Find the magnitude of the horizontal force exerted by the lower hinge

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SUMMARY

The discussion focuses on calculating the horizontal force exerted by the lower hinge of a symmetrical door with a mass of 27.5 kg, height of 5.44 m, and width of 2.46 m. The center of mass is located at the midpoint of the door. The relevant equations include torque equilibrium, specifically the condition that the sum of clockwise torques equals the sum of counter-clockwise torques. The participant attempts to apply these principles but expresses uncertainty regarding the two points of rotation created by the hinges.

PREREQUISITES
  • Understanding of torque and rotational equilibrium
  • Familiarity with the concept of center of mass
  • Knowledge of basic physics equations, particularly Fd=Fd
  • Ability to apply trigonometric functions in physics problems
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  • Study the principles of torque in multi-dimensional systems
  • Learn how to calculate forces at hinges in static equilibrium
  • Explore the application of trigonometric functions in rotational dynamics
  • Investigate real-world examples of door mechanics and hinge forces
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Physics students, mechanical engineers, and anyone interested in understanding the mechanics of doors and hinge forces in static equilibrium scenarios.

angel120
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Homework Statement


A symmetrical door of mass 27.5 kg is 5.44 m high and 2.46 wide. The hinges are placed at the very top and very bottom of the door. Find the magnitude of the horizontal force exerted by the lower hinge.


Homework Equations


Fd=Fd
sum tau clockwise = sum tau counter-clockwise

The Attempt at a Solution


The center of mass is dead center of the door, since it's symmetrical. However, we have only done rotation for "1-dimensional" things like a table. I know that when you apply something like this to a table, you put the counterclockwise on one side, then the clockwise on the other. Since the door is vertical and has 2 points of rotation (the 2 hinges) I'm not sure how to work this out...
 
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angel120 said:

Homework Statement


A symmetrical door of mass 27.5 kg is 5.44 m high and 2.46 wide. The hinges are placed at the very top and very bottom of the door. Find the magnitude of the horizontal force exerted by the lower hinge.


Homework Equations


Fd=Fd
sum tau clockwise = sum tau counter-clockwise

The Attempt at a Solution


The center of mass is dead center of the door, since it's symmetrical. However, we have only done rotation for "1-dimensional" things like a table. I know that when you apply something like this to a table, you put the counterclockwise on one side, then the clockwise on the other. Since the door is vertical and has 2 points of rotation (the 2 hinges) I'm not sure how to work this out...


Gravity acts upon the center of mass of an object. So, I'm assuming the following is legit.:

\sum\tau=0. Taking the bottom hinge as the axes of rotation, we find:
\tau_g=\vec{F}_g\times\vec{r}=m\vec{g}\vec{r}\sin(\theta), where \theta=0.735 rad. Now, what next?
 

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