# Magnitude of F about line CD (Mechanics, 3D vector system)

I am having trouble with this question:

I will write out the question below to make it easier for other people googling the question to find it:

2/139 If the magnitude of the moment of F about line CD is 50 Nm, determine the magnitude of F.

I know that $$\vec M=\vec r \times \vec F$$

What I have tried:

Calculating $$\vec M$$ as following:

$$\vec {CD} = (0.4, 0, -0.2)$$ please note how I set up coordinate system (x and y is in bottom and z upward according to right-hand rule)

$$|\vec{CD}|$$=$$\frac{1}{\sqrt{0.4^2+0.2^2}}=2.2$$

$$\vec M = 50 \times 2.2 (0.4, 0, -0.2)$$=$$(44, 0, -22)Nm$$

I now set $$\vec F=(f_1, f_2, f_3)$$

then try to find some vector $$\vec r$$. I have tried $$\vec CA$$ and $$\vec DA$$ but no matter what vector I choose I end up with something like $$0.2f_1=44$$ $$0.4f_1=0$$ after cross product, which is an impossible system of equation to solve. What am I doing wrong? If $$\vec r$$ is wrong, please explain why. I am having a hard time wrapping my head around how $$\vec F$$ can even make a rotation around $$\vec CD$$ in the first place but I guess this is more of a theoretical question...

• Remember also that $|\vec{M}|=|\vec{r}\times\vec{F}|= |\vec{r}||\vec{F}|\sin{\theta}$, where $\theta$ is the angle between the vectors. This will make life much easier – Triatticus Jan 28 '19 at 12:56
• @Triatticus why do I need that for this problem? Or do you mean more general? – user183956 Jan 28 '19 at 13:03
• Then $\frac{|M|}{|r|\sin{\theta}}=|F|$ – Triatticus Jan 28 '19 at 13:12
• Try this \begin{aligned}M=\left| \overrightarrow {r}\times \overrightarrow {F}\right| =\left| r\right| \cdot \left| F\right| \cdot \sin \left( \varphi \right) \\ \overrightarrow {r}\cdot \overrightarrow {F}=\left| r\right| \cdot \left| F\right| \cdot \cos \left( \varphi \right) \end{aligned} – Eli Jan 28 '19 at 13:14

Your issue is that you know the direction of the vector $$\boldsymbol{F}$$ and therefore $$f_1$$, $$f_2$$ and $$f_3$$ are not independent. Use just one unknown $$F$$ along the correct direction and the problem becomes solvable.

Consider the line about which the moment is measured with the direction

$$\boldsymbol{e} = \frac{ \boldsymbol{r}_D - \boldsymbol{r}_C }{ \| \boldsymbol{r}_D - \boldsymbol{r}_C \| } = \frac{ \pmatrix{0.4 \\ -0.2 \\ 0}}{ \tfrac{1}{\sqrt{5}}} = \pmatrix{2/\sqrt{5} \\ -1/\sqrt{5} \\ 0}$$

and the force vector

$$\boldsymbol{F} = \frac{ \boldsymbol{r}_B - \boldsymbol{r}_A }{ \| \boldsymbol{r}_B - \boldsymbol{r}_A \| }\,F = \pmatrix{2/\sqrt{6}\\ 1/\sqrt{6} \\ -1/\sqrt{6}} F$$

The moment vector about point C is

$$\boldsymbol{M} = (\boldsymbol{r}_A - \boldsymbol{r}_C) \times \boldsymbol{F} = \pmatrix{0 \\ -1/5 \\ -1/5} \times \pmatrix{2/\sqrt{6}\\ 1/\sqrt{6} \\ -1/\sqrt{6}} F = \pmatrix{\sqrt{6}/15 \\ -\sqrt{6}/15 \\ \sqrt{6}/15 } F$$

The projection of the moment along the line is supposed to be 50

$$50 = \boldsymbol{e} \cdot \boldsymbol{M} =\pmatrix{2/\sqrt{5} \\ -1/\sqrt{5} \\ 0} \cdot \pmatrix{\sqrt{6}/15 \\ -\sqrt{6}/15 \\ \sqrt{6}/15 } F = \frac{\sqrt{30}}{25} F$$

$$\Rightarrow\; F = \frac{125 \sqrt{30}}{3} = 228.22$$

I used the following coordinate system.

A quick shortcut to this calculation is the vector triple product

$$M = \boldsymbol{e} \cdot ( \boldsymbol{r}_A-\boldsymbol{r}_C) \times \boldsymbol{F}$$

$$50 = \pmatrix{2/\sqrt{5} \\ -1/\sqrt{5} \\ 0} \cdot \left[ \pmatrix{0 \\ -1/5 \\ -1/5} \times \pmatrix{2/\sqrt{6}\\ 1/\sqrt{6} \\ -1/\sqrt{6}} \right] F$$ $$\boxed{ 50 = \frac{\sqrt{30}}{25} F }$$

• I am not sure how you got your value of $\mathbf M$? – Farcher Feb 6 '19 at 11:11
• I used \begin{aligned} \boldsymbol{r}_A & = \pmatrix{0 \\ 0 \\ 0.2} & \boldsymbol{r}_B & = \pmatrix{0.4 \\ 0.2 \\ 0} \\ \boldsymbol{r}_C & = \pmatrix{0 \\ 0.2 \\ 0.4} & \boldsymbol{r}_D & = \pmatrix{0.4 \\ 0 \\ 0.4} \end{aligned} and the formulas above. I might have made a typo somewhere, so I am checking now. – John Alexiou Feb 6 '19 at 15:09
• @Farcher - I edited the answer such that the numbers reflect the equations. I had switched points C and D by mistake. – John Alexiou Feb 6 '19 at 15:23
• @Farcher - Now the expression $$\boldsymbol{M} = (\boldsymbol{r}_A - \boldsymbol{r}_C) \times \boldsymbol{F}$$ returns the value shown in the answer. – John Alexiou Feb 6 '19 at 15:29
• Thank you for going to all that trouble. I made the comment because I thought that I was doing something wrong! – Farcher Feb 6 '19 at 15:32