Point B is at < 4.4, 3.8, 9.1 > ft. A force vector of < 16, 15, 20 > pounds is a
ID: 1550060 • Letter: P
Question
Point B is at < 4.4, 3.8, 9.1 > ft. A force vector of < 16, 15, 20 > pounds is applied at point B.
A.)Determine the x-component of the moment of the force about the origin. (include units with answer)
Hint #1:Cross-product
Hint #2:The 'r' vector is the coordinates of point B
B.)Determine the magnitude of the moment about the origin. (include units with answer)
Hint #1:3D magnitude equation is similar to 2D magnitude equation
Hint #2:sqrt(x^2+y^2+z^2)
C.)Point C is at <-5.7, -3.2, 8.9>ft. Determine the x-component of the moment of the force about the point C. (include units with answer)
Hint #1:The 'r' vector is the vector from C to B
Hint #2:The 'r' vector is B - C (subtract components)
D.)Determine the magnitude of the moment about the point C. (include units with answer)
Explanation / Answer
A)
moment = r x F
M = (4.4i + 3.8j + 9.1 k ) x ( 16i + 15j + 20 k)
M = -60.5 i + 57.6 j + 5.2 k Pound ft
(B)
magnitude = sqrt(60.5^2 + 57.6^2+5.2^2) = 83.7 Pound ft
(C)
moment = r x F
M = (-5.7i - 3.2 j + 8.9 k ) x ( 16i + 15j + 20 k)
M = -197.5 i + 256.4 j - 24.3 k Pound ft
(D)
magnitude = sqrt(197.52 + 256.4^2 + 24.3^2) = 258 Pound ft
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