From your description, and lettng i and j be unit vectors along the +x and +y axes,
A = 12.00 (sin 37 i + cos 37 j)
= 6.00 i + 9.58 j
B = 15.00 (cos 40 i - sin 40 j)
= 11.49 i - 9.64 j
C = 5.00 (-cos 60 i -sin 60 j)
= -2.50 i - 4.33 j
The changes of sign here and there, and the switch from sin to cos for i, are due to your changing definition of the quadrant the vector is in and the changing reference axis (y for A, +x for B and -x for C).
Now that you have defined A, B, and C with unit vectors, the calculation of R = A+B+C and S = C-A-B is easy. Once you have R & S in unit vector notation, switch back to magnitude/angle notation
(a) Express the vectors A, B, and C in the figure below in terms of unit vectors.
(b) Use unit vector notation to find the vectors R = A + B + C and S = C – A – B.
(c) What are the magnitude and directions of vectors R and S?
Well the picture goes and has a vector
A pointing in the 1st quadrant of the graph and between the y axis it has a angle of 37 deg
and a magnitude of 12m
vector B is in the 4th quadrant and the angle from the x axis to the vector is 40 deg
and a magnitude of 15m
vector C is in the 3rd quadrant and measured from the x axis it's angle is 60 deg
and a magnitude of 6.0m
~What are unit vectors? Would that just refer to the m? And does that include the i, j, and k hat? (i [^ on top of i])
How would I find these? like for C it has the angle of 60 deg from the x axis in the 3rd quadrant but if I was finding the x component of C I wouldn't use 60 right for the angle but rather 180+60 = 240 deg.
I'm not sure if that's correct but that has gotten me confused with unit vector.
And for the B vector in the 4th quadrant wouldn't I find the x and y component using not 40 deg but rather -40 deg?
~Thanks~
I really need to solve this by today...
3 answers
Why is
A = 12.00 (sin 37 i + cos 37 j)
= 6.00 i + 9.58 j ??
I got when I plugged it into my calculator in degrees mode
7.22 i + 9.58 j
Well for
R= A+ B + C
R= (7.22i + 9.58j) + (11.49i- 9.64j) + (-2.50i - 4.33j) = (16.21i - 4.39j)m
S= C-A-B
S= (-2.50i - 4.33j)- (7.22i + 9.58j) - (11.49i - 9.64j)=
-2.50i - 4.33j- 7.22i - 9.58j -11.49i + 9.64j= (-21.21i- 4.27j)m
For step
c.) I don't know how to switch back to magnitude/ angle notation...does that include using tan ?
A = 12.00 (sin 37 i + cos 37 j)
= 6.00 i + 9.58 j ??
I got when I plugged it into my calculator in degrees mode
7.22 i + 9.58 j
Well for
R= A+ B + C
R= (7.22i + 9.58j) + (11.49i- 9.64j) + (-2.50i - 4.33j) = (16.21i - 4.39j)m
S= C-A-B
S= (-2.50i - 4.33j)- (7.22i + 9.58j) - (11.49i - 9.64j)=
-2.50i - 4.33j- 7.22i - 9.58j -11.49i + 9.64j= (-21.21i- 4.27j)m
For step
c.) I don't know how to switch back to magnitude/ angle notation...does that include using tan ?
Um PLEASE help me out on this part c..seriously ...I did show what I did for a and b and I just need to know how c would work...
1 answer