C. Geometry and Algebra of Vectors ยท Vectors in Two-Space and Three-SpaceApplications: Forces, Resultants and Equilibrants
C2.3 C1.1About 75 minutes
Learning goals- I can find the resultant and equilibrant of two or more forces, using diagrams, trigonometry or components.
- I can resolve a force into perpendicular components, including along and perpendicular to a ramp.
- I can find the tensions in cables holding an object in equilibrium.
- I can solve force problems in two-space and three-space.
Words to know
force
A push or pull, measured in newtons (N); a vector.
resultant
The vector sum of all the forces acting on an object.
equilibrant
The single force that balances a set of forces: ๐ = โ๐.
equilibrium
The state in which the vector sum of the forces on an object is ๐.
tension
The pulling force in a rope, string or cable, directed along it.
weight
The force of gravity on an object, W = mg, directed down.
Pushes and pulls have directions
Every bridge, tower, crane and hockey stick is designed by people who add forces as vectors. A force is a push or a pull, measured in newtons (N), and it always acts in a direction. When several forces act on one object, their vector sum is the resultant. If the resultant is zero, the forces balance and the object stays at rest (or keeps moving at a constant velocity). This lesson uses everything from the unit so far to solve force problems in two-space and three-space.
Remember from science: the force of gravity on a mass m is its weight, W = mg, where g โ 9.8 N/kg near Earth's surface. A 20 kg sign has a weight of about 20 ร 9.8 = 196 N, pointing straight down. If you took SPH3U, you used free-body diagrams and Newton's laws in the Forces unit.
Resultant and equilibrant
- The resultant ๐ of forces ๐
โ, ๐
โ, โฆ is their sum: ๐ = ๐
โ + ๐
โ + โฏ.
- The equilibrant ๐ is the single force that would balance them: ๐ = โ๐. It has the same magnitude as the resultant and the opposite direction.
- An object is in equilibrium when the sum of all forces on it (including any equilibrant) is ๐. In components, every component of the sum is 0.
When three forces are in equilibrium, drawing them tip to tail makes a closed triangle (the sum is ๐). That is why the sine and cosine laws are so useful for force problems.
Worked exampleExample 1: Two forces at an angle
Forces of 120 N and 80 N act on an object, with an angle of 50ยฐ between them. Find the resultant and the equilibrant.
- Tip to tail, the angle inside the triangle is 180ยฐ โ 50ยฐ = 130ยฐ.
- |๐|ยฒ = 120ยฒ + 80ยฒ โ 2(120)(80) cos 130ยฐ โ 14 400 + 6400 + 12 340.8 = 33 140.8, so |๐| โ 182.0 N.
- Angle between ๐ and the 120 N force: sin ฮฑ = 80 sin 130ยฐ / 182.05 โ 0.3366, ฮฑ โ 19.7ยฐ.
- The equilibrant has the same magnitude and the opposite direction.
Resultant โ 182.0 N at 19.7ยฐ from the 120 N force (toward the 80 N force). Equilibrant โ 182.0 N in the opposite direction.
Worked exampleExample 2: Forces in three-space
Three forces act on a drone: ๐
โ = [3, โ2, 5] N, ๐
โ = [โ1, 4, 2] N and ๐
โ = [2, 1, โ3] N. Find the resultant, its magnitude and the equilibrant.
- Add components: ๐ = [3 โ 1 + 2, โ2 + 4 + 1, 5 + 2 โ 3] = [4, 3, 4].
- |๐| = โ(16 + 9 + 16) = โ41 โ 6.40 N.
- ๐ = โ๐ = [โ4, โ3, โ4] N.
๐ = [4, 3, 4] N, |๐| โ 6.40 N, ๐ = [โ4, โ3, โ4] N.
Tension in cables
When an object hangs from two cables, the knot where they meet is in equilibrium. Three forces act on it: the two tensions ๐โ and ๐โ along the cables, and the weight ๐ straight down. Resolve each force into horizontal and vertical components and set each total to 0. That gives two equations in the two unknown tensions.
Worked exampleExample 3: Tension in two cables
A 20 kg sign hangs from two cables that make angles of 30ยฐ and 45ยฐ with the horizontal ceiling. Find the tension in each cable (g = 9.8 N/kg).
- Weight: W = 20 ร 9.8 = 196 N downward.
- Horizontal balance: Tโ cos 30ยฐ = Tโ cos 45ยฐ, so Tโ = Tโ cos 30ยฐ / cos 45ยฐ โ 1.2247 Tโ.
- Vertical balance: Tโ sin 30ยฐ + Tโ sin 45ยฐ = 196.
