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 = (5^{2} + 10^{2})^{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 = (10^{2} + 4^{2})^{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^{2 }
For the cross product only unaligned vectors give a value.
A x B= (2i x -3j) + (2i x -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) m^{2}
To find the angle between them use the definition of the dot product.
|A| = A = (2^{2} + 6^{2} + 3^{2})^{1/2} = 7 m
|B| = B = (5^{2} + (-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 - 2k. Find the magnitude of D + E and the magnitude of D - E.
D + E = 7i - 5j
So the magnitude is |D + E| = (7^{2} + (-5)^{2})^{1/2} = 8.6
D - E = -i - 3j + 4k
So the magnitude is |D - E| = (1^{2} + (-3)^{2 }+ 4^{2})^{1/2} = 5.1
8. (moderate) If force vector F_{1} has a magnitude of 30 N pointing in the -z direction and force vector F_{2} has a magnitude of 60 N pointing in the +x direction, determine the dot product (F_{1} • F_{2}) and the cross product (F_{1} x F_{2}). How would the answers change if the vectors switched position in the equations?
F_{1} • F_{2} = (F_{1})(F_{2})cos90 = (30)(60)(0) = 0
(The dot product has no value because the vectors are at 90° to each other)
|F_{1} xF_{2}| = (30)(60)sin 90 = 1800 N^{2}
(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 m^{2}).
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 = (2.28^{2} + 0.4^{2})^{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 |