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Show some work for assistance. Post one long question at a time.
1. Consider the two vectors P and Q. The definition of P is P = i × (i × j). The
definition of Q is Q = (i × i) × j .
a. The magnitude of P is equal to the magnitude of Q.
b. The magnitude of P is greater than the magnitude of Q.
c. The magnitude of P is less than the magnitude of Q.
2. We have two vectors that line in the x-y plane. If we take the cross-product of those two
vectors we know that
a. the magnitude of the cross-product will always be equal to the product of the
magnitudes of the two vectors.
b. the magnitude of the cross-product will always be greater than the product of the
magnitudes of the two vectors.
c. the magnitude of the cross-product will always be less than the product of the
magnitudes of the two vectors.
d. the cross-product vector will be perpendicular to the first vector and in the x-y
plane.
e. the cross-product vector will be in either the +z or -z direction.
3. A particle of mass 0.5 kg has velocity 2 m/s. It is moving at constant velocity along a
line that passes 0.5 m away from the origin, as shown in the figure below. Rank the
angular momentum of the particle when it is at the points A, B, and C from smallest to
greatest. Point A is 1.5 m away from the point of closest approach, B is 0.5 m away and
C is located at the point of closest approach.
a. A < B < C
b. A = B = C
c. B > C > A
d. C > B > A
e. C > A > B
4. A force, F = 5i is applied at a point located at a position of R = 2i + 4j . What
torque does this force exert relative to the origin?
a. 5 Nm in the + x-direction
b. 5 Nm in the + z-direction.
c. 10 Nm in the - x-direction
d. 10 Nm in the - z-direction.
e. 20 Nm in the + x-direction
9 answers
1. I am thinking A since the vectors are techinqually the same and multiplication is associative
2. I am pretty sure it is E.
3.since the particle is not attached/drawn to the origin i believe it has no angular momentum and therefore B
4. 5 times 2 is 10 so C or D so I am thinking C
2. I am pretty sure it is E.
3.since the particle is not attached/drawn to the origin i believe it has no angular momentum and therefore B
4. 5 times 2 is 10 so C or D so I am thinking C
4. Torque is the cross product of F and R vectors. It must be in the z (k vector) direction since F and R are in the x-y plane. I would expect the answer to be 20 N*m in the +z direction.
Did you type choice e. direction correctly?
I do not understand the meaning of your square symbols ��.
Did you type choice e. direction correctly?
I do not understand the meaning of your square symbols ��.
the square was is the ^ on top of the i and j vectors.
also does that mean 1-3 is correct?
also does that mean 1-3 is correct?
I did not answer 1-3. I suggest you post them separately for someone else to answer
I did check E and it is 20NM in the -Z
So now I am even more confused.
So now I am even more confused.
The +/- sign depends upon whether the torque is defined by R x F or F x R. Perhaps I got the order wrong.
E is the right answer.
E is the right answer.
20 in -z direction (-20 k) is correct
R x F
i j k
2 4 0
5 0 0
= - 20 k
R x F
i j k
2 4 0
5 0 0
= - 20 k
I had torque as F x R, nor R x F.
Forgot which it is.
Hence the sign change.
Forgot which it is.
Hence the sign change.