes FN, which changes the friction.
| Pulling at θ above horizontal | Pushing at θ below horizontal | |
|---|---|---|
| Normal force | FN = mg − F sinθ | FN = mg + F sinθ |
| Kinetic friction | μk(mg − F sinθ) | μk(mg + F sinθ) |
| Acceleration | a = [F cosθ − μk(mg − F sinθ)]/m | a = [F cosθ − μk(mg + F sinθ)]/m |
Worked example 1: sled with friction
The 15 kg sled from before is now on snow with μk = 0.15. The same 60 N pull acts at 30° above the horizontal. Find the acceleration.
- FN = 147 − 30 = 117 N, so friction = 0.15 × 117 = 17.55 N.
- a = (51.96 − 17.55)/15 = 2.29 m/s².
Worked example 2: pushing vs pulling
A 15 kg box on a floor with μk = 0.15 is pushed with 60 N directed 30° below the horizontal. Find its acceleration and compare with pulling at 30° above.
- FN = 147 + 30 = 177 N; friction = 0.15 × 177 = 26.55 N.
- a = (51.96 − 26.55)/15 = 1.69 m/s², less than the 2.29 m/s² for pulling.
- Pushing down presses the box into the floor and increases friction. Pulling up does the opposite. This is why it is easier to pull a lawnmower or sled than push it.
Worked example 3: constant velocity at an angle
A 40 kg crate is dragged at constant velocity by a rope at 20° above the horizontal. μk = 0.40. Find the tension.
- Constant velocity: ΣFx = 0 and ΣFy = 0.
- x: T cos 20° = μk(mg − T sin 20°), so T(cos 20° + μk sin 20°) = μkmg.
- T = (0.40 × 392)/(0.940 + 0.40 × 0.342) = 156.8/1.077 = 146 N.
Key equations
| Components (angle from horizontal) | Fx = F cosθ, Fy = F sinθ |
|---|---|
| Equilibrium | ΣFx = 0, ΣFy = 0 |
| Normal force, pulled at θ above horizontal | FN = mg − F sinθ |
| Normal force, pushed at θ below horizontal | FN = mg + F sinθ |
| Symmetric hanging load | 2T sinθ = mg |
Practice problems
17 problems, from warm-up to exam level. Answer each one before you look: use the hint if you are stuck, then reveal the worked solution one step at a time. Number answers are marked within 2.5%, so g = 10 m/s² still counts.
Hint
The horizontal side is next to the 25° angle.
Worked solution
- Fx = 80 cos 25° = 72.5 N.
Hint
The vertical side is opposite the angle.
Worked solution
- Fy = 80 sin 25° = 33.8 N.
Hint
Vertical forces balance: FN + T sinθ = mg.
Worked solution
- FN = mg − T sinθ = 196 − 50 sin 40° = 196 − 32.1 = 164 N.
Hint
Only the horizontal component accelerates the sled.
Worked solution
- a = 50 cos 40°/20 = 38.3/20 = 1.92 m/s².
Hint
How does the upward component of the pull change the normal force?
Worked solution
- The rope now has an upward component, which reduces FN = mg − T sinθ.
- Kinetic friction is μkFN, so it decreases. The answer is B.
Hint
Two equal vertical components hold up the weight.
Worked solution
- 2T sin 20° = mg = 147 N.
- T = 147/(2 × 0.342) = 215 N.
Hint
2T sinθ = mg. What happens to T as sinθ grows?
Worked solution
- T = mg/(2 sinθ).
- A larger angle means a larger sinθ, so a smaller tension. The answer is B.
Hint
Horizontal balance gives one tension in terms of the other; vertical balance finishes it.
Worked solution
- x: T1 cos 30° = T2 cos 60°, so T2 = √3 T1.
- y: T1(0.5) + √3 T1(0.866) = 2T1 = 98 N.
- T1 = 49.0 N.
Hint
Which cable is steeper, and so closer to straight up?
Worked solution
- The horizontal parts must be equal: T1 cos 30° = T2 cos 60°.
- cos 60° is smaller, so T2 must be larger. The steeper cable carries more. The answer is B.
Hint
Find FN first, using the vertical forces.
Worked solution
- FN = 147 − 30 = 117 N; friction = 0.15 × 117 = 17.6 N.
- a = (52.0 − 17.6)/15 = 2.29 m/s².
Hint
Pushing down adds to the normal force.
Worked solution
- FN = mg + F sinθ = 245 + 50 = 295 N; friction = 88.5 N.
- a = (86.6 − 88.5)/25 = -0.0759 m/s².
Hint
Compare the vertical component of the force in each case.
Worked solution
- Pushing down adds F sinθ to the normal force; pulling up subtracts it.
- Less normal force means less friction, with the same horizontal component. The answer is B.
Hint
Both ΣFx and ΣFy are zero. Friction depends on the tension through FN.
Worked solution
- T cos 20° = 0.40(mg − T sin 20°).
- T = 0.40 × 392/(cos 20° + 0.40 sin 20°) = 156.8/1.077 = 146 N.
Hint
T = mg/(2 sinθ).
Worked solution
- The larger the angle, the larger sinθ, so the smaller T.
- 70° gives the smallest tension. The answer is D.
Hint
At the limit, static friction is at its maximum and balances the horizontal pull. Remember the rope reduces FN.
Worked solution
- Friction needed: 60 cos 30° = 52.0 N.
- FN = 245 − 60 sin 30° = 215 N.
- μs = 52.0/215 = 0.242.
Hint
Add the two pulls as vectors first.
Worked solution
- The two pulls add to √(10² + 10²) = 14.1 N toward the northeast.
- Friction balances the resultant: 14 N toward the southwest. The answer is A.
