Equilibrium Problem - Balancing Torques

[HoodedFreak]

Homework Statement

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A door 1.00 m wide and 2.00 m high weighs 330 N
and is supported by two hinges, one 0.50 m from the top and the
other 0.50 m from the bottom. Each hinge supports half the total
weight of the door. Assuming that the door’s center of gravity is at
its center, find the horizontal components of force exerted on the
door by each hinge.

Homework Equations

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T = F*l

The Attempt at a Solution

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This problem is really giving me a hard time, because I know that both the horizontal and vertical components of the force of the hinge are exerting torques

So, I start off with

H₁cos(a) + H₂cos(b) = 0

H₂sin(a) + H₂sin(b) - mg = 0

For balancing the torque I choose the top left hinge as a pivot point

Then the torque due to gravity would be mg*l which is just 0.5mg, now I am not quite sure how to find the lever arm for the hinge force. Any suggestions?

 

[billy_joule]

now I am not quite sure how to find the lever arm for the hinge force. Any suggestions?

Have you drawn a diagram?
What is the distance between the hinges?
(You don't need any trig functions to answer the question).

I know that both the horizontal and vertical components of the force of the hinge are exerting torques

That depends on where the axis is you take moments about. You've chosen to use one of the hinges as the axis, this means the vertical forces at the hinges are irrelevant.

 

[HoodedFreak]

Have you drawn a diagram?
What is the distance between the hinges?
(You don't need any trig functions to answer the question).

The distance between the hinges is √2. I've drawn a diagramThat depends on where the axis is you take moments about. You've chosen to use one of the hinges as the axis, this means the vertical forces at the hinges are irrelevant.

Why is it the case that the vertical forces at the hinges are irrelevant. I can see why this is the case for the hinge that you choose as your pivot point, but wouldn't the vertical force at the other hinge still exert a torque since it may have a component that is perpendicular to the distance between the two hinges.

 

[billy_joule]

Look at the door in the room you're in; both hinges lie on the same vertical line, correct? And the vertical forces are, of course, collinear with that line, correct?
So if we take moments about any point on that vertical line (including at either hinge) the lever arm length for the vertical forces is zero, as their line of action passes through that point.
If that's not clear, draw a diagram and we can see where you're going wrong.

 

[HoodedFreak]

Look at the door in the room you're in; both hinges lie on the same vertical line, correct? And the vertical forces are, of course, collinear with that line, correct?
So if we take moments about any point on that vertical line (including at either hinge) the lever arm length for the vertical forces is zero, as their line of action passes through that point.
If that's not clear, draw a diagram and we can see where you're going wrong.

Right, ofcourse, that makes sense. For some weird reason I thought the hinges were on either side of the door

So, now I just have the horizontal force of the bottom hinge, H₂cos(b) * 1 = 0.5mg, and plugging that into the first equation gives us H₁cos(a) = -0.5mg

Thank you for your help