- #1
dark-ryder341
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This is for a first year university question in Physics (Mechanics).
1. Write each vector in terms of the unit vectors i(hat) and j(hat). Use the unit vectors to express the vector C[tex]\rightarrow[/tex] where C[tex]\rightarrow[/tex] = 3.00A[tex]\rightarrow[/tex] - 4.00B[tex]\rightarrow[/tex]. Find the magnitude and direction of C[tex]\rightarrow[/tex].
A[tex]\rightarrow[/tex] = Axi(hat) + Ayj(hat)
Ax = 3.60cos(70deg.)=1.23
Ay = 3.60sin(70deg.)=3.38
A[tex]\rightarrow[/tex] = 1.23i(hat) + 3.38j(hat)
B[tex]\rightarrow[/tex] = Bxi(hat) + Byj(hat)
Bx = -2.4cos(30deg.)=-2.078
By = -2.4sin(30deg.)=-1.2
B[tex]\rightarrow[/tex] = -2.078i(hat) + (-1.2)j(hat)
C[tex]\rightarrow[/tex] = 3.00A[tex]\rightarrow[/tex] - 4.00B[tex]\rightarrow[/tex]
C[tex]\rightarrow[/tex] = 3.00(1.23i(hat) + 3.38j(hat)) - 4.00(-2.0781i(hat) + (-1.2)j(hat))
C[tex]\rightarrow[/tex] = (3.69i(hat) + 10.14j(hat)) - (-8.312i(hat) + (-4.8)j(hat))
I got stuck at this stage; I'm pretty sure up until now I've been doing it right, but now I'm a little confused about combining the like terms (ie. ihat with ihat and jhat with jhat terms). Would I, for instance with the jhat terms, go 10.14-(-4.8) which would give me a large positive number? or would I just go 10.14(-4.8) which equals 5.34, and then subtract it later (ie. -4.622i(hat) - 5.34j(hat))?
Basically, I just need to know what to do from here. I'm pretty sure I know how to solve for the magnitude and direction once I figure out these terms. Thanks very much for any help!
1. Write each vector in terms of the unit vectors i(hat) and j(hat). Use the unit vectors to express the vector C[tex]\rightarrow[/tex] where C[tex]\rightarrow[/tex] = 3.00A[tex]\rightarrow[/tex] - 4.00B[tex]\rightarrow[/tex]. Find the magnitude and direction of C[tex]\rightarrow[/tex].
Homework Equations
A[tex]\rightarrow[/tex] = Axi(hat) + Ayj(hat)
Ax = 3.60cos(70deg.)=1.23
Ay = 3.60sin(70deg.)=3.38
A[tex]\rightarrow[/tex] = 1.23i(hat) + 3.38j(hat)
B[tex]\rightarrow[/tex] = Bxi(hat) + Byj(hat)
Bx = -2.4cos(30deg.)=-2.078
By = -2.4sin(30deg.)=-1.2
B[tex]\rightarrow[/tex] = -2.078i(hat) + (-1.2)j(hat)
The Attempt at a Solution
C[tex]\rightarrow[/tex] = 3.00A[tex]\rightarrow[/tex] - 4.00B[tex]\rightarrow[/tex]
C[tex]\rightarrow[/tex] = 3.00(1.23i(hat) + 3.38j(hat)) - 4.00(-2.0781i(hat) + (-1.2)j(hat))
C[tex]\rightarrow[/tex] = (3.69i(hat) + 10.14j(hat)) - (-8.312i(hat) + (-4.8)j(hat))
I got stuck at this stage; I'm pretty sure up until now I've been doing it right, but now I'm a little confused about combining the like terms (ie. ihat with ihat and jhat with jhat terms). Would I, for instance with the jhat terms, go 10.14-(-4.8) which would give me a large positive number? or would I just go 10.14(-4.8) which equals 5.34, and then subtract it later (ie. -4.622i(hat) - 5.34j(hat))?
Basically, I just need to know what to do from here. I'm pretty sure I know how to solve for the magnitude and direction once I figure out these terms. Thanks very much for any help!