M2/1 Statics of rigid bodies
Maths and statistics discussion, revision, exam and homework help.
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M2/1 Statics of rigid bodies
Heres the question, all it asks for is to find the sum of the moments about P
http://imgur.com/GPHsS
Only thing I'm really struggling on is for the anti-clock moment, why is it 3cos55? why is it not 3sin55??
Any tips on these sorts of questions? haven't managed to find much on the web
EDIT: Also in this one, why is it 3 x4cos55?? I see no 3N force going clockwise??
http://imgur.com/8aa1vLast edited by Jetblast; 23-05-2012 at 21:07. -
Re: M2/1 Statics of rigid bodiesYou need the perpendicular distance of the line of the action of the force. The line in red on the left, which is the same length as the line in red on the right.(Original post by Jetblast)
Heres the question, all it asks for is to find the sum of the moments about P
http://imgur.com/GPHsS
Only thing I'm really struggling on is for the anti-clock moment, why is it 3cos55? why is it not 3sin55??
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Re: M2/1 Statics of rigid bodiesThey've got their clockwise and anticlockwise mixed up.(Original post by Jetblast)
EDIT: Also in this one, why is it 3 x4cos55?? I see no 3N force going clockwise??
http://imgur.com/8aa1v
First line is for anticlockwise forces,
Second line is for clockwise force. -
Re: M2/1 Statics of rigid bodiesThanks for the reply, but I still don't understand. Yes, the lines are the same length, but how does that help? I'm confused as its virtually the opposite as to how I've learnt to solve forces (cosx on the side which has the angle and sinx on the other side with F in the middle at angle x)(Original post by ghostwalker)
You need the perpendicular distance of the line of the action of the force. The line in red on the left, which is the same length as the line in red on the right. -
Re: M2/1 Statics of rigid bodiesWhat technique have you been taught for finding moments?(Original post by Jetblast)
Thanks for the reply, but I still don't understand. Yes, the lines are the same length, but how does that help? I'm confused as its virtually the opposite as to how I've learnt to solve forces (cosx on the side which has the angle and sinx on the other side with F in the middle at angle x)
You can either find the perpendicular distance of the line of action of the force (which is what the solid line in red is, in the diagram).
Or, resolve the force into components radial and tangential to the line connecting the point of action of the force to the pivot.
I would be surprised if it's the latter, as it's somewhat messier to deal with, IMHO.
And with the former, there is no requirement to resolve the force into components. -
Re: M2/1 Statics of rigid bodiesPerpendicular distance and force.(Original post by ghostwalker)
What technique have you been taught for finding moments?
You can either find the perpendicular distance of the line of action of the force (which is what the solid line in red is, in the diagram).
Or, resolve the force into components radial and tangential to the line connecting the point of action of the force to the pivot.
I would be surprised if it's the latter, as it's somewhat messier to deal with, IMHO.
And with the former, there is no requirement to resolve the force into components. -
Re: M2/1 Statics of rigid bodiesGood.(Original post by Jetblast)
Perpendicular distance and force.
So, you're not resolving the force into components, hence the cos/sin you were talking about doesn't apply here.
You're just interested in finding the perpendicular distance; either of the two solid lines in red, which are adjacent to the 55 degree angles, and hence the use of "4 cos 55". -
Re: M2/1 Statics of rigid bodiesNo, no, I am resolving into components, clock and anti-clock.(Original post by ghostwalker)
Good.
So, you're not resolving the force into components, hence the cos/sin you were talking about doesn't apply here.
You're just interested in finding the perpendicular distance; either of the two solid lines in red, which are adjacent to the 55 degree angles, and hence the use of "4 cos 55".
4 is the distance from P to where the 3N is applied. What is making me crazy is why its 3cos55 and not 3sin55
I think, we need to find F in the diagram below. A=55
The solution says 3cos55=F but I only see that it has to be 3sin55=F (I think)
http://imgur.com/SdscQ
I'm so confused -
Re: M2/1 Statics of rigid bodiesOK, I think I can see your confusion.(Original post by Jetblast)
No, no, I am resolving into components, clock and anti-clock.
4 is the distance from P to where the 3N is applied. What is making me crazy is why its 3cos55 and not 3sin55
I think, we need to find F in the diagram below. A=55
The solution says 3cos55=F but I only see that it has to be 3sin55=F (I think)
http://imgur.com/SdscQ
I'm so confused
The force you are resolving is the 3N force and that is your F. Here's a revised diagram. -
Re: M2/1 Statics of rigid bodies(Original post by ghostwalker)
OK, I think I can see your confusion.
