Question

The force of magnitude F-95 lb acts along the edge of the triangular plate. Determine the moment of F about point O. The moment is positive if counterclockwise, negative if clockwise. 26" 0 16" F 95 lb Answer: MO b-in.

Answer 1

Force at A in vector form is;

$\vec{F}_A = -\left ( \frac{26}{\sqrt{26^2+16^2}}\times95 \right )\hat{i}-\left ( \frac{16}{\sqrt{26^2+16^2}}\times95 \right )\hat{j}$

$\Rightarrow \vec{F}_A = -80.908\hat{i}-49.789\hat{j}$

Position vector of point-A;

$\vec{r}_{OA} = 26\hat{i}$

Moment about O will be;

$\vec{M}_O = \vec{r}_{OA}\times\vec{F}_A$

$\Rightarrow \vec{M}_O = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 26 & 0 & 0\\ -80.908& -49.789 & 0 \end{vmatrix}$

$\Rightarrow \vec{M}_O = \hat{k}[(26)(-49.789)-0]$

$\Rightarrow \vec{M}_O =-1294.5 \hat{k}$

$\Rightarrow M_O = -1294.5\;\;lb.in\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...(Answer)$

...(Answer)

NOTE: USE NEGATIVE SIGN