# 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 $\phi = \theta = {60}^{\circ}$ (not sure if this is right). I'm not too sure what causes the force N. By taking moments about the hinge, I got $F = 19.62 N$, but this apparently is not the correct answer. Thank you very much in advance!

Mar 15, 2018

You are almost got it !! See below.

$F = 9.81 \text{ N}$

#### Explanation:

The trap door is $4 \text{ kg}$ uniformally distributed. Its length is $l \text{ m}$.

So the center of mass is at $\frac{l}{2}$.

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 $\frac{l}{2}$ from the hinge.

So you have a moment relation like this:

${m}_{\text{perp}} \times g \times \frac{l}{2} = F \times l$

$2 \times 9.81 \times \frac{1}{2} = F$

or $\textcolor{g r e e n}{F = 9.81 \text{ N}}$