How Is Torque Calculated on a Hinged Door Due to Its Weight?

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SUMMARY

The torque exerted by a uniform door weighing 53.0 N, measuring 0.9 m in width and 2.5 m in height, is calculated about a horizontal axis perpendicular to the door at a corner. The formula used is τ = r(Fsinθ), where 'r' is the distance from the pivot to the center of mass, 'F' is the force (weight of the door), and θ is the angle of force application. The center of mass for the door is located at its midpoint, which is crucial for determining the correct distance for torque calculation. The correct approach involves using the distance from the pivot to the center of mass, which is 0.45 m for this door.

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Homework Statement



A uniform door weighs 53.0 N and is 0.9 m wide and 2.5 m high. What is the magnitude of the torque due to the door's own weight about a horizontal axis perpendicular to the door and passing through a corner?


Homework Equations




torque is equal to the perpendicular component of the force with the shortest distance between the roation axis and the point of application of the force

and

τ=r(Fsinθ)



The Attempt at a Solution




530*0.9*2.5*(sin90)


i really don't know where to begin, my teacher hasn't taught this yet, even though the webassign is due tomorrow. am i using the wrong formulas?
 
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trainumc said:

Homework Statement



A uniform door weighs 53.0 N and is 0.9 m wide and 2.5 m high. What is the magnitude of the torque due to the door's own weight about a horizontal axis perpendicular to the door and passing through a corner?

Speaking generally you need to figure the distance of the Center of Mass as the point that you would have its weight act through relative to the desired axis of rotation.
 

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