Hint
Only the vertical part of the push holds the frame up; the horizontal part is balanced by the wall's normal force.
Worked solution
- Vertical: F sin 50° = mg = 19.6 N.
- F = 19.6/0.766 = 25.6 N.
Free-response practice
Original questions in the style of the AP exam's free-response section. Write a full answer on paper, with words as well as equations, then open the scoring guide and tick each point you earned.
Mathematical Routines (MR) style · 8 points · about 14 minutes
A student pulls a sled of mass M across level, snowy ground with a rope at angle θ above the horizontal. The tension in the rope is FT, and the coefficient of kinetic friction between the sled and the snow is μ.
- (a) Draw and label a free-body diagram for the sled.
- (b) Derive an expression for the acceleration of the sled in terms of M, FT, θ, μ and physical constants.
- (c) Calculate the acceleration when M = 18.0 kg, FT = 75.0 N, θ = 35° and μ = 0.2.
- (d) A classmate says the student should pull with the rope horizontal, because then the whole force goes into moving the sled forward. Explain whether a horizontal pull always gives a larger acceleration.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
Dot with four forces: weight Mg down, normal force FN up (shorter than Mg), tension FT at θ above horizontal toward the front, kinetic friction backward.
Part (b)
Vertical: FN = Mg − FT sinθ. Horizontal: FT cosθ − μFN = Ma. So a = [FT cosθ − μ(Mg − FT sinθ)]/M.
Part (c)
FN = 176 − 75 sin 35° = 133 N. a = (75 cos 35° − 0.20 × 133)/18 = (61.4 − 26.7)/18 = 1.93 m/s².
Part (d)
Not always. A horizontal rope gives the largest forward component, but it also gives the largest normal force and so the most friction. Tilting the rope up a little loses only a small amount of forward pull (cosθ is close to 1 for small angles) but reduces friction by μFT sinθ. With friction present, a small upward angle gives a bigger acceleration than a horizontal pull.
Your mark: 0 of 8
Qualitative/Quantitative Translation (QQT) style · 7 points · about 10 minutes
A sign of mass m hangs at rest from a knot. Cable 1 pulls on the knot at 30° above the horizontal to the left, and cable 2 pulls at 60° above the horizontal to the right.
- (a) Without calculating, state which cable has the greater tension and explain your reasoning.
- (b) Derive expressions for the tensions T1 and T2 in terms of m and g.
- (c) Explain how your answers to (b) are consistent with (a). Then explain why the tensions add up to more than the sign's weight.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
Cable 2 (at 60°). The horizontal parts of the two tensions must cancel. Cable 2 is steeper, so a smaller fraction of its tension is horizontal; to have the same horizontal part, its tension must be larger.
Part (b)
x: T1 cos 30° = T2 cos 60°, so T2 = √3 T1. y: T1 sin 30° + T2 sin 60° = mg gives T1(1/2 + 3/2) = mg. T1 = mg/2 and T2 = (√3/2)mg ≈ 0.87mg.
Part (c)
√3/2 ≈ 0.87 is larger than 1/2, so T2 > T1, as predicted. The tensions are not parallel to the weight: only their vertical parts hold the sign up, and their horizontal parts cancel each other. So the tensions themselves (0.5mg + 0.87mg = 1.37mg) can add to more than mg.
Your mark: 0 of 7
Common mistakes
- Writing FN = mg when a force has a vertical component. Always find FN from ΣFy.
- Mixing up sin and cos because the angle is measured from the vertical, not the horizontal.
- Using the whole pull F, not F cosθ, as the force that accelerates an object along the floor.
- Thinking the two tensions in a symmetric hanging load each equal half the weight. They are larger, because only their vertical parts support it.
- Drawing both the force and its components on the free-body diagram. Draw the force on the FBD, and components separately.
- Forgetting that the friction depends on the tension when a rope is angled, so T appears on both sides of the equation.
Frequently asked questions
When do I use sin and when do I use cos?
Use cos for the component next to (adjacent to) the angle, and sin for the component opposite it. If the angle is measured from the horizontal, Fx = F cosθ. If it is measured from the vertical, Fx = F sinθ. Draw the right-angled triangle each time.
Do I need to know sin and cos values without a calculator?
No. You can use a calculator on the whole AP Physics 1 exam. It helps to remember that sin 30° = 0.5, cos 60° = 0.5, and that the 3-4-5 triangle has angles of about 37° and 53°, because many questions are built on them.
What is the difference between static equilibrium and dynamic equilibrium?
Both have zero net force (and, from Unit 5, zero net torque). Static equilibrium means the object is at rest. Dynamic equilibrium means it moves at constant velocity. The equations are the same.
Why does pulling a sled at an angle sometimes give more acceleration than pulling horizontally?
The upward part of the pull reduces the normal force, which reduces friction. For small angles you lose very little forward pull, so the reduction in friction wins. Without friction, a horizontal pull is always best.
Can a rope ever be perfectly horizontal while holding up a weight?
No. A perfectly horizontal rope has no vertical component, so it cannot hold anything up. As the rope gets flatter, the tension needed rises toward infinity, which is why stretched cables sag.
Is torque part of equilibrium in this lesson?
Not yet. Here every force acts at a single point (a knot or a small object), so only the forces need to balance. Extended objects like beams and ladders also need the torques to balance; that is Unit 5 in AP Physics 1.
How do I check my answer for a two-cable problem?
Put your tensions back into both equations. Also check the steeper cable has the larger tension and that each tension is at least as big as its own vertical component.
Keep going
Related guides and practice:
- How to draw free-body diagrams
- Torque and rotational equilibrium
- AP Physics 1 forces summary
- AP Physics 1 forces practice problems
All lessons in the AP Physics 1 forces and energy course · AP Physics 1 past papers