Not sure how to solve this one:
A particle at the corner P of a cube is acted on by three forces each of magnitude F, acting along the three diagonals of the three adjacent faces of the cube which pass through P. Find the resultant force.
I have got so far:
Best Answer
Let's think in terms of Cartesian coordinates, where the directions of the cube edges correspond to the coordinate axes. From the picture, we see that all of the components are equal to $\frac{F}{\sqrt{2}}$. This makes sense, because $\left(\frac{F}{\sqrt{2}}\right)^2 + \left(\frac{F}{\sqrt{2}} \right)^2 = F^2$.
When the components are summed together, each of the $i,j$ and $k$ components appears exactly twice in the sum. Therefore, the sum is $$ 2\times \left(\frac{F}{\sqrt{2}}\right)\hat{i} + 2 \times \left(\frac{F}{\sqrt{2}}\right) \hat{j} + 2\times \left(\frac{F}{\sqrt{2}}\right) \hat{k} = \sqrt{2}F (\hat{i} + \hat{j} + \hat{k} ) $$ which has length $$ \sqrt{2}F \times \sqrt{3} = \sqrt{6}F $$