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?

Issaac physics

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

Answer:

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"}#