M2 Statics (Edexcel)

Could someone explain how to take moments about A?

I don't understand the answer given in the solution bank and am really confused :frown:

thank you!

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Reply 1

Ayman!

Original post by Katiee224

Could someone explain how to take moments about A?

I don't understand the answer given in the solution bank and am really confused :frown:

thank you!

If you take moments about A, you're only concerned with the perpendicular distance to each force. Look at which forces are perpendicular to A - I can see 1 m for F, 1 m for 5g and 2 m for F once again.

Reply 2

Katiee224

Original post by aymanzayedmannan

How can you see 1m for F? Can I travel to the right or vertically below A and it won't matter so long as the force is perpendicular?

Also for the 5g force, do I just imagine the line of action passes through the top 2m length and then solve it?

Thanks :smile:

Reply 3

Ayman!

Original post by Katiee224

Take a look at the force F acting at point C - the perpendicular distance from A is actually 1 m. If you subtend the line of action of this force till it meets the left side of the lamina, you'll see what I mean.

Yes, you do exactly that. Subtend the lines of actions of forces if it helps you understand.

This is in the definition of moments and I feel it's a concept that's quite tricky to understand at first, but gets quite simple to visualise later on. I had problems in this exact question last year!

Reply 4

Katiee224

Original post by aymanzayedmannan

Moments are confusing me so much in m2! they were okay in m1 but now they're impossible :frown:

thanks for the help :biggrin:

Reply 5

Ayman!

Original post by Katiee224

Moments are confusing me so much in m2! they were okay in m1 but now they're impossible :frown:

thanks for the help :biggrin:

Feel free to ask me if you need help with mechanics!

Reply 6

Katiee224

Original post by aymanzayedmannan

Feel free to ask me if you need help with mechanics!

Hehe while I got you, could you explain this question to me, like some of the forces are at a 45 degree angle to A and the centre of mass is vertical and they have ignored this and solved it like the last one, i'm so confused :frown:

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Reply 7

Ayman!

Original post by Katiee224

Alright, remember what I said about needing the perpendicular distance from pivot to the line of action to the force? They've done this with the weight now. They've resolved the weight into it's horizontal and vertical components so that one component is perpendicular to PS and the other is parallel to it. This way, when you take moments about P, one component's line of action is perpendicular to P, while the other is negated as it has no turning effect. Try resolving the weight on your diagram.

Do you get what they did with the reaction at the hinge? :smile:

Reply 8

Katiee224

Original post by aymanzayedmannan

So if I resolve the weight I get 6gcos45 perpendicular to PS and the other resolved force parallel to PS, but they haven't included this in their answer?

Reply 9

Ayman!

Original post by Katiee224

So if I resolve the weight I get 6gcos45 perpendicular to PS and the other resolved force parallel to PS, but they haven't included this in their answer?

What the solution bank has done is not resolve the force and directly found the perpendicular distance by using pythagoras'. You could do this as well, it makes the working much simpler. Resolving forces should also work. Would you like me to show you both methods on paper?

Reply 10

Katiee224

Original post by aymanzayedmannan

omg that would help so much, thank you much, if i saw it written i would understand allot easier :colondollar:

Reply 11

Ayman!

Original post by Katiee224

omg that would help so much, thank you much, if i saw it written i would understand allot easier :colondollar:

I've realised that the resolving forces method would be very long winded - probably best if we just use the solution bank's way. They've first found PR by pythagoras', and since the COM (I labeled it "G") is at the centre of the lamina, PG would be exactly half of PR. The line of action of the weight directly passes through PG, so it would be the perpendicular distance from the point P for the weight.

As for the other forces - the line of action of F meets S, right? So the perpendicular distance from P would be PS.

Also, the line of action of 2F meets Q so the perpendicular distance would be PQ.

Now it's just a matter of anticlockwise = clockwise moments!

Reply 12

Katiee224

Original post by aymanzayedmannan

is at the centre of the lamina, PG would be exactly half of PR. The line of action of the weight directly passes through PG, so it would be the perpendicular distance from the point P for the weight.

As for the other forces - the line of action of F meets S, right? So the perpendicular distance from P would be PS.

Also, the line of action of 2F meets Q so the perpendicular distance would be PQ.

Now it's just a matter of anticlockwise = clockwise moments!

Thank you so much for the help! you are a lifesaver!! so say the 6g weight was in another position inside the square, as long as you workout the perpendicular length from P you could then work out the moment?

Reply 13

Ayman!

Original post by Katiee224

Yes, precisely! A little trigonometry is all you need. :smile:

Reply 14

Katiee224

Original post by aymanzayedmannan

Yes, precisely! A little trigonometry is all you need. :smile:

ok hehe thanks you are soooo helpful :biggrin:

Reply 15

Katiee224

Original post by aymanzayedmannan

Yes, precisely! A little trigonometry is all you need. :smile:

hey could you really quickly tell me how i can take moments in this example as i don't see how they arrived at their answer, like how they resolve the 2mg force :colondollar:

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Reply 16

Ayman!

Original post by Katiee224

hey could you really quickly tell me how i can take moments in this example as i don't see how they arrived at their answer, like how they resolve the 2mg force :colondollar:

You could resolve it along the rod so that one component is perpendicular to the rod, or you could once again find the perpendicular distance from the pivot to the force using trigonometry by subtending the line of action of the force.

Reply 17

Ayman!

Original post by Katiee224

hey could you really quickly tell me how i can take moments in this example as i don't see how they arrived at their answer, like how they resolve the 2mg force :colondollar:

The hypotenuse is a, the angle between the adjacent side and the hypotenuse is 60. Easy to find the perpendicular distance from here!

Reply 18

Katiee224

Original post by aymanzayedmannan

The hypotenuse is a, the angle between the adjacent side and the hypotenuse is 60. Easy to find the perpendicular distance from here!

I understand this, but i'm confused how to solve from the point where the lines of action of R and P meet like in the example :frown:

we need an equation where F is the subject :biggrin:

Reply 19

Ayman!

Original post by Katiee224

I understand this, but i'm confused how to solve from the point where the lines of action of R and P meet like in the example :frown:

we need an equation where F is the subject :biggrin:

There is no need to take moments about that point, I feel it unnecessarily complicates things. Taking moments about B would also give you the answer you require.

The Edexcel M2 textbook introduces this strange method of taking moments about where the two forces meet, but to be honest, I've never used it while doing past papers.