"
Set 5 Problem number 8
What vector of magnitude 3.2 must be added to the
velocity vector A = < 5.28 m/s, 6.82 m/s> in order to obtain a vertical vector R?
Answer by giving the magnitude and angle of the vector to be added.
- (Note on notation: <u,v> stands for a vector
whose x component is u and whose y component is v.)
If the resultant vector is to be vertical, then its
x component will be 0.
- So we must find a vector which when added to <
5.28 m/s, 6.82 m/s>, results in a vector whose x component is 0.
- Clearly then the x component of the added vector
will have to be - 5.28 m/s, since this is the only way to cancel out the x component of
the original vector and end up with x component 0.
The vector being added must have magnitude 3.2
m/s.
- We can use this fact to find its y component.
- If y stands for the y component of the added vector,
the Pythagorean Theorem tells us that
- (- 5.28 m/s) ^ 2+y ^ 2 = ( 3.2 m/s) ^ 2, or 27.87
m/s ^ 2 + y ^ 2 = 10.24 m/s ^ 2.
- We can solve this equation for y to obtain y =
`sqrt( 10.24 m/s ^ 2 - 27.87 m/s ^ 2) = 10.23 m/s.
The vector being added therefore has components -
5.28 m/s and 10.23 m/s, and is represented <- 5.28 m/s, 10.23 m/s>
- The magnitude and angle of this vector are easily
found to be 3.2 m/s, as required, and 117.29 degrees.
"