Question

A uniform rectangular trapdoor of mass `m=4.0kg` is hinged at one end. It is held open, making an angle `theta=60^@` to the horizontal, with a force magnitude `F` at the open end acting perpendicular to the trapdoor. Find the force on the trapdoor?

This diagram was given as a hint. I was working based on `phi=theta=60^@` (not sure if this is right). I'm not too sure what causes the force N. By taking moments about the hinge, I got `F=19.62N`, but this apparently is not the correct answer.

Thank you very much in advance!

Answer 1

You are almost got it !! See below.

`F= 9.81 " N"`

Explanation:

The trap door is `4 " kg"` uniformally distributed. Its length is `l " m"`.

So the center of mass is at `l/2`.

The inclination of the door is `60^o`, which means the component of the mass perpendicular to the door is:

`m_{"perp"} = 4 sin30^o = 4 xx 1/2 = 2 " kg"`
This acts at distance `l/2` from the hinge.

So you have a moment relation like this:

`m_{"perp"} xx g xx l/2 = F xx l`

`2 xx 9.81 xx 1/2 = F`

or `color(green){F= 9.81 " N"}`