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WHAT IS THE FORCE F REQUIRED AT EQUILIBRIUM OF THE GATE?


2 years ago

Venkat
273 Points
							WHAT IS THE FORCE F REQUIRED AT EQUILIBRIUM OF THE GATE?

2 years ago
Venkat
273 Points
							sorry the attachm

2 years ago
Venkat
273 Points
							THE CORrect option is (b) = 0 $\\F_{horizontal}=0.5(9.81)(2)(2)=19.62kN$ It will act at   $1-\frac{2}{3}=\frac{1}{3}$    from the hinge $\\F_{vertical}=\pi \times \frac{(1)^2}{2}(9.81)=\frac{9.81\pi}{2}kN$ It will act at     $\\\frac{4}{3\pi}$    from the hinge Take moments anout hinge.$\\19.62\times\frac{1}{3}-\frac{9.81\pi}{2}\times \frac{4}{3\pi}=F\times1 \\ \\\frac{19.62}{3}-\frac{19.62}{3}=F \\ \\F=0$

2 years ago
Aswanth
18 Points
							 THE CORrect option is (b) = 0 $\\F_{horizontal}=0.5(9.81)(2)(2)=19.62kN$ It will act at   $1-\frac{2}{3}=\frac{1}{3}$    from the hinge $\\F_{vertical}=\pi \times \frac{(1)^2}{2}(9.81)=\frac{9.81\pi}{2}kN$ It will act at     $\\\frac{4}{3\pi}$    from the hinge Take moments anout hinge.$\\19.62\times\frac{1}{3}-\frac{9.81\pi}{2}\times \frac{4}{3\pi}=F\times1 \\ \\\frac{19.62}{3}-\frac{19.62}{3}=F \\ \\F=0$

one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions