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A Hч T Hx 80N 200N 700N Balancing torque about the left end, {cz - (1x700) - (x 200) - 16X80) + (67 sin 60%) = 0 [T = 340 N Hx o 171 N Efx = Hx T cas 60° 700-200-80=0 Ety = Hy + T sin 60° Hy = 683 N CS Scanned with CamScanner
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