with step by step explanation Question 4 18 pts Vector A has a magnitude of 18 a

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with step by step explanation

Question 4 18 pts Vector A has a magnitude of 18 and is at 180 degrees counter-clockwise from the +x axis, vector B has a magnitude of 4 and is at 28 degrees counter-clockwise from the +x axis, vector C has a magnitude of 12 and is at 149 degrees counter-clockwise from the +x axis, and vector D has a magnitude of 3 and is at 216 degrees counter-clockwise from the +x axis. If vector R = A-B + C-D, what is the angle of vector R measured counter-clockwise from the x axis in degrees?

Explanation / Answer

Suppose given that Vector is R and it makes angle theta CCW with +x-axis, then it's components are given by:

Rx = R*cos theta

Ry = R*sin theta

Using above rule:

A = 18 m & angle = 180 deg with -ve x-axis

Ax = 18*cos 180 deg = -18

Ay = 18*sin 180 deg = 0

A = -18 i + 0 j

B = 4 m & angle = 28 deg with -ve x-axis

Bx = 4*cos 28 deg = 3.53

By = 4*sin 28 deg = 1.88

B = 3.53 i + 1.88 j

C = 12 m & angle = 149 deg with -ve x-axis

Cx = 12*cos 149 deg = -10.29

Cy = 12*sin 149 deg = 6.18

C = -10.29 i + 6.18 j

D = 3 m & angle = 216 deg with -ve x-axis

Dx = 3*cos 216 deg = -2.43

Dy = 3*sin 216 deg = -1.76

D = -2.43 i - 1.76 j

Now We need

R = A - B + C - D

R = (-18 i + 0 j) - (3.53 i + 1.88 j) + (-10.29 i + 6.18 j) - (-2.43 i - 1.76 j)

R = (-18 - 3.53 - 10.29 + 2.43) i + (0 - 1.88 + 6.18 + 1.76) j

R = -29.39 i + 6.06 j

Now Angle of R CCW with +x-axis will be

R = 180 - arctan (6.06/29.39) = 180 - 11.65

R = 168.35 deg CCW with +x-axis

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