Physics C Mechanics

Practice Problems: Vectors Solutions

1. (easy) Vector A represents 5.0 m of displacement east.  If vector B represents 10.0 m of displacement north, find the addition of the two displacements (R).
R = (52 + 102)1/2 =
11 m
tanθ = 10/5 = 2
θ = 63°

2. (easy) Determine the x and y components of a displacement whose magnitude is 30.0 m at a 23° angle from the x-axis.
x-comp = 30cos23 = 28 m
y-comp = 30sin23  = 12 m

3. (moderate) A car moves 150.0 m at a 63° "north of east" (this simply means 63° from the x-axis).  It stays at rest for a while then moves 300 m at 34° "south of west" (this means 214° from the x-axis.)  Find the total displacement of the car.

 vector mag (m) angle x-comp (m) y-comp (m) 1 150 63 68 134 2 300 214 -248 -167 R 183 190 -180 -33

4. (easy) Two forces are being exerted on an object, but in different directions. For example, you and a friend might both be pulling on strings attached to a single block of wood. Find the magnitude and direction of the resultant force in the following circumstances.

a) The first force has a magnitude of 10 N and acts east. The second force has a magnitude of 4 N and acts west.
F = 6N, east
b) The first force has a magnitude of 10 N and acts east. The second force has a magnitude of 4 N and acts north.
The forces are at right angles. Use the Pythagorean Theory.
F = (102 + 42)½ = 11 N
tanθ = 4/10
θ = 22º

5. (moderate) Find the equilibrant force for the system of forces described here:
Force A: 20 N at 20°
Force B: 40 N at 230°

 Force mag (N) angle x-comp (N) y-comp (N) A 20 20 18.8 6.8 B 40 230 -25.7 -30.6 EQ 24.8 74 6.9 23.8 R 0 0 0

6.
(moderate) Vector A represents a displacement in meters expressed in unit vector notation as
A = 2i + 6j + 3k
Vector B represents a second displacement.
B = 5i - 3j – 2k
Find the dot product of the two vectors, the cross product of the two vectors, and the angle between them.
The dot product: Only unit vectors in the same direction give a value.
A • B = 10 - 18 - 6 = -14 m
For the cross product only unaligned vectors give a value.
A x B= (2x -3j) + (2x -2k) + (6j x 5i) + (6j x -2k) + (3k x 5i) + (3k x -3j)
A x B= -6k + 4j - 30k - 12i + 15j + 9i
A x B= (-3i + 19j - 36k) m2
To find the angle between them use the definition of the dot product.
|A| = A = (22 + 62 + 32)1/2 = 7 m
|B| = B = (52 + (-3)2 + (-2)2)1/2 = 6.2 m
A
• B = -14 = ABcosθ = (7)(6.2)cosθ
cosθ = 0.32

θ = 109°

7. (moderate) Vector D = 3i - 4j + 2k and vector: E = 4i - j - 2kFind the magnitude of D + E and the magnitude of  D - E.
D + = 7i - 5j
So the magnitude is |D + E| = (72 + (-5)2)1/2 = 8.6
D - = -i - 3j + 4k
So the magnitude is |D - E| = (12 + (-3)+ 42)1/2 = 5.1

8. (moderate) If force vector F1 has a magnitude of 30 N pointing in the -z direction and force vector F2 has a magnitude of 60 N pointing in the +x direction, determine the dot product (F1 • F2) and the cross product (F1 F2). How would the answers change if the vectors switched position in the equations?
F1 • F2 = (F1)(F2)cos90 = (30)(60)(0) = 0
(The dot product has no value because the vectors are at 90° to each other)

|F1 xF2| = (30)(60)sin 90 = 1800 N2
(This is the magnitude of the cross product)

To determine the direction, use the RHR:
Fingers toward -z
Curl toward +x
Thumb points toward -y.
(The cross product points downward on the screen)
The dot product answer would not change if the vectors switched positions, but the cross product would have the opposite direction (+y) while maintaining the same magnitude (1800 m2).

9. (moderate) Two displacements with magnitudes of 10 m and 12 m can be combined to form resultant vectors with many different magnitudes. Which of the following magnitudes can result from these two displacents? 22 m, 2 m, 30.9 m, 15.6 m. For the possible resultants, what angle exists between the original displacements?

If the displacements are parallel (0° angle between them): R = 22 m
If the displacements are antiparallel (180° angle between them): R = 2 m
If the displacements are perpendicular (90° angle between them): R = 15.6 m
Resultants greater than 22 m are not possible.

10. (moderate) A bicycle tire (Radius = R = 0.4 m) rolls along the ground (with no slipping) through three-quarters of a revolution. Consider the point on the tire that was originally touching the ground. How far has it displaced from its starting position?
The point moves three-fourths of the circumference (this is the distance traveled by the center of the system) plus 1 radius distance along ground (beyond where the center point moves) in x-direction. and 1 radius up in the y-direction.
Resultant = [(0.75)π(2R) + R]i + Rj
Resultant
= 2.28i + 0.4j

Resultant magnitude = (2.282 + 0.42)½ = 2.31 m

11. (moderate) A student carries a lump of clay from the first floor (ground level) door of a skyscraper (on Grant Street) to the elevator, 24 m away. She then takes the elevator to the 11th floor. Finally, she exits the elevator and carries the clay 12 m back toward Grant Street. Determine the total displacement for the clay if each floor is 4.2 m above the floor below.

 Displacement Magnitude (m) Angle (°) x-comp (m) y-comp (m) 1 24 0 24 0 2 42 90 0 42 3 12 180 -12 0 R 44 74 12 42