Static Equilibrium Door question: Calculate Horizontal Forces

[jamesbiomed]

Homework Statement

A door, essentially a uniform rectangle of height 2.00 m, width 0.87 m, and weight 130.0 N, is supported at one edge by two hinges, one 35.1 cm above the bottom of the door and one 164.9 cm above the bottom of the door. Calculate the horizontal components of the forces on the two hinges.

Lower Hinge: Magnitude, direction
Upper Hinge: Magnitude, direction

Homework Equations

COM for uniform mass should be right in the middle (coordinates .435, 1)
Through pythagorean theorem, D distance from COM to Lower hinge: (1-.351)=.649 (.649^2+.435^2)^1/2=.7813 m

Same for D to upper hinge.

M=W/g=130/9.81=13.25 N

I com= 1/2mh^2
Moment of inertia for door using parallel axis theorem:
1/2 mh^2+ md^2

In static equilibrium, all forces must add up to zero.

The Attempt at a Solution

First I tried calculating the Moment for each hinge, and that didn't work.

Conceptually, I know that the the Weight should be counterbalanced in some way by the moment of inertia on the door by the hinges. (I don't think the normal force on the hinges factors in here)

You can't get a horizontal force from the weight of the door, and I'm having trouble deriving one from the Moment.

Sorry there isn't more work here---a hint or a push in the right direction would be greatly appreciated. Thanks.

 

[pgardn]

Homework Statement

A door, essentially a uniform rectangle of height 2.00 m, width 0.87 m, and weight 130.0 N, is supported at one edge by two hinges, one 35.1 cm above the bottom of the door and one 164.9 cm above the bottom of the door. Calculate the horizontal components of the forces on the two hinges.

Lower Hinge: Magnitude, direction
Upper Hinge: Magnitude, direction

Homework Equations

COM for uniform mass should be right in the middle (coordinates .435, 1)
Through pythagorean theorem, D distance from COM to Lower hinge: (1-.351)=.649 (.649^2+.435^2)^1/2=.7813 m

Same for D to upper hinge.

M=W/g=130/9.81=13.25 N

I com= 1/2mh^2
Moment of inertia for door using parallel axis theorem:
1/2 mh^2+ md^2

In static equilibrium, all forces must add up to zero.

The Attempt at a Solution

First I tried calculating the Moment for each hinge, and that didn't work.

Conceptually, I know that the the Weight should be counterbalanced in some way by the moment of inertia on the door by the hinges. (I don't think the normal force on the hinges factors in here)

You can't get a horizontal force from the weight of the door, and I'm having trouble deriving one from the Moment.

Sorry there isn't more work here---a hint or a push in the right direction would be greatly appreciated. Thanks.

Im thinking if you look at one hinge, its going to have a force that can be broken into two components horizontal and vertical. I am thinking of using the lower hing as the pivot, as if the upper hinge was absent. If the ground was not present the whole door would rotate if the lower hinge worked like a pivot point. Or I guess you could use the upper hinge and pretend like there was no door jamb or wall and the door would rotate.

So now you use the upper hinge to balance the lower hinge, or vice versa

try this approach...

 

[jamesbiomed]

I think I see what you mean, and it makes sense. I'll try that and post again with the results.

 

[jamesbiomed]

So I tried the method:

If the pivot pt is at .351, I plugged in .7813mg sin 56.2

I got r= .7813 from the distance between center of mass and hinge. I got angle 56.2 from arctan y/x (in between hinge and COM)

Doing a "plug and chug" with these didn't work.

Then I tried cutting the mg in half, since half will be supported by the other pivot point.

This didn't work either. Could you give me a hint on finding the x and y force? I think that is where I'm going wrong

 

[pgardn]

I have not put pencil to paper yet so just thinking.

So if we have a door with the hinges on the left hand side of the door...

Seems like the top hinge would get pulled to the right by the door trying to rotate clockwise, so the hinge would have to apply a force to the left. But since the door does not rotate counter clockwise, the bottom hinge would apply a force to the right and it must be equal to the pull left (forces in the x-direction).

Now for the y direction... I am having trouble seeing the top hinge bear as much weight as the bottom hinge. Oh shucks, they don't want this in the problem as I read it?

