ATMO 303 Practice Excercise 08.26 Problem #4 Please help me understand 4. A force F of 6 units acts through the point P = (2, 3, 1) in the direction of the vector (4, 1, 2).Find the moment of the force about the point Q = (1, 2, 1).

the moment formula M= r×F here r = vector PQ r = OQ→-OP→ = (-1,-1,0) And the F vector is 6 times…

Answered: A force F of 6 units acts through the…

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ATMO 303 Practice Excercise 08.26 Problem #4

Please help me understand

4. A force F of 6 units acts through the point P = (2, 3, 1) in the direction of the vector (4, 1, 2).
Find the moment of the force about the point Q = (1, 2, 1).

Expert Solution

Step 1

the moment formula $M = r \times F$

here r = vector PQ

r = $\vec{O Q} - \vec{O P}$

= $(- 1, - 1, 0)$

And the F vector is 6 times the unit vector along the line of force of action.

$6 \frac{4, 1, 2}{\sqrt{4^{2} + 2^{2} + 1^{2}}} = 6 (\frac{4}{\sqrt{21}}, \frac{1}{\sqrt{21}}, \frac{2}{\sqrt{21}})$

So the Moment vector must be

$M = 6 (\frac{- 2}{\sqrt{21}}, \frac{2}{\sqrt{21}}, \frac{3}{\sqrt{21}}) = (\frac{- 12}{\sqrt{21}}, \frac{12}{\sqrt{21}}, \frac{18}{\sqrt{21}})$

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A force F of 6 units acts through the point P = (2, 3, 1) in the direction of the vector (4, 1, 2). Find the moment of the force about the point Q = (1, 2, 1).