Static Equilibrium of door and hinge

[Wellesley]

Homework Statement

A door made of a uniform piece of wood measures 1 m by 2 m and has a mass of 18 kg. The door is entirely supported by two hinges, one at the bottom corner and one at the top corner. Find the force (magnitude and direction) that the door exerts on each hinge. Assume that the vertical force on each hinge is the same.

Homework Equations

Torque=Force*Distance

The Attempt at a Solution

I've got the magnitudes of the hinges solved, and they coincide with what the answer is (99N).

What baffles me is the direction of the force of the door on the hinge. The book's answer is: for the top hinge, 99N 27^o right of vertical, and for the bottom hinge, 99N, 27^o left of vertical.

Basically, is this answer for a door with the hinges on the right or left? I think that this answer refers to the hinges on the right side of the door (I can go into the logic of my assumption if needed, but I'm not sure it's right). If this is the case, would hinges on the left side have forces of 99N 27^o left of vertical, and for the bottom hinge, 99N, 27^o right of vertical?

Conceptually, this problem is very confusing, and any help would be greatly appreciated. Thanks!

 

[Doc Al]

Basically, is this answer for a door with the hinges on the right or left? I think that this answer refers to the hinges on the right side of the door (I can go into the logic of my assumption if needed, but I'm not sure it's right).

Why do you think the hinges are on the right? (Note that the question asks for the force of the door on the hinge, not the hinge on the door.)

And if you move the hinges to the other side, a force that was pointing to the left of vertical will now point to the right.

 

[Wellesley]

Why do you think the hinges are on the right? (Note that the question asks for the force of the door on the hinge, not the hinge on the door.)

And if you move the hinges to the other side, a force that was pointing to the left of vertical will now point to the right.

First off, I did the problem with the hinges on the left.
When I summed the torque, I chose the bottom hinge as the axis of rotation. Since the force of gravity due to the door has a negative (or counterclockwise) direction, the force (in the x-direction) of the upper hinge has to be positive (or clockwise).

I think the book solved the problem with the hinges on the right. Basically, I'm getting confused on the directions of the components of the force exerted by the door on the hinge.

If the hinges were on the left side, would the force diagrams for the force exerted by the door on the hinge (in component form) look like this?


Top ---> \leftarrow\uparrow


Bottom---> \rightarrow\uparrow


Thanks.

 

[Doc Al]

If the hinges were on the left side, would the force diagrams for the force exerted by the door on the hinge (in component form) look like this?


Top ---> \leftarrow\uparrow


Bottom---> \rightarrow\uparrow

I believe you are confusing the force exerted by the hinge on the door with the force exerted by the door on the hinge. (Those forces are equal but opposite, of course.) Note that you show upward forces in both cases. Does that make sense? (Does the door push up or down on the hinges?)

 

[Wellesley]

I believe you are confusing the force exerted by the hinge on the door with the force exerted by the door on the hinge. (Those forces are equal but opposite, of course.)

You are correct. I'm trying to find the force that the door exerts on the hinge, but I'm unsure of how that force would be sketched out in a diagram. I did the calculations numerically (magnitude), but I'm trying to understand the direction component of the forces. Would the rough diagram I posted in the last post correspond this force (Force that the door exerts on the hinge)?

Note that you show upward forces in both cases. Does that make sense? (Does the door push up or down on the hinges?)

Doesn't the vertical forces of both hinges have to equal the force of gravity?
i.e...H_1y+H_2y=F_g

 

[Doc Al]

You are correct. I'm trying to find the force that the door exerts on the hinge, but I'm unsure of how that force would be sketched out in a diagram. I did the calculations numerically (magnitude), but I'm trying to understand the direction component of the forces.

I don't know what you mean when you say you did the calculation 'numerically'. Why not just analyze the force components acting on the door and apply the conditions for equilibrium?

Would the rough diagram I posted in the last post correspond this force (Force that the door exerts on the hinge)?

No, as I thought I had made clear. You are showing the forces acting on the door, not on the hinge. That's fine, but the question asks for the opposite: the forces on the hinges. (Use Newton's 3rd law to relate them.)

Doesn't the vertical forces of both hinges have to equal the force of gravity?
i.e...H_1y+H_2y=F_g

The net force on the door must be zero. Your equation describes the forces exerted by the hinges on the door.

 

[Wellesley]

I don't know what you mean when you say you did the calculation 'numerically'. Why not just analyze the force components acting on the door and apply the conditions for equilibrium?

I had to work this problem backwards. I knew the answer, and figured out how the magnitude of the force was calculated, but I was unsure whether the "27^o right of vertical" for the top hinge corresponded to hinges on the right side or left side of the door. When I tried to figure this out myself, I got confused with the action-reaction pair.

No, as I thought I had made clear. You are showing the forces acting on the door, not on the hinge. That's fine, but the question asks for the opposite: the forces on the hinges. (Use Newton's 3rd law to relate them.)

If I did this, and I summed the torque, around the bottom hinge, I get the top hinge's force component (x-direction) to be negative. Shouldn't it be the opposite of my first diagram, and be positive (moving to the right)?

 

[Doc Al]

If I did this, and I summed the torque, around the bottom hinge, I get the top hinge's force component (x-direction) to be negative.

When you analyze the forces on the door, you get the the x-component of the force exerted by the top hinge on the door to be negative.

If you want the forces on the hinge you need to take the opposite.

Shouldn't it be the opposite of my first diagram, and be positive (moving to the right)?

Your diagram shows the forces exerted by the hinge on the door. The forces exerted on the hinges would be the opposite of those.

 

[Wellesley]

When you analyze the forces on the door, you get the the x-component of the force exerted by the top hinge on the door to be negative.

If you want the forces on the hinge you need to take the opposite.

Your diagram shows the forces exerted by the hinge on the door. The forces exerted on the hinges would be the opposite of those.

Thanks Doc Al for being patient with me. It gets confusing (at least for me) when the question asks for forces from a different reference point. But it makes much more sense now. Thanks again!