Question Find the components of a force P along two directions making angles 45 and 30° with it on opposite sides.
I have no idea about its figure.
I solved using this but got wrong answer.
On
Problems with Dot Product Approach
In the asker's work they present their solution as taking the dot product to find the components. This approach is logical if you are trying to only see the projection of $\vec{P}$ along $\hat{F_1}$ and $\hat{F_2}$ independently or if $\hat{F_1}$ and $\hat{F_2}$ were perpendicular. For this problem this is not the case and if the question is one of decomposition the dot product approach leads to an issue where components of $\vec{P}$ are double counted.
$$ (\vec{P}\cdot\hat{F_1})\hat{F_1}+(\vec{P}\cdot\hat{F_2})\hat{F_2} = |P|\left<\frac{1}{2}-\frac{\sqrt{3}}{4},\frac{1}{2}+\frac{3}{4},0\right>\neq\vec{P}=|P|\left<0,1,0\right>$$
A Possible Approach
Let's start small and see if we can work it out shall we:
What we know: $$\begin{array}{rccc}\vec{P}=\langle & 0,&P,&0\rangle\\\vec{F_1}=\langle & \sin(45^\circ)F_1,&\cos(45^\circ)F_1,&0\rangle\\\vec{F_2}=\langle & -\sin(30^\circ)F_2,&\cos(30^\circ)F_2,&0\rangle\end{array}$$
What we want to know: $F_1, F_2$
Do you think you can setup equations and solve this?
Sounds like we need another hint: Look component wise:
In the $x$ direction we have that
$$\sin(45^\circ)F1 - \sin(30^\circ)F2= 0$$
In the $y$ direction we have:
$$\cos(45^\circ)F1 + \cos(30^\circ)F2 = P$$
We thus have two linear equations and two unknowns. Can you solve it from here?
The problem is about vector decomposition, where $\vec{P}$ is decomposed into $\vec{F_1}$ and $\vec{F_2}$,
$$\vec{P} = \vec{F_1}+ \vec{F_2}$$
Geometrically, they form a parallelogram as shown in the diagram. Use the cosine rule to establish,
$$F_2^2=P^2+F_1^2-2PF_1\cos45°$$ $$F_1^2=P^2+F_2^2-2PF_2\cos30°$$
The solution to this set of equations are,
$$F_1 = \frac{\sqrt{2}P}{1+\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})P$$ $$F_2 = \frac{2P}{1+\sqrt{3}}=(\sqrt{3}-1)P$$
Alternatively, one could also set up the equations using the sine rule as well,
$$\frac{F_2}{F_1} = \frac{\sin45°}{\sin 30°}$$ $$ F_1\cos 45° + F_2\cos 30° = P$$
which is easier to solve since they are linear and gives the same results.