Static Equilibrium Door Problem

[elianaphys]

Homework Statement

A door made of a uniform piece of wood measures 1 m by 2 m and has a mass of 18kg. The door is entirely supported by two hinges, one at the bottom corner and one at the top corner. Find the force (magnitude and direction) that the door exerts on each hinge. Assume that the vertical force on each hinge is the same.

Homework Equations

The Attempt at a Solution

.5m*18kg*g-2m*F2=0
9mkg*g-2m*F2=0
F2=9mkg*g/2m
F2=44.145N

This is the incorrect answer. The correct answer is 99N at 27 degrees right of vertical at the top and at 27 degrees left of vertical at the bottom. I think I am having so much trouble getting to this correct answer because I keep finding the force that the hinges exert on the door rather than the force that the DOOR exerts on the HINGES. I'm very stuck.

FYI, I'm taking this through an online course and have thus have no instructor--just me and a textbook! Any help is sincerely appreciated!

 

[Xisune]

You have to take into account torque as well, this is why they gave you the dimensions of the door.

 

[Chestermiller]

You have to take into account torque as well, this is why they gave you the dimensions of the door.

He did take into account the torque, as is apparent from his equations. However, he only included the horizontal component of force, and not the vertical component. He needs to include the vertical component in order to calculate the overall magnitude of the force.

 

[azizlwl]

Homework Statement

A door made of a uniform piece of wood measures 1 m by 2 m and has a mass of 18kg. The door is entirely supported by two hinges, one at the bottom corner and one at the top corner. Find the force (magnitude and direction) that the door exerts on each hinge. Assume that the vertical force on each hinge is the same.

Homework Equations

The Attempt at a Solution

.5m*18kg*g-2m*F2=0
9mkg*g-2m*F2=0
F2=9mkg*g/2m
F2=44.145N

This is the incorrect answer. The correct answer is 99N at 27 degrees right of vertical at the top and at 27 degrees left of vertical at the bottom. I think I am having so much trouble getting to this correct answer because I keep finding the force that the hinges exert on the door rather than the force that the DOOR exerts on the HINGES. I'm very stuck.

FYI, I'm taking this through an online course and have thus have no instructor--just me and a textbook! Any help is sincerely appreciated!

You got F2 which is pulling the door the right/left depending on the layout.
There's another force exerted on the hinge by the weight of the door.

There are 3 ways of distributing the weight of the door on the hinges.
1. Hanging from the top, no force exerted on lower hinge
2. Pivotal at the botton hinge, no force exerted on the upper hinge.
3. Equallly shared.

 

[Nickie]

I would first like to clarify the concept of static equilibrium. In this problem, static equilibrium refers to the state in which the door is not moving or rotating, and all the forces acting on it are balanced. This means that the sum of all the forces in the horizontal and vertical directions must be equal to zero.

Now, let's address the issue you mentioned about finding the force that the door exerts on the hinges rather than the other way around. This is a common mistake and can be easily resolved by applying Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. This means that the force exerted by the door on the hinge is equal in magnitude but opposite in direction to the force exerted by the hinge on the door.

To solve this problem, we can use the principle of moments, which states that for an object to be in static equilibrium, the sum of the moments (torques) acting on it must be equal to zero. In this case, we can take moments about the bottom hinge, as it is the point where the door rotates.

Using the equation for moments, we can write:

F1 x 2m - (18kg x 9.8m/s^2 x 1m/2) = 0

Where F1 is the force exerted by the top hinge on the door. Solving for F1, we get:

F1 = 88.2N

Since the vertical forces on each hinge are equal, the magnitude of the force exerted by the bottom hinge on the door is also 88.2N. However, the direction of this force is opposite to that of the top hinge, as mentioned earlier. Therefore, the force exerted by the bottom hinge on the door is 88.2N at 27 degrees left of vertical.

In conclusion, the force that the door exerts on each hinge is 88.2N at 27 degrees right of vertical at the top hinge and 88.2N at 27 degrees left of vertical at the bottom hinge. It is important to note that the direction of the force is always relative to the hinge, not the door. I hope this explanation helps you understand the problem better. Good luck with your studies!