.
Note that the second part of the equation involves the inverse tangent function. By convention, the function tan−1 returns values between π/2(90◦) and −π/2 (−90◦). You must be careful when finding the θ of vectors with an angle outside of this range. Specifically, for vectors with vx < 0, you must add π (180◦) to tan−1(ry/rx) to obtain the correct θ.
Unit vector notation
As discussed above, we can think of a vector ~v = (vx, vy, vz) as a command to “go a distance vxin the x-direction, a distance vy in the y-direction, and vz in the z-direction.”
To write this set of commands more explicitly, we can use multiples of the vectors ˆı, ˆ, and ˆk. These are the unit vectors pointing in the x, y, and z directions, respectively:
ˆı = (1, 0, 0), = (0, 1, 0),ˆ and ˆk = (0, 0, 1).
Any number multiplied by ˆı corresponds to a vector with that number in the first coordinate. For example, 3ˆı ≡ (3, 0, 0). Similarly, 4ˆ ≡ (0, 4, 0) and 5ˆk ≡ (0, 0, 5).
In physics, we tend to perform a lot of numerical calculations with vectors; to make things easier, we often use unit vector notation:
vxˆı + vyˆ + vzˆk ⇔ ~v ⇔ (vx, vy, vz).
The addition rule remains the same for the new notation:
2ˆı + 3ˆ
It’s the same story repeating all over again: we need to add ˆıs with ˆıs, and ˆs with ˆs.
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Examples
Simple example
Compute the sum ~s = 4ˆı+5∠30◦. Express your answer in the length-and-direction notation.
Since we want to carry out an addition, and since addition is performed in terms of components, our first step is to convert 5∠30◦ into component notation: 5∠30◦= 5 cos 30◦ˆı+ 5 sin 30◦ˆ = 5
√3 2ˆı+52ˆ.
We can now compute the sum:
~s = 4ˆı + 5
√3
2 ˆı +52 = (4 + 5ˆ
√3
2 )ˆı + (52)ˆ . The x-component of the sum is sx= (4 + 5
√3
2 )and the y-component of the sum is sy = (52). To express the answer as a length and a direction, we compute the length k~sk = q
s2x+ s2y = 8.697 and the direction tan−1(sy/sx) = 16.7◦. The answer is ~s = 8.697∠16.7◦. Vector addition example
You’re heading to physics class after a “safety meeting” with a friend, and are looking forward to two hours of finding absolute amazement and awe in the laws of Mother Nature. As it turns out, there is no enlightenment to be had that day because there is going to be an in-class midterm. The first question involves a block sliding down an incline. You look at it, draw a little diagram, and then wonder how the hell you are going to find the net force acting on the block. The three forces acting on the block are ~W = 300∠− 90◦, ~N = 260∠120◦, and ~Ff = 50∠30◦.
You happen to remember the net force formula:
XF = ~~ Fnet= m~a [ Newton’s 2nd law ].
You get the feeling Newton’s 2nd law is the answer to all your trou-bles. You sense this formula is certainly the key because you saw the keyword “net force” when reading the question, and notice “net force”
also appears in this very equation.
The net force is the sum of all forces acting on the block:
F~net=XF = ~~ W + ~N + ~Ff.
All that separates you from the answer is the addition of these vectors.
Vectors have components, and there is the whole sin/cos procedure for decomposing length-and-direction vectors into their components.
If you have the vectors as components you’ll be able to add them and find the net force.
Okay, chill! Let’s do this one step at a time. The net force must have an x-component, which, according to the equation, must equal the sum of the x-components of all the forces:
Fnet,x= Wx+ Nx+ Ff,x
= 300 cos(−90◦) + 260 cos(120◦) + 50 cos(30◦)
=−86.7.
Now find the y-component of the net force using the sin of the angles:
Fnet,y= Wy+ Ny+ Ff,y
= 300 sin(−90◦) + 260 sin(120◦) + 50 sin(30◦)
=−49.8.
Combining the two components of the vector, we get the final answer:
F~net≡ (Fnet,x, Fnet,y)
= (−86.7, −49.8) = −86.7ˆı − 49.8ˆ
= 100∠209.9◦.
Bam! Just like that you’re done, because you overstand them vectors!
Relative motion example
A boat can reach a top speed of 12 knots in calm seas. Instead of cruising through a calm sea, however, the boat’s crew is trying to sail up the St-Laurence river. The speed of the current is 5 knots.
If the boat travels directly upstream at full throttle 12ˆı, then the speed of the boat relative to the shore will be
12ˆı− 5ˆı = 7ˆı,
since we have to “deduct” the speed of the current from the speed of the boat relative to the water.
If the crew wants to cross the river perpendicular to the current flow, they can use some of the boat’s thrust to counterbalance the current, and the remaining thrust to push across. In what direction should the boat sail to cross the river? We are looking for the di-rection of ~v the boat should take such that, after adding in the
ve-locity of the current, the boat moves in a straight line between the two banks (the ˆ direction).
2.1 VECTORS 97
A picture is necessary: draw a river, then draw a triangle in the river with its long leg perpendicular to the current flow. Make the short leg of length 5. We will take the up-the-river component of the speed ~v to be equal to 5ˆı, so that it cancels exactly the −5ˆı flow of the river. Finally, label the hypotenuse with length 12, since this is the speed of the boat relative to the surface of the water.
From all of this we can answer the question like professionals. You want the angle? Well, we have that 12 sin(θ) = 5, where θ is the angle of the boat’s course relative to the straight line between the two banks.
We can use the inverse-sin function to solve for the angle:
θ = sin−1 5 12
= 24.62◦.
The across-the-river component of the velocity can be calculated using vy = 12 cos(θ) = 10.91, or from Pythagoras’ theorem if you prefer vy=pk~vk2− v2x=√
122− 52= 10.91.