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

AI Thread Summary
To find the horizontal force exerted by the lower hinge of a symmetrical door, the center of mass is identified at the midpoint due to the door's symmetry. The torque equation is applied, using the bottom hinge as the pivot point, and the gravitational force acting on the center of mass is considered. The torque due to gravity is calculated with the formula τ_g = F_g × r, where F_g is the weight of the door and r is the distance to the center of mass. The discussion highlights the challenge of applying rotational dynamics to a two-hinge system, emphasizing the need for a clear understanding of torque balance. Ultimately, the solution requires careful consideration of forces and torques to determine the horizontal force at the lower hinge.
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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