Calculating Torques on a Door with Hinges

[darkduck]

Homework Statement

Consider a do or, hanging on two hinges (s ee figure). The do or is L tall and
d wide, and the hinges are placed at heights L/4 (point A) and 3 L/4 (point
B) from the floor. The door is uniform, and so its centre of mass is in the
middle of the door (point C)

b) What is the torque from each of the 3 forces around point A?

Homework Equations

[/B]

The Attempt at a Solution

I'm not qute sure what I'm going to do. Are the forces from the two hinges pointing upwards or 45 degrees north east?

When I'm going to find the torques around point A, should i just use T=fx ? The torque to point B will then be straight forward, but for find the torque to point C i have to decomposed the direction 45 degrees north east?

 

[haruspex]

Are the forces from the two hinges pointing upwards or 45 degrees north east?

They will have an upwards component, but until you've done the algebra you won't know what the horizontal components are or what direction the net force is in. What equations can you write?

When I'm going to find the torques around point A, should i just use T=fx ?

That depends on what T, f and x are ;). Seriously, nobody can answer that until you define them.

 

[darkduck]

Is the forces right on that drawing?When I'm going to calculate the torque around A, the torque of A will be 0 (because the arm is 0). Is it right that when I shall find the two others torques around i have to use the arm times the force?

 

[darkduck]

When I'm going to find the torque around A from B:

The length to B from A = 3L/4 - L/4 = L/2

T=L/2*Cosx*B

is this correct?

 

[nasu]

What is B here? The force at B? What is x?

 

[darkduck]

What is B here? The force at B? What is x?

B is the force at B. X is the angle, because it's unknown...

 

[nasu]

Cos x may be or not correct, depending on which angle you call x in your drawing. The general formula for torque is a cross product, so the magnitude is written in terms of the sin of the angle between the two vectors. But if you use the complement of the angle, it may be cos.
You also can write the torque by using the components of B. It will be simpler. Anyway you will need to write the balance of forces and teh balance of torques by using components.

 

[haruspex]

B is the force at B. X is the angle, because it's unknown...

Further to nasu's reply, you should not assume the angle of the net force is the same at both hinges. Just put in four unknowns for the horizontal and vertical components of force at each hinge.

 

[vebjoern]

but shouldn't the forces from hing A and B have opposite directions in the x-axis to get static equilibrium?

 

[nasu]

Yes, they should.

 

[haruspex]

but shouldn't the forces from hing A and B have opposite directions in the x-axis to get static equilibrium?

Yes, the horizontal components must be equal and opposite, but you don't yet know whether the vertical components are equal. It is likely from the symmetry of the hinge positions that they will be, but it needs to emerge from the equations.

 

[vebjoern]

is it wrong to write that the Torque from hing A to point C = -d/2*Mg ? where d/2 would be the lever-arm and (-) because of the clockwise rotation ? ?

 

[PhanthomJay]

is it wrong to write that the Torque from hing A to point C = -d/2*Mg ? where d/2 would be the lever-arm and (-) because of the clockwise rotation ? ?

Watch terminology. The torque about A from the vertical force at C is not the same as the torque about C from the vertical and horizontal forces at A. In fact, it is not possible to determine the vertical force at A from the equilibrium equations. Let's not lose sight of what the original problem is asking: what is the torque about A from each of the three forces C_y, B_x, and B_y? (The sum of which add to 0).

 

[vebjoern]

the Torque around A from the force B_y should then be equal to 0,because of that the vectorproduck = 0, and therefor the Torque around A from the forces could be that
r1*FBx*sinθ = r2*FGysinΦ
where r1 should be ½*L and r2 should be √((¼*l)²+(½*d)²) ?

 

[PhanthomJay]

the Torque around A from the force B_y should then be equal to 0,because of that the vectorproduck = 0,

yes

and therefor the Torque around A from the forces could be that
r1*FBx*sinθ = r2*FGysinΦ

well OK using cross product rule

where r1 should be ½*L

yes, and where θ is 90 degrees and thus sin θ is 1

and r2 should be √((¼*l)²+(½*d)²) ?

well, yes, but where sin Φ = 1/2 d/r2, so r2 cancels out , no need to calculate it.

When breaking up forces into x and y components, instead of using the cross product rule, note that the moment of a force or force component about a point is the product of that force or force component times the perpendicular distance from its line of action to the point. Makes things a lot simpler. The moment of the gravity force at C about A has already been correctly calculated somewhere in the above posts.