The force you are resolving is the 3N force and that is your F. Here's a revised diagram.
so A isn't 55?? *goes back to GCSE maths
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How about this one, how would the anti-clock's be resolved?
http://imgur.com/iSOVh
Thanks for your help, I would have literally had nightmares if I didn't understand by the time I went to sleep! I have difficulties on simple maths while I am able to do complex maths
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Re: M2/1 Statics of rigid bodiesCome on. Have a go. Choose one of the forces, mark the line for the direction in which you need to resolve it, mark the angle you need to work out, and then start working out some angles. Work from what you know, adding a step at a time.(Original post by Jetblast)
so A isn't 55?? *goes back to GCSE maths
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How about this one, how would the anti-clock's be resolved?
http://imgur.com/iSOVh
Thanks for your help, I would have literally had nightmares if I didn't understand by the time I went to sleep! I have difficulties on simple maths while I am able to do complex maths
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Re: M2/1 Statics of rigid bodiesOk, I tried it, but am a little off the answer(Original post by ghostwalker)
Come on. Have a go. Choose one of the forces, mark the line for the direction in which you need to resolve it, mark the angle you need to work out, and then start working out some angles. Work from what you know, adding a step at a time.
Clockwise: 15sin50=11.5
Anti: 6sin18+12sin72=13.27
13.27-11.5=1.77 Anti-clock
I'm thinking I'm right as they are 1 degree off with the angles for the anti-clock moments -
Re: M2/1 Statics of rigid bodiesYou want to quote to the same degree of accuracy for each part, so the first one should be 11.49. Aside from that, you're correct, and your angles are correct.(Original post by Jetblast)
Ok, I tried it, but am a little off the answer
Clockwise: 15sin50=11.5
Anti: 6sin18+12sin72=13.27
13.27-11.5=1.77 Anti-clock
I'm thinking I'm right as they are 1 degree off with the angles for the anti-clock moments
It's not very impressive that 2 out of the 3 examples have errors in given solution. Where are these from? -
Re: M2/1 Statics of rigid bodiesThese questions are from the edexcel M2 modular textbook.(Original post by ghostwalker)
You want to quote to the same degree of accuracy for each part, so the first one should be 11.49. Aside from that, you're correct, and your angles are correct.
It's not very impressive that 2 out of the 3 examples have errors in given solution. Where are these from?
I remember spending hours trying to figure them out, it was the first exercise so i thought that i might be wrong but actually most of the answers were wrong, this exercise also has questions with wrong data(the lengths marked on the given shape are wrong).
This thread contains another question from the same exercise of the book, using the sine/cosine rule it can be deduced that the marked length/angle are wrong. -
Re: M2/1 Statics of rigid bodiesMakes one wonder, if Edexcel sat one of their own exams, would they pass?(Original post by raheem94)
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Re: M2/1 Statics of rigid bodies(Original post by ghostwalker)
Makes one wonder, if Edexcel sat one of their own exams, would they pass?
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Re: M2/1 Statics of rigid bodiesProbably not. It really puts a lot of pressure on us who waste time thinking about what is wrong when we're actually right.(Original post by ghostwalker)
Makes one wonder, if Edexcel sat one of their own exams, would they pass?
A number of other question solutions were wrong throughout the rest of the chapter. As oppose to the previous chapter, where I done a whole chapter in a day, I haven't even done half of this chapter yet
Heres another question I'm confused about even though its an example. Care to explain please? I don't understand it from the taking of the moments
http://imgur.com/AdIz9Last edited by Jetblast; 24-05-2012 at 22:00. -
Re: M2/1 Statics of rigid bodiesTaking moments about A; in the first moments equation, the LHS is the sum of the clockwise moments (weight of ladder plus mass), and the RHS is the sum of the anticlockwise moments. If you were writing out all the steps it would be "P" on the RHS, but since F=P, we can put F.(Original post by Jetblast)
Probably not. It really puts a lot of pressure on us who waste time thinking about what is wrong when we're actually right.
A number of other question solutions were wrong throughout the rest of the chapter. As oppose to the previous chapter, where I done a whole chapter in a day, I haven't even done half of this chapter yet
Heres another question I'm confused about even though its an example. Care to explain please? I don't understand it from the taking of the moments
http://imgur.com/AdIz9
Does that cover it? If not, can you be more specific. -
Re: M2/1 Statics of rigid bodiesAh thanks. In my stupor last night i didn't realise moments were taken from B(Original post by ghostwalker)
Taking moments about A; in the first moments equation, the LHS is the sum of the clockwise moments (weight of ladder plus mass), and the RHS is the sum of the anticlockwise moments. If you were writing out all the steps it would be "P" on the RHS, but since F=P, we can put F.
Does that cover it? If not, can you be more specific.
Thanks again
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