# Door Torque Question - Solved Question

1. Oct 11, 2007

### Sabellic

Preemptive note: I got this question from a textbook that showed some SOLVED PROBLEMS. So, the torque problem is already solved. My misunderstanding is how they arrived at the problem and in such case, it may be more of a trigonometric problem (misunderstanding on my part).

1. The problem statement, all variables and given/known data

http://img402.imageshack.us/img402/4899/door1tb4.jpg [Broken]

As shown, hinges A and B hold a uniform, 400-N door in place. If the upper hinge happens to support the entire weight of the door, find the forces exerted on the door at both hinges. The width of the door is exactly h/2 where h is the distance between the hinges.

2. Relevant equations
As this is a torque question, the obvious equation would be t=rF sin x where x is the acute angle between the lines-of-action of r and F. Since the door is in equilibrium, we are measuring the downward torque on the hinges. (The door stays closed at all times; we are not measuring the torque at the hinges as the door opens or closes).

Remember, this question comes from a textbook that already SOLVED the problem and it was written as:

+(h)(F)(sin 90.0) - (h/4)(400N)(sin 90.0) = 0

3. The attempt at a solution

As I got this question from a textbook, and it was already solved, my problem lies more on HOW they got the conclusion.

Here was what THEY wrote:

The forces acting on the door are shown in above drawing. Only a horizontal force acts at B, because the upper hinge is assumed to support the doors weight. Let us take torques about point A as axis:

+(h)(F)(sin 90.0) - (h/4)(400N)(sin 90.0) = 0

from which F=100N. We also have:

F - Frh = 0 (horizontal torque)
Frv - 400N = 0 (vertical torque)

We find from these that Frh = 100N and Frv = 400N.

For the resultant reaction force Fr on the hinge at A, we have:

Fr= the square root of 400^2 + 100^2 = 412N

The tangent of the angle that Fr makes with the negative x-direction is Frv/Frh and so the angle is arctan 4.00=76 degrees.

I have a problem with how they arrived at the h/4 in the above equation. I will write it again here, and highlight the part in green:

+(h)(F)(sin 90.0) - (h/4)(400N)(sin 90.0) = 0

Because the value of r (the radial distance between the axis and point of application) can be derived from the location of hinge A to the point of application. We would then use the Pythagorean Theorem:

c^2 = the square root of a^2 + b^2

...so.......

c^2 = the square root of (h/4^2 + h/2^2)

which later works out to:

c^2 = the square root of (5h^2/16)

Ultimately, my problem is:

I dont know how we derive c=h/4 from c^2 = the square root of (5h^2/16)
So, I guess its more of a mathematical misunderstanding.

Last edited by a moderator: May 3, 2017
2. Oct 11, 2007

### ptr

The centre of mass of the door is at the point halfway between the two ends of the width of the door, and the width of the door is h/2 and half of that is $$\frac{h}{4}$$. There is a force applied in another orientation that also produces a moment about the same point.

3. Oct 11, 2007

### dvdqnoc

I didnt read your entire thread, but I have come across this question before. The best way to approach this problem is to get a calculator or book, and hold it with two fingers at where the top hinge would be holding a door.

You will notice the book is hanging at an angle. Now with your finger from your other hand, push the side of the book to make it stand upright. This finger is like your second hinge. Now, your fingers will tell you the type of force applied onto the door from the hinges.

Hope that helps.

4. Oct 11, 2007

### Sabellic

Hello. Thank you so much for your replies. But Im still a bit lost.

Let me supplement my initial statement.

The force acting on hinge B goes in a counter-clockwise direction, while the one at hinge A goes in a clockwise direction. Both of their countering forces act in unison to keep the door upright.

The question takes hinge A as the axis. Therefore, when we do the initial Torque equation we eliminate the forces acting through Hinge A. So, this is how I arrived:

rF1 (sin x) - rF2 (sin x) = 0 (since the system is in equilibrium)

If we take A as the axis:

+[(h)(sin 90)(F)] - [(c)(sin x)(400N)] = 0

http://img503.imageshack.us/img503/2853/doornew2ld5.jpg [Broken]

I can totally understand the significance of h/4 being the horizontal distance of the centre of gravity of the door from the hinges. HOWEVER, how can that be a value of r in rF sin x???

r is defined as the radial distance from the axis to the point of application of the force. That means the second r has to go from the chosen axis (hinge A) to the point of application of the force 400N (which is shown from the diagram to be at the centre of the door, where I labelled D). There must be something else I`m not getting.

Last edited by a moderator: May 3, 2017