 

[vebjoern]

Thanks for a good answer! :)

 

[iamnotsmart]

yes well OK using cross product ruleyes, and where θ is 90 degrees and thus sin θ is 1 well, yes, but where sin Φ = 1/2 d/r2, so r2 cancels out , no need to calculate it.

When breaking up forces into x and y components, instead of using the cross product rule, note that the moment of a force or force component about a point is the product of that force or force component times the perpendicular distance from its line of action to the point. Makes things a lot simpler. The moment of the gravity force at C about A has already been correctly calculated somewhere in the above posts.

This may be a couple of years back, but I somehow have the same problem in my hand right now. I sat it up just like all of you, the torque around A needs to be equal 0, where the ones that give torque is the x-component of the force from hinge A and the perpendicular distance of r-vector from A to C (aka d/2) times Fg. However in the next task, we are asked to do the same around B, and I literally get the same thing again. The next task is setting up two more equations when you the door doesn't fly sideways or upwards and so on, which is just that x-component of hinge force A and B equals 0. Both of the vertical components from A and B needs to cancel out gravity force. The last task, we were asked to solve for the complete hinge force vectors when we know that the vertical components of A and B are the same, aka 2Hinge(Y)=Fg. I have no idea how to find this as the equation of F_a_x-F_b_x=0 is useless when we already know when calculating torques around A and B that x-component from A and B is the same. What do I even do next? Sorry for this long post, but I have been sitting here, trying everything and frustrated :/

 

[haruspex]

we were asked to solve for the complete hinge force vectors when we know that the vertical components of A and B are the same, aka 2Hinge(Y)=Fg. I have no idea how to find this

I am not quite clear what you are told and and what you are struggling with.
You are told to assume the two vertical reaction components are equal, yes? So you can calculate those vertical components.
If you mean the x components, you already have those from the moments balance, no?

x-component from A and B is the same.

If they were exactly the same the door would accelerate sideways.

 

[iamnotsmart]

I am not quite clear what you are told and and what you are struggling with.
You are told to assume the two vertical reaction components are equal, yes? So you can calculate those vertical components.
If you mean the x components, you already have those from the moments balance, no?

If they were exactly the same the door would accelerate sideways.

b) What is the torque from each of the 3 forces around point A? c) What is the torque from each of the 3 forces around point B?

d) For the door not to rotate, the sum of all the torques must be zero, around any point. Write down the corresponding equations.

e) For the door to neither fly sideways, upwards or downwards, what other two equations can you write down for the forces?

f) We will assume that the upwards components of the two hinge forces, FyA and FyB, are the same. Solve for the complete hinge force vectors.

 

[iamnotsmart]

I am not quite clear what you are told and and what you are struggling with.
You are told to assume the two vertical reaction components are equal, yes? So you can calculate those vertical components.
If you mean the x components, you already have those from the moments balance, no?

If they were exactly the same the door would accelerate sideways.

I mean the absolute value of x-component of A and B is the same, but they act in opposite direction so they cancel out each other? And that's why the door doesn't move sideways? Anyways that's what I found out in b), c) and d), that they are the same, but in opposite direction. However in e), we were asked to write the equation where x of A and X of b is equal to 0 when we already knew it in the previous questions? And how do I use all of these equations to solve for f)?

 

[haruspex]

b) What is the torque from each of the 3 forces around point A? c) What is the torque from each of the 3 forces around point B?

d) For the door not to rotate, the sum of all the torques must be zero, around any point. Write down the corresponding equations.

e) For the door to neither fly sideways, upwards or downwards, what other two equations can you write down for the forces?

f) We will assume that the upwards components of the two hinge forces, FyA and FyB, are the same. Solve for the complete hinge force vectors.

Ok, but this leaves me baffled as to where you are stuck. What is "this" in "I have no idea how to find this"?

 

[iamnotsmart]

Ok, but this leaves me baffled as to where you are stuck. What is "this" in "I have no idea how to find this"?

Question f), I don't really understand what we have to find or how to do it... Are we going to find an expression of the hinge force or what? I don't really know lol

 

[haruspex]

the absolute value of x-component of A and B is the same, but they act in opposite direction so they cancel out each other?

Right.

in e), we were asked to write the equation where x of A and X of b is equal to 0 when we already knew it in the previous questions

Yes. In 2D statics there are only three independent equations available. Mostly people write two linear force equations, usually for two directions at right angles, one one moment equation. But one linear and two moments also works. You have been asked to write four, so one of them will be redundant.

how do I use all of these equations to solve for f)?

