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Thread: 3D Equilibrium of a rigid body

  1. #1
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    3D Equilibrium of a rigid body

    I am stuck on this question


    I generally know how to solve the problem but the force is throwing me off. Since it is parallel to the XY plane there will no be k component of the force. But how should I about breaking the 970N force into i,j components? Thanks.
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  2. #2
    Senior Member TriKri's Avatar
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    The distance from $\displaystyle O$ to the point where $\displaystyle \vec{F}$ takes effect is $\displaystyle \vec{r} = (250,\ 200,\ -150)\ mm$. $\displaystyle \vec{F}$ is in the x-y plane, so $\displaystyle \vec{F}=(F_x,\ F_y,\ 0)$. The momentum will then be

    $\displaystyle \vec{M}=\vec{r}\times\vec{F}$

    Now, we probably want to charge the support at $\displaystyle O$ as much as possible, so we set $\displaystyle |\vec{F}| = 970\ N$. Besides, we want $\displaystyle \vec{r}$ and $\displaystyle \vec{F}$ to be orthogonal to each other, so

    $\displaystyle \vec{r}\cdot\vec{F} = 0 \Leftrightarrow 250 F_x + 200 F_y = 0 \Leftrightarrow F_y = -\frac{5}{4}F_x$

    $\displaystyle F_z = 0$, so $\displaystyle \vec{F}$ is already in the x-y plane.

    So we could parametrisize $\displaystyle \vec{F}$ and write $\displaystyle \vec{F} = t\cdot(4,\ -5,\ 0)$, implying that $\displaystyle |\vec{F}| = \sqrt{41}\ |t|$, which means that $\displaystyle \vec{F}$ could have basically any non-negative length.

    Now we know that there exist $\displaystyle \vec{F}$ with the in the x-y plane with the length 970 N, orthogonal to $\displaystyle \vec{r}$. Let's use that in our momentum formula:

    $\displaystyle \vec{M} = \vec{r}\times\vec{F} \Rightarrow$

    $\displaystyle |\vec{M}| = |\vec{r}\times\vec{F}| = |\vec{r}|\cdot|\vec{F}|\cdot\sin(\alpha) = \sqrt{250^2+200^2+(-150)^2}\ mm\ \cdot\ 970\ N\cdot\ \sin(\alpha) =$ $\displaystyle \frac{485}{\sqrt{2}}\ Nm\cdot\sin(\alpha)$, where $\displaystyle \alpha$ is the angle between $\displaystyle \vec{r}$ and $\displaystyle \vec{F}$

    Now, since $\displaystyle \vec{r}$ and $\displaystyle \vec{F}$ are orthogonal, $\displaystyle \sin(\alpha) = 1$, so $\displaystyle |\vec{M}| = \frac{485}{\sqrt{2}}\ Nm$, which should be the maximum magnitude of the momentum.





    Alternativelly, you could turn $\displaystyle \vec{F}$ into poolar coordinates:

    $\displaystyle F_x=970\ N\cdot\cos(\varphi)$
    $\displaystyle F_y=970\ N\cdot\sin(\varphi)$
    $\displaystyle F_z=0$

    perform the cross product between $\displaystyle \vec{r}$ and $\displaystyle \vec{F} = 970\ N\cdot(\cos(\varphi),\ \sin(\varphi),\ 0)$ to get the momentum, extract the magnitude of the momentum (the vector lenth), and analyse the expression to se which $\displaystyle \varphi$ that gives the greatest magnitude of the momentum or to directly extract the greatest value of the expression.
    Last edited by TriKri; Nov 3rd 2008 at 10:14 AM.
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  3. #3
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    I got the answer, thanks man
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  4. #4
    Senior Member TriKri's Avatar
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    I'm studuing mechanics myself right now.
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