The displacement vectors A and B shown in the figure below both have magnitudes

Question

The displacement vectors A and B shown in the figure below both have magnitudes of 2.08 m. The direction of vector A is theta = 41.6 degree. (a) Find A + B graphically. magnitude direction degree counterclockwise from the +x axis (b) Find A - B graphically. magnitude m direction degree counterclockwise from the +x axis (c) Find B - A graphically. magnitude m direction degree counterclockwise from the +x axis (d) Find A - 2B graphically. magnitude m direction degree counterclockwise from the +x axis

Explanation / Answer

here: A = (L cos , L sin ) , B= (0 , L)

a) A+ B = L cos i + (L sin +L)j = 1.56i + (1.38+2.08)j = 1.56i +3.46j

magnitude = sqrt[(1.56^2 + 3.46^2)] = 3.79

direction = arctan(3.46/1.56) = 65.73 degree

b)

A- B = L cos i + (L sin -L)j = 1.56i + (1.38-2.08)j = 1.56i - 0.7j

magnitude = sqrt[(1.56^2 + 0.7^2)] = 1.71

direction = arctan(0.7/1.56) +360= 384.16 degree

c)

B - A= (L - L sin )j - L cos i  = (2.08 - 1.38)j - 1.56i= -1.56i + 0.7j

magnitude = sqrt[(1.56^2 + 0.7^2)] = 1.71

direction = arctan(0.7/1.56) +360 = 384.16 degree

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