Question #250193

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.


1
Expert's answer
2021-10-12T16:07:19-0400
P1=(P1cos[tan1(600300)],P1sin[tan1(600300)])\vec P_1=\bigg(P_1\cos[\tan^{-1}(\dfrac{600}{300})], P_1\sin[\tan^{-1}(\dfrac{600}{300})]\bigg)

P2=(P2cos[tan1(400300)],P2sin[tan1(400300)])\vec P_2=\bigg(P_2\cos[\tan^{-1}(\dfrac{400}{300})], -P_2\sin[\tan^{-1}(\dfrac{400}{300})]\bigg)

P3=(P3,0)\vec P_3=\bigg(-P_3, 0\bigg)

R=(Rcos[30°],Rsin[30°])\vec R=\bigg(R\cos[30\degree], R\sin[30\degree]\bigg)

R=P1+P2+P3\vec R=\vec P_1+\vec P_2+\vec P_3

cos[tan1(600300)]=cos[tan1(2)]=15\cos[\tan^{-1}(\dfrac{600}{300})]=\cos[\tan^{-1}(2)]=\dfrac{1}{\sqrt{5}}

sin[tan1(600300)]=sin[tan1(2)]=25\sin[\tan^{-1}(\dfrac{600}{300})]=\sin[\tan^{-1}(2)]=\dfrac{2}{\sqrt{5}}

cos[tan1(400300)]=cos[tan1(43)]=35\cos[\tan^{-1}(\dfrac{400}{300})]=\cos[\tan^{-1}(\dfrac{4}{3})]=\dfrac{3}{5}

sin[tan1(400300)]=sin[tan1(43)]=45\sin[\tan^{-1}(\dfrac{400}{300})]=\sin[\tan^{-1}(\dfrac{4}{3})]=\dfrac{4}{5}

cos[30°]=32,sin[30°]=12\cos[30\degree]=\dfrac{\sqrt{3}}{2}, \sin[30\degree]=\dfrac{1}{2}

40(32)=15P1+35P22040(\dfrac{\sqrt{3}}{2})=\dfrac{1}{\sqrt{5}}P_1+\dfrac{3}{5}P_2-20

40(12)=25P145P2+040(\dfrac{1}{2})=\dfrac{2}{\sqrt{5}}P_1-\dfrac{4}{5}P_2+0



15P1=203+2035P2\dfrac{1}{\sqrt{5}}P_1=20\sqrt{3}+20-\dfrac{3}{5}P_2

15P1=10+25P2\dfrac{1}{\sqrt{5}}P_1=10+\dfrac{2}{5}P_2

203+2035P2=10+25P220\sqrt{3}+20-\dfrac{3}{5}P_2=10+\dfrac{2}{5}P_2

P2=203+30P_2=20\sqrt{3}+30

15P1=10+25(203+30)\dfrac{1}{\sqrt{5}}P_1=10+\dfrac{2}{5}(20\sqrt{3}+30)

P1=815+125P_1=8\sqrt{15}+12\sqrt{5}

P1=45(23+3) kNP_1=4\sqrt{5}(2\sqrt{3}+3) \ kN

P2=10(23+3) kNP_2=10(2\sqrt{3}+3)\ kN


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