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?

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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!

1 Answer
Mar 15, 2018

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"}