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Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Force and Velocity Vector Problems

Forces and velocities are vectors, so the word problems in this unit are vector addition in disguise. The skill is turning the story into a triangle; after that it is the cosine law, the sine law, or Pythagoras.

How It Works

Resultant force. Several forces on one object act like a single force, their sum. Add them tip to tail. With two forces of size F1F_1 and F2F_2 at angle θ\theta (tails together), ∣R⃗∣2=F12+F22+2F1F2cos⁡θ|\vec{R}|^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta. The plus sign appears because the triangle's inside angle is 180∘−θ180^\circ - \theta.

Equilibrant. The force that balances the others, so the object stays still. It has the length of the resultant and points the opposite way: E⃗=−R⃗\vec{E} = -\vec{R}.

Components. A force FF at angle θ\theta to the horizontal has a horizontal part Fcos⁡θF\cos\theta and a vertical part Fsin⁡θF\sin\theta. Only the part along the motion moves the object forward.

Relative velocity. A boat or plane moves through water or air that is itself moving. The velocity you see from the ground is the sum: ground velocity == velocity through the air or water ++ wind or current. The heading is where the nose points; the track is where the craft actually goes.

Two kinds of question. If you are given the heading, add the vectors to find the ground velocity. If you are given the direction you want to travel, the heading is the unknown: the nose has to point partly into the wind so the sideways push cancels.

Crossing a river. Pointing straight across is fastest, and the current carries you downstream. The time depends only on the speed across, because the current adds nothing in that direction.

Always draw the triangle and label the known sides and angles before you pick a law.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

  1. Easy

    Forces of 3030 N and 4040 N act on an object at right angles. Find the resultant and the equilibrant.
    Show solution
    1. Right angle, so ∣R⃗∣=302+402=50|\vec{R}| = \sqrt{30^2 + 40^2} = 50 N.
    2. The angle from the 3030 N force is tan⁡−1(4030)≈53.1∘\tan^{-1}\left(\frac{40}{30}\right) \approx 53.1^\circ.
    3. The equilibrant has the same size, opposite direction.

    Answer: Resultant 5050 N at 53.1∘53.1^\circ to the 3030 N force; equilibrant 5050 N in the opposite direction

  2. Medium

    A sled is pulled with a force of 200200 N along a rope at 25∘25^\circ above the horizontal. Find the horizontal and vertical components.
    Show solution
    1. Horizontal: 200cos⁡25∘≈181.3200\cos 25^\circ \approx 181.3 N.
    2. Vertical: 200sin⁡25∘≈84.5200\sin 25^\circ \approx 84.5 N.

    Answer: About 181.3181.3 N horizontally and 84.584.5 N vertically

  3. Medium

    Forces of 1212 N and 1818 N act on an object with 50∘50^\circ between them. Find the resultant.
    Show solution
    1. The triangle's inside angle is 180∘−50∘=130∘180^\circ - 50^\circ = 130^\circ.
    2. ∣R⃗∣2=144+324−2(12)(18)cos⁡130∘≈745.7|\vec{R}|^2 = 144 + 324 - 2(12)(18)\cos 130^\circ \approx 745.7, so ∣R⃗∣≈27.3|\vec{R}| \approx 27.3 N.
    3. Sine law for the angle θ\theta from the 1212 N force: sin⁡θ=18sin⁡130∘27.31≈0.505\sin\theta = \frac{18\sin 130^\circ}{27.31} \approx 0.505, so θ≈30.3∘\theta \approx 30.3^\circ.

    Answer: About 27.327.3 N, at 30.3∘30.3^\circ to the 1212 N force

  4. Medium

    A boat heads straight across a river 120120 m wide at 44 m/s through the water. The current flows at 33 m/s. Find the boat's ground velocity, the crossing time and how far downstream it lands.
    Show solution
    1. The heading and the current are perpendicular: speed 42+32=5\sqrt{4^2 + 3^2} = 5 m/s, at tan⁡−1(34)≈36.9∘\tan^{-1}\left(\frac{3}{4}\right) \approx 36.9^\circ downstream of straight across.
    2. Time depends on the speed across only: 1204=30\frac{120}{4} = 30 s.
    3. Downstream drift: 3×30=903 \times 30 = 90 m.

    Answer: 55 m/s at 36.9∘36.9^\circ downstream; 3030 s; it lands 9090 m downstream

  5. Test level

    A plane with an airspeed of 400400 km/h must fly due north. A 6060 km/h wind blows from the west. Find the heading and the ground speed.
    Show solution
    1. The wind pushes east, so the nose must point west of north by an angle θ\theta that cancels it.
    2. Air velocity is the hypotenuse (400400), the wind is the side opposite θ\theta (6060): sin⁡θ=60400=0.15\sin\theta = \frac{60}{400} = 0.15, so θ≈8.6∘\theta \approx 8.6^\circ.
    3. Ground velocity is the side left over: 4002−602=156 400≈395.5\sqrt{400^2 - 60^2} = \sqrt{156\,400} \approx 395.5 km/h.

    Answer: Heading 8.6∘8.6^\circ west of north; ground speed about 395.5395.5 km/h

Learn It on Martian Lab

Martian Lab teaches this in the Applications of Vectors: Force lesson of its MCV4U Geometric Vectors unit. Watch it, then take the Geometric Vectors practice tests to see if it stuck.

Watch the Applications of Vectors: Force lesson free

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Free Geometric Vectors practice test, with solutions

More Geometric Vectors topics: Adding and Subtracting Vectors · Cartesian Vectors in 2D and 3D

More MCV4U topics: Evaluating Limits · Chain Rule · Optimization Problems · Related Rates · Curve Sketching Steps · Dot Product and Projections · Cross Product, Area and Torque · Equations of Planes · Intersection of Three Planes · Continuity · Derivative from First Principles · Power, Product and Quotient Rules · Velocity and Acceleration · Increasing, Decreasing and Critical Points · Concavity and the Second Derivative Test · Derivatives of e^x and a^x · Derivatives of Sin, Cos and Tan · Equations of Lines in 3D · Intersection of Lines and Planes · Distance from a Point to a Line or Plane

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