Forces on Hinges HW Statement: F&Tau Calculation

[IMGOOD]

Homework Statement

A door 2.30 m high and 1.30m wide, has a mass of 13.0 kg. A hinge 0.40 m from the top and another hinge 0.40 m from the bottom each support half the door's weight. Assume that the center of gravity is at the geometrical center of the door, and determine the horizontal and vertical force components exerted by each hinge on the door.

Homework Equations

$$F_{net} = 0$$
$$\tau_{net} =0$$
$$\tau = Fd\sin(\theta)$$

The Attempt at a Solution

I stared at the problem for a long while but I don't even know how I should get started on this. Any hints would be appreciated.

 

[PhanthomJay]

Homework Statement

A door 2.30 m high and 1.30m wide, has a mass of 13.0 kg. A hinge 0.40 m from the top and another hinge 0.40 m from the bottom each support half the door's weight. Assume that the center of gravity is at the geometrical center of the door, and determine the horizontal and vertical force components exerted by each hinge on the door.

Homework Equations

$$F_{net} = 0$$
$$\tau_{net} =0$$
$$\tau = Fd\sin(\theta)$$

The Attempt at a Solution

I stared at the problem for a long while but I don't even know how I should get started on this. Any hints would be appreciated.

Locate the center of gravity of the door, calculate its weight, and you may apply all of its weight vertically through that point. Now sum torques about the lower hinge, and see what you get, noting that there is a vertical and horizonatl force at each hinge acting on the door.

 

[IMGOOD]

So, is this correct or not?

If you imagine a door that is located to the left of the hinges then
1) There are two forces on the lower hinge, one pointing directly to the left and the other pointing directly upwards.
2) There are also two forces on the upper hinge, one is pointing directly to the right, and the other pointing directly upwards.

 

[AlephZero]

Yes those are the 4 forces on the hinges.

BTW it doesn't matter if you guess wrong (e.g. left or right in this question) which direction a force acts in. If the direction is wrong the magnitude will just come out as a negative number.