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`        A door 1 m wide and 2 m high weights 280 N and is supported by two hinges, one 0.5 m from the top and the other is 0.50 m from the bottom. Each hinge supports half the total weight of the door. Assuming that the doors center of gravity is at its center, find the horizontal components of forces exerted on the door by each hinge.`
one year ago

Arun
22998 Points
```							ay.  The wall end can in straight forward words transmit everyday forces (horizontal), at the same time as the floor end can transmit both everyday (vertical) and frictional (horizontal) forces.  The frictional stress and the traditional stress from the wall end might want to be equivalent and opposite by technique of definition.  The question is, what are the torques from the different forces, and what's a competent position to degree them from? enable's degree from the floor end of the ladder.  that's (6.0m * sin 22° = 2.25 m) from the wall.  the load of the ladder is 17.0 kg * g, and is a million.one hundred twenty 5 m from the point of length; the torque is therefore 19.one hundred twenty 5 g N-m.  the guy on the ladder is a million.688 m from the point of length, and has a weight of seventy six g N; the torque is therefore 128.25 g N-m.  The sum of the torques is 147.375 g N-m. This torque might want to be resisted by technique of the traditional stress from the wall.  the point of contact with the wall is (6.0 m * cos 22°=5.56m) from the point of reference, so the traditional stress might want to be 147.375 g N-m / 5.fifty six m = 25.8 g N. because the contact with the wall is frictionless, each and every of the load is exerted on the contact with the floor.  This sums to ninety 3 kg * g.  The minimum coefficient of friction for the ladder to stay positioned is: 25.8 g N / ninety 3 g N = 0.277.
```
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