This is where I am puzzled that you are stuck. You know the horizontal components and you have two equations for the vertical components.

 

[haruspex]

Question f), I don't really understand what we have to find or how to do it... Are we going to find an expression of the hinge force or what? I don't really know lol

All they are asking for is the horizontal and vertical components of each hinge force. If you have been taught a vector notation, like ##x\hat i+y\hat j##, then use that.

 

[iamnotsmart]

Right.

Yes. In 2D statics there are only three independent equations available. Mostly people write two linear force equations, usually for two directions at right angles, one one moment equation. But one linear and two moments also works. You have been asked to write four, so one of them will be redundant.

This is where I am puzzled that you are stuck. You know the horizontal components and you have two equations for the vertical components.

I don't know what the task is asking for. As you can see, English is not my primary language, pretty hard to transfer everything I have learned earlier in another language to English right now.

And how do you manage to get two equations for vertical? b) and c) gives an equation in d) where the sum of torque around point A and B equal to 0. Only the x-components matter here as far as I am concerned, cause they are the one perpendicular to radius vector. e) gives another equation for the x-components and one for the vertical components canceling out gravity force. And I don't know anything about hat or something like that lol.

 

[haruspex]

how do you manage to get two equations for vertical

One from the force balance, sum=mg, and the one they gave you, that the two are equal. You already combined these to obtain the answer: 2 * F_y=mg.
So you already have all the answers. You know the vertical and horizontal components at each hinge. All that remains is deciding the form in which to write answer.
If you have not been taught a vector notation then that leaves two possibilities. Either just list components (F_Ax=... etc.) or calculate the magnitude and direction of each.

 

[iamnotsmart]

One from the force balance, sum=mg, and the one they gave you, that the two are equal. You already combined these to obtain the answer: 2 * F_y=mg.
So you already have all the answers. You know the vertical and horizontal components at each hinge. All that remains is deciding the form in which to write answer.
If you have not been taught a vector notation then that leaves two possibilities. Either just list components (F_Ax=... etc.) or calculate the magnitude and direction of each.

The problem is that I am not allowed to just write it as F_y=mg/2 because m is not listed in the task, therefore Fg is still unknown... And I have no idea how to find Fg as using the other equations will make Fg cancel out anyways (I might have choked though, but I tried several times and still can't find out how to find Fg expressed by the terms given in the task.)

 

[haruspex]

m is not listed in the task

So how were you able to answer the earlier parts? You need the weight of the door to find the torques.

 

[iamnotsmart]

So how were you able to answer the earlier parts? You need the weight of the door to find the torques.

I found an expression of the torque from Fg around point A and B, however I just defined it as Fg, thinking I'll manage to find an expression for Fg later in the task. For example I found out that torque around point A is: (L/2)*F_B_x - (d/2)*Fg = 0 where L/2 is the distance from A to B and F_B_x is the x-component of B that is perpendicular to the distance from A to B. (d/2)*Fg is the torque from gravity around point A as far as I am concerned, I just take the radius vector component that is perpendicular to the gravity force (I was told by the teacher that F*r*sin(theta) is the same as taking either the force component that is perpendicular to r times r, or the component of r that is perpendicular to F times F.) As you see, I only express it as Fg, thinking I would be able to find an expression of Fg later to solve for the vertical component of the hinge forces.

 

[haruspex]

I found an expression of the torque from Fg around point A and B, however I just defined it as Fg, thinking I'll manage to find an expression for Fg later in the task. For example I found out that torque around point A is: (L/2)*F_B_x - (d/2)*Fg = 0 where L/2 is the distance from A to B and F_B_x is the x-component of B that is perpendicular to the distance from A to B. (d/2)*Fg is the torque from gravity around point A as far as I am concerned, I just take the radius vector component that is perpendicular to the gravity force (I was told by the teacher that F*r*sin(theta) is the same as taking either the force component that is perpendicular to r times r, or the component of r that is perpendicular to F times F.) As you see, I only express it as Fg, thinking I would be able to find an expression of Fg later to solve for the vertical component of the hinge forces.

Ok, but without a given mass or weight for the door there is no hope of finding any torques or forces. Are you quite sure there is nothing given? Any diagram?

 

[iamnotsmart]

Ok, but without a given mass or weight for the door there is no hope of finding any torques or forces. Are you quite sure there is nothing given? Any diagram?

So the professor said he forgot to say that we can express our Fg with the mass of door m, therefore Fg isn't unknown anymore and I can express everything with the terms m,g,d and l now, thanks for your time :)