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

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To calculate the torque on a hinged door due to its weight, the center of mass must be identified, as this is where the weight acts. The door weighs 53.0 N and has dimensions of 0.9 m wide and 2.5 m high. The torque is determined using the formula τ = r(Fsinθ), where r is the distance from the pivot point to the center of mass. For a uniform door, the center of mass is located at its midpoint, which is 0.45 m from the hinge along the width. Understanding these principles is essential for solving the problem accurately.
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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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