Answer to Question #250193 in Math for weiouuu

Question

The force R is the resultant of the forces P1, P2, and P3 acting on the rectangular plate. Find P1 and P2 if 𝑅 = 40 kN and 𝑃3 = 20 kN.

Answer 1

$$\vec P_1=\bigg(P_1\cos[\tan^{-1}(\dfrac{600}{300})], P_1\sin[\tan^{-1}(\dfrac{600}{300})]\bigg)$$

$$\vec P_2=\bigg(P_2\cos[\tan^{-1}(\dfrac{400}{300})], -P_2\sin[\tan^{-1}(\dfrac{400}{300})]\bigg)$$

$$\vec P_3=\bigg(-P_3, 0\bigg)$$

$$\vec R=\bigg(R\cos[30\degree], R\sin[30\degree]\bigg)$$

$$\vec R=\vec P_1+\vec P_2+\vec P_3$$

$$\cos[\tan^{-1}(\dfrac{600}{300})]=\cos[\tan^{-1}(2)]=\dfrac{1}{\sqrt{5}}$$

$$\sin[\tan^{-1}(\dfrac{600}{300})]=\sin[\tan^{-1}(2)]=\dfrac{2}{\sqrt{5}}$$

$$\cos[\tan^{-1}(\dfrac{400}{300})]=\cos[\tan^{-1}(\dfrac{4}{3})]=\dfrac{3}{5}$$

$$\sin[\tan^{-1}(\dfrac{400}{300})]=\sin[\tan^{-1}(\dfrac{4}{3})]=\dfrac{4}{5}$$

$$\cos[30\degree]=\dfrac{\sqrt{3}}{2}, \sin[30\degree]=\dfrac{1}{2}$$

$$40(\dfrac{\sqrt{3}}{2})=\dfrac{1}{\sqrt{5}}P_1+\dfrac{3}{5}P_2-20$$

$$40(\dfrac{1}{2})=\dfrac{2}{\sqrt{5}}P_1-\dfrac{4}{5}P_2+0$$

$$\dfrac{1}{\sqrt{5}}P_1=20\sqrt{3}+20-\dfrac{3}{5}P_2$$

$$\dfrac{1}{\sqrt{5}}P_1=10+\dfrac{2}{5}P_2$$

$$20\sqrt{3}+20-\dfrac{3}{5}P_2=10+\dfrac{2}{5}P_2$$

$$P_2=20\sqrt{3}+30$$

$$\dfrac{1}{\sqrt{5}}P_1=10+\dfrac{2}{5}(20\sqrt{3}+30)$$

$$P_1=8\sqrt{15}+12\sqrt{5}$$

$$P_1=4\sqrt{5}(2\sqrt{3}+3) \ kN$$

$$P_2=10(2\sqrt{3}+3)\ kN$$