How Do Torque and Hinge Placement Affect a Door's Stability?

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The discussion focuses on calculating the forces at the top and bottom hinges of a door, given its dimensions and mass. The vertical components of the forces can be determined using the equilibrium of forces in the y-direction. To find the horizontal components, the summation of torques around each hinge must equal zero, requiring knowledge of the distance from the hinge to the force application point. The torque is calculated using the weight of half the door and the distance of 0.30m from each hinge. Visualizing the problem with a diagram is recommended to aid understanding of torque and hinge dynamics.
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I'm having quite a bit of trouble with this problem. It seems easy, but I just can't seem to get the idea.

The Problem said:
A door has a height of 2.1m along a y-axis that extends vertically upward and a width of 0.91 m along an x-axis that extends outward from the hinged edge of the door. A hinge 0.30 m from the top and a hinge 0.30 m from the bootom each support half the door's mass, which is 27kg. In unit-vector notation, what are the forces on the door at a) the top hinge and b) the bottom hinge.

I'm able to calculate the vertical components of the forces by using the summation of forces in the y direction equal to zero. The horizantal components, I'm pretty sure I need to use summation of torques, but there are only two forces! The force from each hinge in the horizantal direction! How would I go about solving htis?
 
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You are given the vertical components of the force the hinges exert on the door. To find the horizontal components, take each hinge as a pivot point and realize that the torques about that point must add to zero.
 


It sounds like you are on the right track with using the summation of torques to solve this problem. However, in order to calculate the torque, you will need to know the distance from the hinge to the point where the force is applied. In this case, the distance is given as 0.30m.

To find the torque at the top hinge, you will need to use the force and distance in a cross product. The force will be the weight of half the door (27kg/2 = 13.5kg) multiplied by the acceleration due to gravity (9.8m/s^2) and the distance will be 0.30m. This will give you the torque in the horizontal direction, which can then be broken down into its x and y components using trigonometry.

Similarly, for the bottom hinge, you will use the same force and distance, but in the opposite direction (since the force will be acting in the opposite direction at the bottom hinge). Again, you can use trigonometry to break down the torque into its x and y components.

Remember, torque is a rotational force, so it is important to consider the distance from the hinge to the point where the force is applied. I would recommend drawing a diagram to visualize the problem and make it easier to solve.

I hope this helps you understand the concept of torque and how it relates to door hinges. Keep practicing and don't be afraid to ask for help if you are still having trouble. Good luck!
 
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