Equilibrium Question: Determining Forces on a Hinged Door | Homework Help

[VanKwisH]

Homework Statement

A door 2.3m high and 1.3m wide, has a mass of 13.0kg. A hinge 0.4m from the top and another hinge 0.4m from the bottom each support has the doors weight. Assume that the center of gravity at the geometrical center of the door, and determine the horizontal and vertical force components exerted by each hinge on the door.

Homework Equations

The Attempt at a Solution

I made a picture of what it should look like ...
[img=http://img82.imageshack.us/img82/5474/91305910xw5.th.png]
I have Fg of the door which is
Fg = 13 * 9.8
= 127.4N
and I know that the forces of the hinges are equal but what do i do in order to solve it

 

[anathema]

?Hello,

To solve this problem, you can use the principle of moments. This principle states that for a body to be in equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point.

In this case, we can choose the point of rotation to be at the bottom hinge. This means that the anticlockwise moments will be positive and the clockwise moments will be negative.

First, let's calculate the total moment due to the weight of the door. Since the center of gravity is at the geometrical center, the weight of the door can be represented by a single force acting at this point. This force will have a magnitude of 127.4N (as you correctly calculated) and its line of action will be vertical. Therefore, the moment due to this force will be:

M = Fg * d = 127.4 * 0.4 = 50.96 Nm

Next, we need to calculate the moments due to the forces exerted by the hinges. Since the forces are acting at a distance of 0.4m from the point of rotation, the moments due to these forces will be:

M1 = F1 * d = F1 * 0.4

M2 = F2 * d = F2 * 0.4

Since we know that the forces exerted by the hinges are equal, we can write:

F1 = F2 = F

Therefore, the total moment due to the forces exerted by the hinges will be:

M = M1 + M2 = F * 0.4 + F * 0.4 = 2F * 0.4

Now, using the principle of moments, we can write:

M = M1 + M2 = M

50.96 = 2F * 0.4

F = 50.96 / 2 * 0.4 = 63.7N

Finally, we can calculate the vertical and horizontal components of the forces exerted by the hinges using trigonometry.

The vertical component will be equal to the weight of the door, which is 127.4N.

The horizontal component will be equal to:

Sinθ = 0.4 / 2.3

θ = Sin^-1(0.4 / 2.3) =

 

[Moetasim]

?

I would approach this problem by first identifying all the known variables and then using basic principles of physics to determine the forces on the hinged door. From the given information, we know the following:

- The door has a mass of 13.0kg and a height of 2.3m and a width of 1.3m.
- There are two hinges, each located 0.4m from the top and bottom of the door.
- The center of gravity of the door is at its geometrical center.

Using this information, we can first calculate the weight of the door, which is equal to its mass multiplied by the acceleration due to gravity (9.8m/s^2). This gives us a weight of 127.4N.

Next, we can use the principle of moments to determine the forces on the hinges. The principle of moments states that for an object to be in equilibrium, the sum of the clockwise moments must be equal to the sum of the counterclockwise moments.

In this case, the door is in equilibrium, so the sum of the forces acting on it must be equal to zero. This means that the counterclockwise moment (caused by the weight of the door) must be equal to the clockwise moment (caused by the forces on the hinges).

To solve for the forces on the hinges, we can set up the following equations:

Clockwise moment = Force x Distance
Counterclockwise moment = Weight x Distance

Since the door is in equilibrium, these two moments must be equal, so we can set the equations equal to each other:

Force x Distance = Weight x Distance

Solving for the force of each hinge, we get:

Force = Weight / Distance

Plugging in the values, we get:

Force = 127.4N / 0.4m
= 318.5N

Therefore, the horizontal and vertical forces exerted by each hinge on the door are both 318.5N. This means that each hinge is supporting half of the weight of the door (127.4N) and also providing an equal amount of force to keep the door in equilibrium.

In conclusion, by using basic principles of physics, we can determine the forces on a hinged door and understand how the hinges are supporting the weight of the door.