Forces on a Door Homework: Solution Attempt

[Darkalyan]

Homework Statement

http://docs.google.com/Doc?id=d277r7r_56tbhcw4cf

The Attempt at a Solution

So, I don't know how to draw this out by hand, so I'll explain what I thought I should do. I thought I'd pretend the door was only at its center of mass, making it a point particle. Then, I could find the torque produced by door's mass and Earth's gravity on each of the doors, decompose the torque into its X and Y vectors, and go along my merry way. However, because there are 2 hinges, do I assume they both hold up 1/2 the door's weight? Plus, is the general idea even correct in the first place? Any help would be appreciated.

-Thanks

 

[tiny-tim]

Hi Darkalyan!

So, I don't know how to draw this out by hand, so I'll explain what I thought I should do. I thought I'd pretend the door was only at its center of mass, making it a point particle.

Yes, that's right … that preserves all the information.

Then, I could find the torque produced by door's mass and Earth's gravity on each of the doors, decompose the torque into its X and Y vectors, and go along my merry way.

No … the torque is in the z-direction … you can't decompose it!

However, because there are 2 hinges, do I assume they both hold up 1/2 the door's weight? Plus, is the general idea even correct in the first place?

Nooo … the correct idea is to notice that you have two unknown forces, so take moments about the point of application of one of them (a hinge), and you'll get an easy equation for the other.

 

[Darkalyan]

Hm, okay. When you say take 'moments' about the point for one of them, should I take the moment of inertia of the door about one of the hinges, using the CM as the door? Or, should I take the angular momentum of the door? But, the angular momentum is 0 because the door doesn't move, so I'm not sure which one to do. Finally, when I take moments, should I take an X direction moment of inertia and a Y direction moment of inertia separately? Basically I'm confused on what exactly the taking 'moments' means :(

 

[tiny-tim]

moments of forces

Hm, okay. When you say take 'moments' about the point for one of them, should I take the moment of inertia of the door about one of the hinges, using the CM as the door? Or, should I take the angular momentum of the door? But, the angular momentum is 0 because the door doesn't move, so I'm not sure which one to do. Finally, when I take moments, should I take an X direction moment of inertia and a Y direction moment of inertia separately? Basically I'm confused on what exactly the taking 'moments' means :(

Hi Darkalyan!

Angular momentum and moment of inertia are about momentum and moment of momentum.

I'm talking about moment of forces.

You can resolve forces in a particular direction, or you can take moments of forces about a particular point (in 3D, an axis).

In this case, take moments.

 

[Darkalyan]

Okay, I'm still a little confused about what exactly a moment of force is. So, for example, the top hinge. The moment of force in the y diriection would be MG*(W/2), correct? And, in the X direction, there is no moment of force because there is no force in the X direction. Do I have the right idea by what a moment of force is?

 

[tiny-tim]

… Do I have the right idea by what a moment of force is?

Nope … completely wrong!

Here's an extract from the PF Library entry on moments …

Every force F has a Moment about any point P.

To find the Moment, draw R, the point of application of the force, and L, the line of application of the force, and draw the perpendicular line PQ from P to L (so both Q and R lie on L).

For the Moment of a velocity, R is the position of the centre of mass, and L is the line of the velocity.

Then the Moment of F about P is the vector written "r x F" (pronounced "r cross F"), where r is the position vector PR. Its direction is perpendicular to both L and the line PR (and PQ). And its magnitude is PQ times F.

r is sometimes called the lever arm.

Note that if P is on the line L, then P = Q, so PQ = 0, so the Moment of the force is 0.

In nearly all exam problems, everything is in the same plane (the plane of the examination paper!), so all the Moment vectors are vertically out of the page.

In other words, they're all parallel to each other, so we can forget that they're vectors, and treat them simply as numbers, F times PQ.

We can take Moments about any point, so we always choose whatever point makes the calculations easiest.

Usually, it's the point of application of an unknown force, so that the Moment of that force is 0, making the equation shorter!

Now take moments (of all the forces) about one hinge, and then (since there is no angular momentum) put the total equal to zero!

 

[Darkalyan]

Ugh. I still don't get what exactly a moment of force is, so I'm going to try another way, okay? :(. Okay, so I got the horizontal forces is F * (L-2d) = Mg(W/2), and the force is the same for both the upper and lower hinge, just opposite in direction. For the upper hinge, the force is towards the hinge itself, and for the bottom hinge the force is directed away from the hinge.

However, for the vertical forces, I'm still not 100% sure. I think it's just that both hinges exert a force of Mg/2 upward, thus cnacelling out gravity's downward force. Does that work now? I'm really, really sorry I don't get the whole moment of force thing, I looked it up online and with all the perpendicularity and the cross product (? huh?) I was terribly confused, so I just resorted to forces. Does this make sense now?

 

[tiny-tim]

I'm really, really sorry I don't get the whole moment of force thing, I looked it up online and with all the perpendicularity and the cross product (? huh?) I was terribly confused, so I just resorted to forces. Does this make sense now?

