Weight of a 1.02m x 2.05m Door Supported by Two Hinges

[guille1]

A door of width 1.02 m and height 2.05 m weighs 289 N and is supported by two hinges, one 0.600 m from the top and the other 0.600 m from the bottom. Each hinge supports half the total weight of the door.

Assuming that the door's center of gravity is at its center, find the horizontal components of force exerted on the door by each hinge.

 

[berkeman]

A door of width 1.02 m and height 2.05 m weighs 289 N and is supported by two hinges, one 0.600 m from the top and the other 0.600 m from the bottom. Each hinge supports half the total weight of the door.

Assuming that the door's center of gravity is at its center, find the horizontal components of force exerted on the door by each hinge.

Draw the FBD and then what do you do?

 

[tomcue]

I would approach this problem by first calculating the total weight of the door, which is given as 289 N. Next, I would divide this weight by two to determine the weight supported by each hinge, which is 144.5 N.

To find the horizontal components of force exerted by each hinge, I would use the basic principles of physics, specifically the laws of equilibrium. Since the door is not moving, the sum of the horizontal forces acting on it must be equal to zero.

Using the concept of moments, I would determine the distance from the hinges to the center of gravity of the door, which is half of its width, or 0.51 m. This distance multiplied by the weight supported by each hinge (144.5 N) gives us the moment of force exerted by each hinge, which is 73.695 Nm.

Since the door is in equilibrium, the sum of the clockwise moments must be equal to the sum of the counterclockwise moments. Therefore, the horizontal component of force exerted by the hinge at the top can be calculated as 73.695 Nm divided by its distance from the center of gravity (0.600 m), which gives us 122.825 N. Similarly, the horizontal component of force exerted by the hinge at the bottom can be calculated as 73.695 Nm divided by its distance from the center of gravity (1.45 m), which gives us 50.83 N.

In conclusion, the door is supported by two hinges, each exerting a horizontal force of 122.825 N and 50.83 N respectively. This information can be useful in ensuring that the hinges are strong enough to support the weight of the door and in designing proper door hardware.

 

[nlightner]

I would approach this problem by first determining the total weight of the door and then dividing it by two to find the weight supported by each hinge. In this case, the total weight of the door is 289 N, so each hinge is supporting 144.5 N.

Next, I would use the principles of torque to calculate the horizontal components of force exerted by each hinge. Torque is defined as the product of force and the distance from the pivot point. In this case, the pivot point is the center of the door, so the distance from the pivot point to each hinge is 0.600 m.

Using the formula for torque, T = F x d, we can calculate the horizontal component of force exerted by each hinge as follows:

For the top hinge:
T = 144.5 N x 0.600 m = 86.7 Nm

For the bottom hinge:
T = 144.5 N x 0.600 m = 86.7 Nm

Therefore, each hinge is exerting a horizontal force of 86.7 N on the door. These forces are equal and opposite, resulting in a stable equilibrium for the door.

It is important to note that this calculation assumes that the door's center of gravity is at its center. If the center of gravity is not at the center, the horizontal forces exerted by each hinge may differ. Additionally, other factors such as the material and construction of the door may also affect the forces exerted by the hinges.