1. **Problem statement:** We are given the line of action of a force with equation $y=2x$ and a point $B(-3,2)$. We need to find the length of the perpendicular dropped from point $B$ to the line $y=2x$. 2. **Formula for distance from a point to a line:** The distance $d$ from a point $(x_0,y_0)$ to a line $Ax+By+C=0$ is given by: $$d=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$$ 3. **Rewrite the line equation:** The line is $y=2x$, which can be rewritten as: $$2x - y = 0$$ So, $A=2$, $B=-1$, and $C=0$. 4. **Substitute point coordinates:** For point $B(-3,2)$: $$d=\frac{|2(-3) -1(2) + 0|}{\sqrt{2^2 + (-1)^2}} = \frac{|-6 - 2|}{\sqrt{4 + 1}} = \frac{|-8|}{\sqrt{5}} = \frac{8}{\sqrt{5}}$$ 5. **Simplify the distance:** Rationalize the denominator: $$d= \frac{8}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{8\sqrt{5}}{5}$$ **Final answer:** The length of the perpendicular from point $B(-3,2)$ to the line $y=2x$ is: $$\boxed{\frac{8\sqrt{5}}{5}}$$