Hi Darkalyan!

ok … the moment of the weight about one hinge is Mg times the horizontal distance, W/2 (horizontal, because the weight is vertical).

the moment of the vertical component at one hinge about the other hinge is zero, because the horizontal distance is zero.

the moment of the horizontal component at one hinge about the other hinge is that component times the vertical distance, L - 2d (vertical, because this component of force is horizontal).

Is that any better?

 

[Darkalyan]

Hm, I think I get it. So, I get the moment of the weight about a hinge is the weight times the horizontal distance. It's kind of like how torque is force times lever arm, but in this case the lever arm is just the horizontal distance to the weight. I understand the vertical component of one hinge about the other is 0 because there's no horizontal distance for the weight to act on, thus it's like there's no lever arm. When you say the moment of the horizontal component at one hinge about the other hinge is 'that component' times the vertical distance, what is that component?
The way I'm understanding the sentence, it sounds like moment=moment*(L-2d).
That doesn't really sound right, so I feel like I'm missing something.

Sorry, I'm really having some dificulty wrapping my head around this whole 'moment' thing, and the math behind it, and how it relates to this problem.

Finally, because the vertical component at one hinge is 0, it's 0 at both hinges. Thus, to find the horizontal moment of force, I just take MGW/2 = (force)*(L-2d) and solve for force? If I do that, what would the force mean? Physically, would it be the force inward on, let's say, the upper hinge, and correspondingly the bottom hinge produces an outward force to keep the door upright?

 

[tiny-tim]

Hi Darkalyan!

Hm, I think I get it. So, I get the moment of the weight about a hinge is the weight times the horizontal distance. It's kind of like how torque is force times lever arm, but in this case the lever arm is just the horizontal distance to the weight. I understand the vertical component of one hinge about the other is 0 because there's no horizontal distance for the weight to act on, thus it's like there's no lever arm.

That's right.

When you say the moment of the horizontal component at one hinge about the other hinge is 'that component' times the vertical distance, what is that component?
The way I'm understanding the sentence, it sounds like moment=moment*(L-2d).
That doesn't really sound right, so I feel like I'm missing something.

I think you're confusing moment and component.

The force, F say, at one hinge is at an angle θ …

F has a horizontal component H and a vertical component V …

we could say that the moment (torque) of F about the other hinge is f times the perpendicular distance from the other hing to the line of F … that would involve cosθ …

but it's easier to say F = H + V, and deal with H and V separately.

When I say "the moment of the horizontal component at one hinge about the other hinge is 'that component' times the vertical distance" I mean that the moment of the force H is H times (L - 2d).

Finally, because the vertical component at one hinge is 0, it's 0 at both hinges.

That's right.

Thus, to find the horizontal moment of force, I just take MGW/2 = (force)*(L-2d) and solve for force? If I do that, what would the force mean? Physically, would it be the force inward on, let's say, the upper hinge, and correspondingly the bottom hinge produces an outward force to keep the door upright?

Nooo … there's no such thing as "horizontal moment".

Moments are circular … they can be clockwise or anti-clockwise, but they can't be up or down or left or in any direction.

In this case, the moment of the weight about either hinge is clockwise, so the moment of either hinge about the other hinge must be anti-clockwise and equal.

That's all!

 

[Darkalyan]

Aaaah! Thank you so much, that last post, for some reason, really cleared up things for me. So, the force has 2 components, horizontal and vertical, but there's really only 1 moment. However, the moment of the vertical component is 0 (about the other hinge) because there's no lever arm. Thus, the moment of the total force is equal to the moment of the horizontal force. We know the moment of the total force is MGW/2, and the moment of the horizontal force is H(L-2d), and we equate them to each other to get that H is (MgW)/(2L-4d). This force is pointing to the right for the bottom hinge, and to the left for the top hinge, because that's the way in which the moment of the force turns out to be anti clockwise.

Okay, so finally, in terms of the original question. We've figured out the horizontal force is H, and in the bottom hinge it's to the right and the top hinge is to the left. The Vertical forces acting on the hinges would simply be gravity acting downward, and the normal force acting upward. Since there's 2 hinges, each hinge has a normal force of Mg/2 directed upwards?

 

[tiny-tim]

Okay, so finally, in terms of the original question. We've figured out the horizontal force is H, and in the bottom hinge it's to the right and the top hinge is to the left. The Vertical forces acting on the hinges would simply be gravity acting downward, and the normal force acting upward. Since there's 2 hinges, each hinge has a normal force of Mg/2 directed upwards?

Perfect!

(oh … except it's not a normal force … I know normal is usually vertical, but in this case the side of the door is vertical, so the normal force would be horizontal, wouldn't it? going to bed now … :zzz:)

 

[Darkalyan]

Okay, that was some really amazing help you've given me. I just wanted to thank you for being such ana wesome tutor. You were so totally insanely useful, it was amazing. I am so happy I've found this website, and I am SO becoming a physics major when in college.