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Sabellic

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__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).## Homework Statement

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.

## Homework 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

## 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 door`s 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) - (

__(400N)(sin 90.0) = 0__

*h*/4)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 don`t know how we derive c=h/4 from c^2 = the square root of (5h^2/16)

So, I guess it`s more of a mathematical misunderstanding.

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