# Moments question A level

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#1
https://isaacphysics.org/questions/t...2-5c61bba54eb7

I am struggling to understand this moments question. I get that you have to work out the moments at F and Weight(mg) and that they are equal to each other.

for mg i got the equation: moment=mg(xcos60) where x is half the length of the trapdoor.

for the force I worked out the moment as F(2xsin60), when I use these equations I get the wrong answer so I'm not sure what I am doing wrong but I think it is the equation for F but I'm not sure.
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1 month ago
#2
(Original post by Emily~3695)
https://isaacphysics.org/questions/t...2-5c61bba54eb7

I am struggling to understand this moments question. I get that you have to work out the moments at F and Weight(mg) and that they are equal to each other.

for mg i got the equation: moment=mg(xcos60) where x is half the length of the trapdoor.

for the force I worked out the moment as F(2xsin60), when I use these equations I get the wrong answer so I'm not sure what I am doing wrong but I think it is the equation for F but I'm not sure.
The question says that the force, F, is perpendicular (90deg) to the trap door. You have assumed it is a vertical force.
Now using that, do you get the right answer?
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#3
(Original post by Stonebridge)
The question says that the force, F, is perpendicular (90deg) to the trap door. You have assumed it is a vertical force.
Now using that, do you get the right answer?
Yes, I got the correct answer by saying F was the perpendicular force but isn't the moment the perpendicular distance from the pivot because In that case I thought the angle between F and the pivot would be 60 so you would need to use trig to get the perpendicular component of the force, is this wrong then?
Last edited by Emily~3695; 1 month ago
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1 month ago
#4
(Original post by Emily~3695)
Yes, I got the correct answer by saying F was the perpendicular force but isn't the moment the perpendicular distance from the pivot because In that case I thought the angle between F and the pivot would be 60 so you would need to use trig to get the perpendicular component of the force, is this wrong then?
It's only mg,the weight, that needs the angle because it's vertical and the trap door isn't horizontal.
The F force is already perpendicular to the trap door and so F times the length of the door is the moment.
The length of the trap door is the perpendicular distance you want.
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#5
(Original post by Stonebridge)
It's only mg,the weight, that needs the angle because it's vertical and the trap door isn't horizontal.
The F force is already perpendicular to the trap door and so F times the length of the door is the moment.
The length of the trap door is the perpendicular distance you want.
Oh ok that makes sense thank you 😊
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