- Substitute: 0.5 Tโ + (1.2247 Tโ)(0.7071) = 196, so 1.3660 Tโ = 196 and Tโ โ 143.5 N.
- Then Tโ โ 1.2247 ร 143.48 โ 175.7 N.
- Check with the closed triangle and the sine law: Tโ = 196 sin 135ยฐ / sin 105ยฐ โ 143.5 N โ (the angle between ๐โ and ๐โ is 180ยฐ โ 30ยฐ โ 45ยฐ = 105ยฐ).
Tโ โ 143.5 N (30ยฐ cable) and Tโ โ 175.7 N (45ยฐ cable). The steeper cable carries more of the load.
Resolving a force on a ramp
A box on a ramp feels its weight ๐ straight down. It's useful to split ๐ into two perpendicular components: one along the ramp (which tends to slide the box down) and one into the ramp (which the ramp pushes back against). If the ramp makes an angle ฮธ with the horizontal, then
along the ramp: W sin ฮธ and into the ramp: W cos ฮธ
(The angle between ๐ and the "into the ramp" direction is also ฮธ, because both are rotated by ฮธ from the vertical and horizontal.)
Worked exampleExample 4: A box on a ramp
A 50 kg box rests on a ramp inclined at 20ยฐ. Find the components of its weight along and perpendicular to the ramp.
- W = 50 ร 9.8 = 490 N.
- Along the ramp: 490 sin 20ยฐ โ 167.6 N.
- Perpendicular to the ramp: 490 cos 20ยฐ โ 460.4 N.
- Check: โ(167.6ยฒ + 460.4ยฒ) โ 490 โ
โ 167.6 N down the slope and โ 460.4 N into the ramp. For the box to stay at rest, friction must push 167.6 N up the slope.
A routine for force problems
- Sketch the object and draw every force as an arrow from one point (a free-body diagram).
- Choose axes. Horizontal/vertical usually works; for ramps, use along/perpendicular to the ramp.
- Resolve each force into components, with signs.
- For a resultant, add the components. For equilibrium, set each component sum to 0 and solve.
- Check: does the answer make sense? A steeper cable should carry more weight; the equilibrant should point roughly opposite to the forces.
Just for funEngineers like to say a bridge stays up because all of its forces agree to cancel out. It's the one place where a resultant of zero is exactly what everyone wants.
Hands-on activityTwo-string lift
You need: a 1 L plastic bottle full of water with the cap on tightly (its weight is about 9.8 N), two pieces of string about 60 cm long, and a protractor. A small luggage scale or spring scale is optional. Work over a carpet or the grass, and keep your feet clear in case the bottle slips.
- Tie both strings around the neck of the bottle. Hold one string in each hand and lift the bottle with both strings vertical. Each hand holds about half the weight.
- Slowly move your hands apart so the strings make about 60ยฐ, then 30ยฐ, then 10ยฐ with the horizontal. Notice how much harder you must pull.
- Predict each tension with T = W รท (2 sin ฯ), where ฯ is the angle each string makes with the horizontal. (This comes from the vertical balance 2T sin ฯ = W.)
- If you have a spring scale, put it in one string and compare.
What to look for: predicted tensions are about 5.7 N at 60ยฐ, 9.8 N at 30ยฐ and 28.2 N at 10ยฐ. As the strings get closer to horizontal, the tension grows very quickly. That's why a clothesline or power line that is pulled almost straight carries a huge tension.
Your turn
1Find the resultant of 6 N [E] and 8 N [N], and the equilibrant.
|๐| = โ(36 + 64) = 10 N. Angle from north toward east: tanโปยน(6/8) โ 36.9ยฐ. ๐ = 10 N [N 36.9ยฐ E]; equilibrant 10 N [S 36.9ยฐ W].
2Two forces of 25 N each act on an object with 60ยฐ between them. Find the magnitude of the resultant.
Triangle angle 120ยฐ: |๐|ยฒ = 625 + 625 + 625 = 1875, |๐| = 25โ3 โ 43.3 N, pointing along the bisector (30ยฐ from each force).
3Forces of 50 N and 70 N act with 110ยฐ between them. Find the magnitude of the resultant and its angle with the 50 N force.
Triangle angle 70ยฐ: |๐|ยฒ = 2500 + 4900 โ 7000 cos 70ยฐ โ 5005.9, |๐| โ 70.8 N. sin ฮฑ = 70 sin 70ยฐ / 70.75 โ 0.9297, ฮฑ โ 68.4ยฐ.
4What is the weight of a 15 kg mass?
15 ร 9.8 = 147 N downward.
5Find the equilibrant of the force [12, โ5] N and its magnitude.
[โ12, 5] N, magnitude โ(144 + 25) = 13 N.
