The Physics
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Opus in profectus

Vector Addition & Subtraction

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Practice

practice problem 1

Two people are pushing a disabled car. One exerts a force of 200 N east, the other a force of 150 N east. What is the net force exerted on the car? (Assume friction to be negligible.)

solution

The forces point in the same direction, so they add up.

 
ΣF = F1 + F2
ΣF = 200 N + 150 N
ΣF = 350 N
 

The two original forces are east, so the resultant is east.

ΣF = 350 N east

No tricks here. Some problems are easy to solve.

practice problem 2

Two soccer players kick a ball simultaneously from opposite sides. Red #3 kicks with 50 N of force while Blue #5 kicks with 63 N of force. What is the net force on the ball?

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?

 
ΣF = Fred − Fblue
ΣF = 50 N − 63 N
ΣF = −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.

ΣF = 13 N away from Blue #5

We could also write…

ΣF = 13 N toward Red #3

That would be a good answer, too. Neither of these is more correct than the other.

practice problem 3

An airplane heads due north at 100 m/s through a 30 m/s cross wind blowing from the east to the west. Determine the resultant velocity of the airplane (relative to due north).

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…

v2 =  v2plane + v2wind  
 
v2 =  (100 m/s)2 + (30 m/s)2  
 
v =  104 m/s  
 

…and the tangent to find its direction…

tan θ =  opposite  =  vwind
adjacent vplane
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

A mountain climbing expedition establishes a base camp and two intermediate camps, A and B. Camp A is 11,200 m east of and 3,200 m above base camp. Camp B is 8,400 m east of and 1,700 m higher than Camp A. Determine the displacement between base camp and Camp B.

solution

Add vectors in the same direction with "ordinary" addition.

x =  11,200 m + 8,400 m
x =  19,600 m
y =  3,200 m + 1,700 m
y =  4,900 m

Add vectors at right angles with a combination of pythagorean theorem for magnitude…

r =  √(x2 + y2)
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.