The magnitude of force P is 908.3N
Forces in cables:
[tex]\vec{F}_{AC} = F_{AC}\left [ \frac{\vec{AC}}{|AC|} \right ] = (300)\left [ \frac{1.6\hat{j}+1.2\hat{k}}{\sqrt{1.6^2+1.2^2}} \right ]\Rightarrow \vec{F}_{AC} =240\hat{j}+180\hat{k}[/tex]........(i)
And,
[tex]\vec{F}_{AE} = F_{AE}\left [ \frac{\vec{AE}}{|AE|} \right ] = F_{AE}\left [ \frac{-0.4\hat{i}+1.6\hat{j}-0.86\hat{k}}{\sqrt{0.4^2+1.6^2+0.86^2}} \right ]\Rightarrow \vec{F}_{AE} =-0.2151F_{AE}\hat{i}+0.8602F_{AE}\hat{j}-0.4624F_{AE}\hat{k}[/tex]
And,
[tex]\vec{F}_{AD} = F_{AD}\left [ \frac{\vec{AD}}{|AD|} \right ] = (P)\left [ \frac{1.3\hat{i}+1.6\hat{j}+0.4\hat{k}}{\sqrt{1.3^2+1.6^2+0.4^2}} \right ]\Rightarrow \vec{F}_{AD} =0.619P\hat{i}+0.7619P\hat{j}+0.1905P\hat{k}[/tex]
Now,
[tex]\vec{W} = -W\hat{j}[/tex]
By equilibrium of x-forces (i components):
[tex]\sum F_x = 0\Rightarrow -0.2151F_{AE}-0.4382P+0.619P = 0\Rightarrow F_{AE} = 0.8405P[/tex]
By equilibrium of z-forces (k components):
[tex]\sum F_z = 0\Rightarrow 180-0.4624F_{AE}+0.1905P = 0\Rightarrow 180-0.4624(0.8405P)+0.1905P = 0\Rightarrow P = 908.3\;\;N[/tex]
Therefore, the magnitude of force P is 908.3N
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A container of weight [tex]W[/tex] is suspended from ring [tex]A[/tex], to which cables [tex]AC[/tex]and [tex]AE[/tex] are attached. A force [tex]P[/tex] is applied to the end [tex]F[/tex] of a third cable that passes over a pulley at [tex]B[/tex] and through ring [tex]A[/tex] and that is attached to a support at [tex]D[/tex]. Knowing that the tension in cable [tex]AC[/tex] of the system is 300N, determine the magnitude of [tex]P[/tex].
