MCV4U Applications: Forces, Wind and Current: Notes and Practice

Key things to remember

1Resultant

R = F_{1} + F_{2} + \dots

; equilibrant

E = - R

2Ground velocity

=

air (or water) velocity

+

wind (or current) velocity.

3A wind FROM the south blows TOWARD the north; a current is named by the way it flows.

4For a heading measured from north, east component

= v sin β

, north component

= v cos β

5Crossing time

= \frac{width}{across component}

; to land directly opposite, aim upstream at

α

with

sin α = \frac{v _{current}}{v _{boat}}

A pilot flying from Toronto to Ottawa does not simply point the nose at Ottawa. The air itself is moving, and the plane is carried along with it, so a plane pointed at Ottawa in a crosswind would drift off course. In the same way, a canoe aimed straight across a river lands downstream, and a crate pulled by two ropes moves in neither rope's direction. In each case the real motion is the VECTOR SUM of everything acting, and the tools from the last three lessons let you compute it.

Resultants and equilibrants

When several forces act on one object, their vector sum is the RESULTANT, the single force that would have the same effect. The force that exactly cancels the resultant is the EQUILIBRANT: same magnitude, opposite direction, $E = - R$ . An object with zero resultant force is in equilibrium (it stays at rest or keeps moving at constant velocity), which is why engineers check that the forces at every joint of a structure add to $0$ .

Convert every direction to a standard angle (or measure every angle from the same axis).
Resolve each vector into components: $x = r cos θ$ , $y = r sin θ$ .
Add all the $x$ components, then add all the $y$ components.
Find the resultant's magnitude with $x^{2} + y^{2}$ and its angle with $tan^{- 1} (y / x)$ , adjusting for the quadrant.
Convert the angle back into the form the question uses (bearing or quadrant bearing) and state magnitude AND direction.

VideoMr. Cioffi treats forces as vectors, finding the resultant of forces acting on an object and the equilibrant that balances it.Watch on YouTube

The vocabulary of navigation

Term	Meaning
Heading	The direction the aircraft or boat is POINTED
Air speed (or water speed)	Speed relative to the air (or water); the magnitude of the velocity the engine provides
Wind or current velocity	How fast and which way the air or water itself is moving
Ground velocity	Air velocity + wind velocity: the actual motion over the ground
Ground speed and track	The magnitude and the direction of the ground velocity

Ground velocity is always the vector sum: the vehicle's velocity relative to the medium, plus the medium's velocity.

Wind comes FROM a direction

Weather reports name the direction a wind blows FROM. A wind from the south blows TOWARD the north, so its vector points north. A wind from the west pushes things east. Currents are the other way round: a current flowing east moves the water east. Draw the arrow the way the air or water actually moves.

A plane in a crosswind

Here is the classic problem. A plane has a heading of N $27°$ E and an air speed of $375$ km/h. The wind blows from the south at $62$ km/h. What are the ground speed and the actual direction of travel? Put east along $x$ and north along $y$ . A heading of N $27°$ E is $27°$ from the $y$ -axis, so the east component uses sine and the north component uses cosine.

Fig 1The air velocity of 375 km/h on a heading of N 27 degrees E and a 62 km/h wind blowing north (from the south), added tip to tail. The ground velocity is about 431 km/h at about N 23 degrees E.

v_{air} = (375 sin 27°, 375 cos 27°) \approx (170.25, 334.13) v_{wind} = (0, 62) v_{ground} \approx (170.25, 396.13)

Components (east, north) in km/h.

So the ground speed is $170.2 5^{2} + 396.1 3^{2} \approx 431.2$ km/h and the track is $tan^{- 1} (\frac{170.25}{396.13}) \approx 23.3°$ east of north, that is, N $23.3°$ E. The wind raises the speed over the ground by about $56$ km/h, and because it pushes straight north it pulls the track about $3.7°$ closer to north than the heading. The triangle method agrees: the angle inside the triangle between the air and wind vectors is $180° - 27° = 153°$ , and the cosine law gives $37 5^{2} + 6 2^{2} - 2 (375) (62) cos 153° \approx 185901,$ whose square root is about $431.2$ .

A small twin-propeller plane just above a runway, its body angled to the runway as it comes in to land. — PhotoA Britten-Norman Islander landing in a crosswind. The pilot points the nose partly into the wind so the plane's actual path follows the runway, which is the gap between heading and ground track that this section calculates.Uli Elch, CC BY-SA 4.0, via Wikimedia Commons

VideoMr. Cioffi's MCV4U lesson on velocity vectors, where an object's actual motion is the vector sum of its own velocity and the velocity of the air or water it moves through.Watch on YouTube

A boat crossing a current

A river current only carries a boat sideways, so the time to cross depends ONLY on the boat's velocity component straight across. Aim straight across and you cross fastest but land downstream. To land directly opposite, aim partly upstream so that the upstream component of the boat's velocity cancels the current exactly; this needs $sin α = \frac{current speed}{boat speed},$ which is only possible when the boat is faster than the current.

Components or triangle?

With two vectors, the cosine and sine laws on the tip-to-tail triangle are quick. With three or more vectors, or awkward angles, components are safer: they never need you to work out an interior angle of the triangle.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

A plane on a heading of N

27°

E has an air speed of

375

km/h. The wind is blowing from the south at

62

km/h. Find the ground speed and the actual direction of travel.

Example 2

Two forces act on a ring:

50

N at a standard angle of

0°

and

80

N at

60°

. Find the resultant and the equilibrant.

Example 3

A river

300

m wide flows east at

2.0

m/s. A boat moves at

5.0

m/s relative to the water. (a) If the boat aims straight across (north), find its ground velocity, the crossing time and how far downstream it lands. (b) In which direction should it aim to land directly opposite, and how long does that crossing take?

Example 4

A pilot wants a track of due north. The air speed is

300

km/h and the wind blows from the west at

60

km/h. What heading should the pilot fly, and what is the ground speed?

Where marks get lost

Drawing the wind the wrong way. A wind FROM the south points north; drawing it pointing south turns a tailwind into a headwind.

Using cosine for the east component of a bearing. For an angle measured from NORTH, east

= v sin β

and north

= v cos β

, the reverse of the standard-angle formulas.

Using the angle between the vectors (tail to tail) as the interior angle of the tip-to-tail triangle. For the plane, the triangle angle is

180° - 27° = 153°

, not

27°

Dividing the river width by the ground speed to find the crossing time. Only the component straight across the river counts.

Giving the ground speed without the track, or the equilibrant with the resultant's direction. A vector answer needs magnitude and direction.

Key things to remember

Worked examples

Where marks get lost

Flashcards 7

Practice