a) Determine the angle between vectors FB and FC. b) Determine magnitude of the
ID: 2993638 • Letter: A
Question
a) Determine the angle between vectors FB and FC.
b) Determine magnitude of the projection of the force FC along a line joining point A and the origin of the coordinate system. Express this as a Certesian vector.
c) Determine magnitude of the projection of the force FC along the line of action of force FB. Express this as a Cartesian vector.
d) Determine the magnitude of the resultant of the two forces, FC and FC. Find the angle between the resultant force and the line joining points A and C.
Explanation / Answer
A(0.5, -1.5, 0)
B(-1.5, -2.5, 2)
C(-1.5, 0.5, 3.5)
Vector AB = (B - A) = (-1.5 - 0.5) i + (-2.5 + 1.5) j + (2 - 0) k
AB = -2 i - j + 2 k
Magnitude of AB = sqrt (2^2 +1^2 + 2^2) = 3
Unit vector along AB = (-2 i - j + 2 k) / 3
Vector AC = C - A = (-1.5 - 0.5) i + (0.5 + 1.5) j + (3.5 - 0) k
AC = -2 i + 2 j + 3.5 k
Magnitude of AC = sqrt (2^2 + 2^2 + 3.5^2) = 4.5
Unit vector along AC = (-2 i + 2 j + 3.5 k) / 4.5
a)
Angle between AB and AC = acos (Unit vector along AB . Unit vector along AC)
= acos [(-2 i - j + 2 k) / 3 . (-2 i + 2 j + 3.5 k) / 4.5]
= acos ((4 - 2 + 7) / 13.5)
= acos (2/3)
= 48.19 deg
b)
Vector OA = 0.5 i - 1.5 j
|OA| = sqrt (0.5^2 + 1.5^2) = 1.58
Unit vector along OA = (0.5 i - 1.5 j) / 1.58
Magnitude of projection of AC along OA = 450*Unit vector along AC. Unit vector along AO
= -450 * (-2 i + 2 j + 3.5 k) / 4.5 . (0.5 i - 1.5 j) / 1.58
= -450*(-2*0.5 -2*1.5) / (4.5*1.58)
= 252.98 N
Thus, in cartesian form,
= 252.98*(-0.5 i + 1.5 j) / 1.58
= -80 i + 240 j
c)
Magnitude of projection of AC along AB = 450*Unit vector along AC. Unit vector along AB
= 450*cos48.19
= 300 N
Thus, in cartesian form
= 300*(-2 i - j + 2 k) / 3
= -200 i - 100 j + 200 k
d)
Resultant = 450*(-2 i + 2 j + 3.5 k) / 4.5 + 600* (-2 i - j + 2 k) / 3
= -600 i + 750 k
Magntiude = sqrt (600^2 + 750^2) = 960.47 N
Angle = acos [(-600 i + 750 k) / 960.47 . (-2 i + 2 j + 3.5 k) / 4.5]
= 27.75 deg
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