 

[jamesbiomed]

yes, but I didn't factor the I in on my second attempt. I was using the rXF definition of torque, not using I*α since α would be zero in this case...

Or is that not correct? Because if we could find angular acceleration we could set

I*α = r X F to get F

I'm looking through my book and I'm not seeing any Static equilibrium problems that use angular accel.

I do agree with what you're saying. The top hinge will have a force in the positive x direction (assuming usual coordinates) and the bottom hinge will have a force in the negative x direction.

How would I factor in ?
Btw, my professor used a similar example in class and used I=1/2 mh^2 for the door then mr^2 for the p.a.th

 

[pgardn]

yes, but I didn't factor the I in on my second attempt. I was using the rXF definition of torque, not using I*α since α would be zero in this case...

Or is that not correct? Because if we could find angular acceleration we could set

I*α = r X F to get F

I'm looking through my book and I'm not seeing any Static equilibrium problems that use angular accel.

I do agree with what you're saying. The top hinge will have a force in the positive x direction (assuming usual coordinates) and the bottom hinge will have a force in the negative x direction.

How would I factor in ?
Btw, my professor used a similar example in class and used I=1/2 mh^2 for the door then mr^2 for the p.a.th

Well one classic problem is a rod perfectly horizontal attached to a wall by a hinge. They ask what is the force on the hinge at the exact moment the rod (at rest) is perfectly horizontal. The rod will obviously start to rotate and then things change. But, Using the The Iα = Fr and then finding α in terms of g and L (length of rod) you can equate this to tangential acceleration of the center of mass and then use this to find the force on the hinge. It turns out the force on the hinge (pivot) is 1/4 Mg of the rod. In this problem we are only concerned with the y-direction. I am trying to apply this to your situation.

I am now going to have to put pencil to paper.

 

[pgardn]

Ok. Organize, I tell myself.

Net torque = zero ;Forces in the horizontal must also balance no net force in the x- direction.

So let's work on the net torque. And use one hinge as the point of rotation. If you don't want to use angles the Fxr = Iα but r is just the distance from a hinge to the center of the door horizontally (or half the width of the door), don't need the angle or pythag and all this.

Now there is another force that works counter clockwise to the one mentioned above so that there is no net torque. That is the force... Let's use the bottom hinge as our point of rotation, that is the force that pulls to the left (key point). These two torques balance out. So you got the torque the top hinge supplies F x (height of the Door).

Is for you to do the numbers now. But I would be interested to see how much weight each hinge bears... I am lazy and don't like punching in numbers, just looking at ideas.

 

[jamesbiomed]

Thank you, that makes sense. Again, I'm going to give it a try.

As for the "thin rod through the axis of rotation" I think that's beyond the scope of the problem.

Thanks again, I'll check, and repost with thanks/questions

 

[pgardn]

Thank you, that makes sense. Again, I'm going to give it a try.

As for the "thin rod through the axis of rotation" I think that's beyond the scope of the problem.

Thanks again, I'll check, and repost with thanks/questions

It is indeed. Just thinking. I need to be careful and organize.

 

[jamesbiomed]

Btw, you helped me yesterday with a different problem, thanks :)


Ok, I'm running into a little problem. I'm using "practice mode which has different figures and shows you answers when you are wrong.

Maybe bc of all the questions I've done already, my brain is having trouble with this:

Are you saying:

"F"= the weight, 130 N? Or equals the I moment of inertia?

If we set FxR=Iσ without the rod part, do we say α=9.81 (height)?

Thanks

 

[pgardn]

Btw, you helped me yesterday with a different problem, thanks :)


Ok, I'm running into a little problem. I'm using "practice mode which has different figures and shows you answers when you are wrong.

Maybe bc of all the questions I've done already, my brain is having trouble with this:

Are you saying:

"F"= the weight, 130 N? Or equals the I moment of inertia?

If we set FxR=Iσ without the rod part, do we say α=9.81 (height)?

Thanks

net torque = 0

So -F(c) x 1/2width + F(cc) x h = 0 The F for the width arm is mg. c is clockwise cc is counter

 

[jamesbiomed]

But what is F(c)? and F(cc)? They're the same, but you can't solve for them. Using I in their place didn't work.

