Answer
Answer
Let's assume that beam is making angle 45° with horizontal then moment about point O
"50(\\frac{15}{2}) \\sqrt{2}sin45\u00b0+0.3N_2(15) \\sqrt{2}cos45\u00b0=N_2(15)"
Force required to keep the beam in equilibrium
"35.7(0.3) +P=50cos45\u00b0\\\\P=24.6N"
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