Force to stabilize a door - torque

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AI Thread Summary
To stabilize a 10kg door with the bottom hinge removed, a force must be applied at the handle located halfway up the door. The torque equation is used, where the force multiplied by the distance from the hinge must equal the weight of the door multiplied by the distance from the center of mass to the hinge. After calculating, it is determined that a force of 50N is required to maintain equilibrium. The confusion around the term "stabilize" is clarified as placing the door in a state of equilibrium. The final calculation confirms that the necessary force to stabilize the door is 50N.
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Homework Statement


A 10kg door is hanging on two hinges-the top hinge is 9in from the top of the door, and the bottom hinge is 9in from the bottom of the door. The width of the door is 36in and the height is 86in.

If the bolt from the bottom hinge is removed, what force would someone have to exert at the door handle (halfway up the height of the door on the far side from the hinge) to stabilize the door?

Homework Equations


Torque= F r sin theta

The Attempt at a Solution


What does it mean by stabilize?
F(36in)=100N(12)
Force=33N? maybe?

What does the problem mean by stabilize? I'm confused
 
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r26h said:
What does it mean by stabilize?

Place in a state of equilibrium.

F(36in)=100N(12)

Can you explain what you're doing here and where these numbers are coming from?
 
Mister T said:
Place in a state of equilibrium.
Can you explain what you're doing here and where these numbers are coming from?

Instead of 12 I think I should be using 18.

So, take the distance from the edge of the door to the handle (36) * F and set it equal to Mg (100N) * (.5*36)
F(36in)=100N(18in)
So then my answer would be 50N.
 

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