1
$\begingroup$

I am having trouble with this question:

enter image description here

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...

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Triatticus Jan 28 '19 at 12:56
  • $\begingroup$ @Triatticus why do I need that for this problem? Or do you mean more general? $\endgroup$ – user183956 Jan 28 '19 at 13:03
  • $\begingroup$ Then $\frac{|M|}{|r|\sin{\theta}}=|F|$ $\endgroup$ – Triatticus Jan 28 '19 at 13:12
  • $\begingroup$ Try this $\begin{aligned}M=\left| \overrightarrow {r}\times \overrightarrow {F}\right| =\left| r\right| \cdot \left| F\right| \cdot \sin \left( \varphi \right) \\ \overrightarrow {r}\cdot \overrightarrow {F}=\left| r\right| \cdot \left| F\right| \cdot \cos \left( \varphi \right) \end{aligned}$ $\endgroup$ – Eli Jan 28 '19 at 13:14
2
$\begingroup$

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.

CSYS

A quick shortcut to this calculation is the vector triple product

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

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I am not sure how you got your value of $\mathbf M$? $\endgroup$ – Farcher Feb 6 '19 at 11:11
  • $\begingroup$ 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. $\endgroup$ – John Alexiou Feb 6 '19 at 15:09
  • $\begingroup$ @Farcher - I edited the answer such that the numbers reflect the equations. I had switched points C and D by mistake. $\endgroup$ – John Alexiou Feb 6 '19 at 15:23
  • $\begingroup$ @Farcher - Now the expression $$ \boldsymbol{M} = (\boldsymbol{r}_A - \boldsymbol{r}_C) \times \boldsymbol{F} $$ returns the value shown in the answer. $\endgroup$ – John Alexiou Feb 6 '19 at 15:29
  • $\begingroup$ Thank you for going to all that trouble. I made the comment because I thought that I was doing something wrong! $\endgroup$ – Farcher Feb 6 '19 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy