Two forces P&Q of magnitude 25N and 10N are acting at a point. The forces P & Q make angle 15 & 45, measured counter clockwise with the horizontal. What is the resultant in magnitude and direction?

 

Two forces P&Q of magnitude 25N and 10N are acting at a point. The forces P & Q make angle 15 & 45, measured counter clockwise with the horizontal. What is the resultant in magnitude and direction?

Solution

Hint: 

p⃗ +q⃗ =r⃗ $\overrightarrow{p}+\overrightarrow{q}=\overrightarrow{r}$

ie. adopting notation convention so that force P has magnitude p$p$ and direction p^$\hat{p}$ … p^$\hat{p}$ makes an angle to the “horizontal” θp${\theta }_p$

The problem is —

Given:

p=25N$p=25N$, θp=15π/180=π/12${\theta }_p=15\pi /180=\pi /12$ (always do angles in radians!)

q=10N$q=10N$, θq=45π/180=π/4${\theta }_q=45\pi /180=\pi /4$

Find r$r$ and θr${\theta }_r$

Strategy: (I am not going to do it for you!)

You can do this by expressing p⃗ $\overrightarrow{p}$ and q⃗ $\overrightarrow{q}$ in horizontal and vertical components, adding components, and the recombining to get r⃗ $\overrightarrow{r}$.

You can also do this using triangle geometry… adding the vectors head-to-tail makes a triangle with one unknown side.

ie the exterior angle going from p$p$ to q$q$ is

ϕ=θq−θp=π/6$\varphi ={\theta }_q-{\theta }_p=\pi /6$

These are coming out to really nice numbers if you know about triangles.

The solution is in the cosine rule:

r2=p2+q2+2pqcosϕ$r^2=p^2+q^2+2pq\cos \varphi$

If we define α=θr−θp$\alpha ={\theta }_r-{\theta }_p$ (the angle between p⃗ $\overrightarrow{p}$ and r⃗ $\overrightarrow{r}$) then you can find α$\alpha$:

q2=r2+p2−2prcosα$q^2=r^2+p^2-2pr\cos \alpha$

(cosine rule again!) … and thus θp${\theta }_p$.

You are all done!

If that is all gibberish to you — go the “components” rout.

If x is the horizontal axis:

rx=pcosθp+qcosθq$r_x=p\cos {\theta }_p+q\cos {\theta }_q$

ry=psinθp+qsinθq$r_y=p\sin {\theta }_p+q\sin {\theta }_q$

r2=r2x+r2y$r^2=r_x^2+r_y^2$

tanθr=ry/rx