Problem 308 The cable and boom shown in Fig. P-308 support a load of 600 lb. Determine the tensile force T in the cable and the compressive for C in the boom.
Solution 308
$C \cos 45^\circ = T \cos 30^\circ$
$C = 1.2247T$
$\Sigma F_V = 0$
$T \sin 30^\circ + C \sin 45^\circ = 600$
$T \sin 30^\circ + (1.2247T) \sin 45^\circ = 600$
$1.366T = 600$
$T = 439.24 \, \text{ lb}$ answer
$C = 1.2247(439.24)$
$C = 537.94 \, \text{ lb}$ answer
Another Solution (By Rotation of Axes)
$T \sin 75^\circ = 600 \sin 45^\circ$
$T = 439.23 \, \text{ lb}$ (okay!)
$\Sigma F_x = 0$
$C = T \cos 75^\circ + 600 \cos 45^\circ$
$C = 439.23 \cos 75^\circ + 600 \cos 45^\circ$
$C = 537.94 \, \text{ lb}$ (okay!)
Another Solution (By Force Polygon)
$\dfrac{T}{\sin 45^\circ} = \dfrac{C}{\sin 60^\circ} = \dfrac{600}{\sin 75^\circ}$
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