### Forces

```
A
The resultant is the sum of two or more vectors. Vectors can
be added by moving the tail of one vector to the head of another
vector without changing the magnitude or direction of the vector.
Note: The red vector R has the same magnitude and direction.
Multiplying a vector by a scalar number changes its length
but not its direction unless the scalar is negative.
V
2V
-V
If two vectors are added at right angles, the magnitude
can be found by using the Pythagorean Theorem
R2 = A2 + B 2 and the angle by
Opp
Tan 
If two vectors are added at any other angle, the magnitude
can be found by the Law of Cosines
and the angle by the Law of Sines
R 2  A2  B 2  2 AB cos
Sin A
Sin B
Sin C


a
b
c
8 meters
62+82=102
36°
6 meters
10 meters
The distance traveled is 14 meters and the displacement
is 10 meters at 36º south of east.
6
t a n1    36
8
A hiker walks 3 km due east, then makes a 30° turn north of east
walks another 5 km. What is the distance and displacement of the
hiker?
The distance traveled is 3 km + 5 km = 8 km
R
5 km
3 km
R2 = 32+52- 2*3*5*Cos 150°
R2 = 9+25+26=60
R = 7.7 km
s in150  s inθ

7.7
5
The displacement is 7.7 km @ 19° north of east
  19
Add the following vectors and determine the resultant.
3.0 m/s, 45 and 5.0 m/s, 135
5.83 m/s, 104
A boat travels at 30 m/s due east across a river that is 120 m wide
and the current is 12 m/s south. What is the velocity of the boat
relative to shore? How long does it take the boat to cross the river?
How far downstream will the boat land?
30 m/s
30 m/s
12 m/s
12 m/s
The speed will be
122  302
= 32. 3 m/s @ 21° downstream.
The time to cross the river will be t = d/v = 120 m / 30 m/s = 4 s
The boat will be d = vt = 12 m/s * 4 s = 48 m downstream.
Examples
Add the following vectors and determine the resultant.
6.0 m/s, 225 + 2.0 m/s, 90
4.80 m/s, 207.9 
Add the following vectors and determine the resultant.
6.0 m/s, 225 + 2.0 m/s, 90
•
•
•
•
R2 = 22 + 62 – 2*2*6*cos 45
R2 = 4 + 36 –24 cos 45
R2 = 40 – 16.96 = 23
R = 4.8 m
sin 45 sin 

4 .8
2
R
17
  17

6m
2m
R = 4.8 m @ 208 
45°
Ax  A cos
Ay  A sin 
A  Ax  Ay
A  A A
2
x
2
y
and
  tan
1
Ay
Ax
Rx   v x
Ry   v y
  tan
1
Ry
Rx
R  R R
2
x
2
y
Vector components is taking a vector
and finding the corresponding
horizontal and vertical components.
Vector resolution
A

Ax
Ay
Ax  A cos
Ay  A sin 
A plane travels 500 km at 60°south of east.
Find the east and south components of its
displacement.
de
de= 500 km *cos 60°= 250 km
60°
ds
500 km
ds= 500 km *sin 60°= 433 km
Vector equilibrium
Maze Game
```