 

[pgardn]

But what is F(c)? and F(cc)? They're the same, but you can't solve for them. Using I in their place didn't work.

mg x 1/2 width = F x h if net torque = 0

The clockwise force is mg, we know that. What we don't know is the horizontal force, F, from the pivot above the pivot on the bottom (which I chose as the point of rotation).

 

[jamesbiomed]

Ok, I see what you're saying. Sorry, this is all tough for me

 

[pgardn]

I made it tougher with my first suggestions.

Caffeine was not available this morning...

 

[jamesbiomed]

So we get mg x .5 width= Fxh => F = (mg x .5w)/height

This doesn't give me the right answer, but its in the ball park. I tried multiplying this by the moment of inertia, and that didn't work.

Maybe the thin rod was meant to be included after all...

 

[jamesbiomed]

And no problem lol, I'm running on loads and its not helping much

 

[jamesbiomed]

It seems in static equil problems we almost always use the sin/cos to give us the vectors for force...

 

[pgardn]

So we get mg x .5 width= Fxh => F = (mg x .5w)/height

This doesn't give me the right answer, but its in the ball park. I tried multiplying this by the moment of inertia, and that didn't work.

Maybe the thin rod was meant to be included after all...

what did you get for F?

 

[pgardn]

oh for gosh sakes...

Look at the original problems... the hinges are not at the corners of the door.

I was assuming the hinges were at the corners.

You think you can do it with if you go back and look where the hinges really are. It changes the number for the length of both moment arms.

No caffeine and an adopted dog barking incessantly at me does not make for good concentration.

 

[ehild]

A drawing makes things much easier. See attachment.
Calculate the torques with respect to the bottom hinge. Torque is force times arm of force, the distance between the line of force from the axis of rotation. For the weight, it is half the width of the door. For the horizontal force the upper hinge exerts, it is the distance between the hinges.

ehild

 

[jamesbiomed]

No problem, I was aware of that, and not saying anything :)

So for arms running perpendicular to moment of I, we use y's right? So
it would be 1--.351 for the lower hinge.

So 130*(.5*.649)=Fx (2)

That gives me Fx=21.1 N which is incorrect. I hope I'm missing something you're saying here. Is moment of inertia involved?

 

[jamesbiomed]

Ok, didn't see that one moment

 

[jamesbiomed]

please ignore earliest reply after you posted, I didn't see picture

 

[pgardn]

A drawing makes things much easier. See attachment.
Calculate the torques with respect to the bottom hinge. Torque is force times arm of force, the distance between the line of force from the axis of rotation. For the weight, it is half the width of the door. For the horizontal force the upper hinge exerts, it is the distance between the hinges.

ehild

yeah we got that now... just a matter of putting the right numbers in. And that's for James.

thanks...

 

[jamesbiomed]

yeah we got that now... just a matter of putting the right numbers in. And that's for James.

thanks...

Thanks very much ehild

pgardn, I think this is what you were getting at, but my skill with this stuff is lacking. Thanks for the help both days.

That gave me the correct answer for the practice. Thanks to both of you, I'm going to study the method and see how so I know it.

Thanks again!

James

 

[pgardn]

Thanks very much ehild

pgardn, I think this is what you were getting at, but my skill with this stuff is lacking. Thanks for the help both days.

That gave me the correct answer for the practice. Thanks to both of you, I'm going to study the method and see how so I know it.

Thanks again!

James

You are a good student who actually does the work. Its no problem helping people like this. I took a look at the problem that I helped you with before and it was actually more difficult.

My bad... props to you for working hard.

 

[ehild]

You are welcome. Next time make a drawing before you start to plug-in data :tongue:

ehild

 

[jamesbiomed]

yeah we got that now... just a matter of putting the right numbers in. And that's for James.

thanks...

You are welcome. Next time make a drawing before you start to plug-in data :tongue:

ehild

Will do

 

[jamesbiomed]

You are a good student who actually does the work. Its no problem helping people like this. I took a look at the problem that I helped you with before and it was actually more difficult.

My bad... props to you for working hard.

Thanks, I'm taking it over the summer so its taking me longer to absorb some things--that's why I come here!

 

[pongo38]

It's simple. Take moments about bottom hinge, using distances perpendicular to the forces. Moment of inertia not required for this problem in simple statics.