6A 30 kg crate sits on a 25ยฐ ramp. Find the components of its weight along and perpendicular to the ramp.
W = 294 N. Along: 294 sin 25ยฐ โ 124.2 N. Perpendicular: 294 cos 25ยฐ โ 266.5 N.
7Forces [2, โ1, 3] N, [โ4, 2, 1] N and [1, 1, โ2] N act on a point. Find the resultant and the equilibrant.
๐ = [โ1, 2, 2] N (|๐| = 3 N); ๐ = [1, โ2, โ2] N.
More practice: warm-up
1What is the SI unit of force?
2How is the equilibrant related to the resultant?
๐ = โ๐: same magnitude, opposite direction.
3Find the resultant of 5 N [E] and 5 N [E].
4Find the resultant of 5 N [E] and 5 N [W].
0 N (the forces balance).
5Find the magnitude of the resultant of 3 N and 4 N acting at right angles.
More practice: on your own
1A 100 N pull on a sled rope acts at 30ยฐ above the horizontal. Find its horizontal and vertical components.
Horizontal 100 cos 30ยฐ โ 86.6 N; vertical 100 sin 30ยฐ = 50 N.
2A 12 kg traffic light hangs from the middle of a cable. Each half of the cable sags 10ยฐ below the horizontal. Find the tension in the cable.
W = 117.6 N. By symmetry both tensions are equal: 2T sin 10ยฐ = 117.6, so T โ 338.6 N, almost three times the weight.
3Three forces keep an object in equilibrium. Two of them are 30 N [E] and 40 N [N]. Find the third.
The third is the equilibrant of the first two: 50 N [S 36.9ยฐ W].
4A 600 N resultant comes from two perpendicular forces. One of them is 360 N. Find the other.
โ(600ยฒ โ 360ยฒ) = โ(360 000 โ 129 600) = โ230 400 = 480 N.
5Forces of 15 N and 20 N have a resultant of 25 N. Find the angle between them.
625 = 225 + 400 + 600 cos ฮธ, so cos ฮธ = 0 and ฮธ = 90ยฐ.
Challenge
1A 25 kg sign hangs from two cables making angles of 40ยฐ and 55ยฐ with the horizontal ceiling. Find both tensions.
W = 245 N. The angle between the cables is 180ยฐ โ 40ยฐ โ 55ยฐ = 85ยฐ. Sine law on the closed triangle: Tโ (40ยฐ cable) = 245 sin 145ยฐ / sin 85ยฐ โ 141.1 N; Tโ (55ยฐ cable) = 245 sin 130ยฐ / sin 85ยฐ โ 188.4 N. Check: horizontal parts 141.1 cos 40ยฐ โ 108.1 and 188.4 cos 55ยฐ โ 108.1 โ
2Forces of 10 N, 12 N and 15 N keep an object in equilibrium. Find the angle between the 10 N and 12 N forces.
Tip to tail they form a triangle with sides 10, 12, 15. The angle inside the triangle between the 10 and 12 sides: cos C = (100 + 144 โ 225)/240 = 19/240, C โ 85.5ยฐ. The angle between the forces (tail to tail) is 180ยฐ โ 85.5ยฐ โ 94.5ยฐ.
3At what ramp angle is the component of a box's weight along the ramp exactly half the weight? What fraction of the weight presses into the ramp then?
W sin ฮธ = ยฝW, so sin ฮธ = ยฝ and ฮธ = 30ยฐ. Into the ramp: W cos 30ยฐ = (โ3/2)W โ 0.866W.
Quiz
1. What is the equilibrant of a force of 50 N [N 30ยฐ E]?
2. Forces of 9 N and 12 N act at right angles. What is the magnitude of the resultant?
3. What is the weight of an 8 kg mass, using g = 9.8 N/kg?
4. A 200 N force acts at 35ยฐ above the horizontal. What is its vertical component?
5. What does it mean for an object to be in equilibrium?
6. Forces [3, 0, โ4] N and [โ3, 5, 4] N act on a point. What is the magnitude of the resultant?
Big idea (C2.3, C1.1): forces are vectors. Their sum is the resultant, and an object is in equilibrium when that sum is zero. Components turn every force problem into a pair (or triple) of simple equations.
How to help: Try the two-string bottle lift together. It is a memorable way to feel why nearly-horizontal cables carry big tensions. Make sure the bottle cap is tight and feet are clear.
Common difficulties:
- Mixing up sine and cosine when resolving (draw the right triangle each time).
- Forgetting to convert a mass in kilograms to a weight in newtons.
- Giving the equilibrant the same direction as the resultant.
- Using the angle with the vertical when the angle with the horizontal is given (or the reverse).