QUESTION 4. Let A and B be the following vectors, A = 4N at 60º and B = 3.5 N at 150º, where “N” stands for Newtons. Determine:
A) A + B B) A – B C) 4A + 3B.
Please I need a right explanation, step by step.
here,
A = 4 N * ( cos(60) i + sin(60) j)
A = 2 N i + 3.46 N j
B = 3.5 N * ( cos(150) i + sin(150) j)
B = - 3.03 N i + 1.75 N j
a)
R = A + B
R = - 1.03 N i + 5.21 N j
the magnitude of R , |R| = sqrt(1.03^2 + 5.21^2) = 5.31 N
direction , theta = arctan(5.21 /( - 1.03))
theta = 101.18 degree counterclockwise from +X axis
b)
R = A - B
R = 5.03 N i + 1.71 N j
the magnitude of R , |R| = sqrt(5.03^2 + 1.71^2) = 5.31 N
direction , theta = arctan(1.71 /(5.03))
theta = 18.8 degree counterclockwise from +X axis
c)
R = 4A + 3B
R = - 1.09 N i + 19.09 N j
the magnitude of R , |R| = sqrt(1.09^2 + 19.09^2) = 19.2 N
direction , theta = arctan(19.09 /( -1.09))
theta = 93.3 degree counterclockwise from +X axis
QUESTION 4. Let A and B be the following vectors, A = 4N at 60º and...
i need help with question 3 and 4 please
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