Vector Addition and Subtraction
Practice
practice problem 1
solution
The forces point in the same direction, so they add up.
∑F = F_{1} + F_{2} ∑F = 200 N + 150 N ∑F = 350 N |
The two original forces are east, so the resultant is east.
∑F = 350 N east
No tricks here. Some problems are just easy to solve.
practice problem 2
solution
The forces point in opposite directions, so they subtract. Another way to think of it: one of the forces is positive and one is negative. Signs are a way to indicate basic directions. I think I'll make the first one positive and the second one negative because, why not?
∑F = F_{red} − F_{blue} ∑F = 50 N − 63 N ∑F = −13 N |
How do we describe this direction? No cardinal directions like north, south, east, or west were provided. Nothing was mentioned about left or right (or even up or down). We arbitrarily assigned negative to the direction Blue #5 was kicking. The answer was negative, so the net force points in the direction that Blue #5 was kicking. Let's call that away from Blue #5.
∑F = 13 N away from Blue #5
We could also write…
∑F = 13 N toward Red #3
That would be a good answer, too. Neither of these is more correct than the other.
practice problem 3
solution
North (the direction the engines are pushing) is perpendicular to west (the direction the wind is pushing). The resultant of these two vectors is the hypoteneuse of a right triangle. We use pythagorean theorem to find its magnitude…
v^{2} = | v^{2}_{plane} + v^{2}_{wind} | |
v^{2} = | (100 m/s)^{2} + (30 m/s)^{2} | |
v = | 104 m/s | |
and the tangent to find its direction…
tan θ = | opposite | = | v_{wind} |
adjacent | v_{plane} | ||
tan θ = | 30 m/s | ||
100 m/s | |||
θ = | 17° | ||
These 17° are on the west side of north, so the final answer is…
v = 104 m/s, 17° west of north
practice problem 4
solution
Add vectors in the same direction with "ordinary" addition.
x = 11,200 m + 8,400 m x = 19,600 m y = 3200 m + 1700 m y = 4900 m |
Add vectors at right angles with a combination of pythagorean theorem for magnitude…
r = √(x^{2} + y^{2}) r = √[(19,600 m)^{2} + (4,900 m)^{2}] r = 20,200 m |
and tangent for direction.
tan θ = | y | = | 4,900 m | |
x | 19,600 m | |||
θ = | 14.0° | |||
Don't forget to answer the question.
Camp B is 20,200 m away from base camp at an angle of elevation of 14.0°.
Oh yeah, and don't forget to make a drawing. I probably should have told you to do that earlier.