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A level maths mechanics

Hi,

Does anyone have any advice about how to improve mechanics questions. Especially moments and similar, I just never know when to take moments, when to resolve to the plane, etc. I have done loads of questions and still don’t feel confident. Thanks.
Reply 1
Original post by Har6547
Hi,

Does anyone have any advice about how to improve mechanics questions. Especially moments and similar, I just never know when to take moments, when to resolve to the plane, etc. I have done loads of questions and still don’t feel confident. Thanks.

It might be a good idea to post a question that you found hard and point out the bit where you get stuck or are unsure about. Then we can guide you or give you tips for similar questions.
Reply 2
Original post by Notnek
It might be a good idea to post a question that you found hard and point out the bit where you get stuck or are unsure about. Then we can guide you or give you tips for similar questions.

I have attached a question that I found part (b) difficult. Looking back it looks obvious, but there were just so many different things I could have done at the time. I resolved perpendicular, but then I took moments and realised that it gave me the same equation. Any help appreciated, thanks.
Reply 3
For the unknowns you dont know T, beta or R. So taking moments about C would be an obvious choice to eliminate both T and beta from the balance equation. If you resolved horizontally, vertically, along the plane, perpendicular to the rope, ... you wouldn't be able to eliminate both at the same time and would need more than one equation to solve simultaneously. Quesiton spotting a bit, you might note that they ask for beta in part b) so its a bit of a hint that its probably not necessary in a). If necessary, roughly write down each force balance equation at the start, and think about which one(s) simply the problem by eliminating unkown terms at the start.

For b) theres not an easy way (see below) to get beta without also considering T, but a bit of foresight would be that resolving vertically and horizontally (for instance) would give Tsin(beta-alpha) = ... and Tcos(beta-alpha) = ... as you know all the rest and then simply divide to get tan and eliminate T. You could do a variety of approaches though (along plane and perpendicular) or another moment equation and one projection as the ms says.

Edit - another way of thinking about this (less algebra/simultaneous equations) is to resolve the forces horizontally and vertically (without T) then reason that to balance, T must be the hypotenuse of the corresponding force triangle with muR being the horizontal leg and mg-R the vertical leg. So tan gives the angle with the horizontal and add alpha to get beta. Similarly if you wanted to find T without beta, resolve horizontally and vertically then use pythagoras. Its equivalent to the algebra using sin^2+cos^2=1 to combine the equations / eliminate beta

Note that for b) Id be tempted to resolve perpendicular to the string as that would give a single equation which would give beta, though there would be a bit more work setting up the angles/equation in the first place.

Guess the key thing is for forces youre not interested in, try resolving perpendicular to them or taking moments about the point at which theyre applied, rather than setting up "lots" of simultaneous equations and eliminating them. Be clear about which unknowns you want to find and which ones are not necessary.

For a few questions, why not work though a problem (rather than looking at the ms too soon) two or three different ways and thinking about the pros/cons of the different approaches. That way, its easier to appreciate the decisions you make at the start where youre trying to think a couple of steps ahead about which approach(es) are sensible.
(edited 10 months ago)
Reply 4
Original post by mqb2766
For the unknowns you dont know T, beta or R. So taking moments about C would be an obvious choice to eliminate both T and beta from the balance equation. If you resolved horizontally, vertically, along the plane, perpendicular to the rope, ... you wouldn't be able to eliminate both at the same time and would need more than one equation to solve simultaneously. Quesiton spotting a bit, you might note that they ask for beta in part b) so its a bit of a hint that its probably not necessary in a). If necessary, roughly write down each force balance equation at the start, and think about which one(s) simply the problem by eliminating unkown terms at the start.

For b) theres not an easy way (see below) to get beta without also considering T, but a bit of foresight would be that resolving vertically and horizontally (for instance) would give Tsin(beta-alpha) = ... and Tcos(beta-alpha) = ... as you know all the rest and then simply divide to get tan and eliminate T. You could do a variety of approaches though (along plane and perpendicular) or another moment equation and one projection as the ms says.

Note that for b) Id be tempted to resolve perpendicular to the string as that would give a single equation which would give beta, though there would be a bit more work setting up the angles/equation in the first place.

Guess the key thing is for forces youre not interested in, try resolving perpendicular to them or taking moments about the point at which theyre applied, rather than setting up "lots" of simultaneous equations and eliminating them. Be clear about which unknowns you want to find and which ones are not necessary.

For a few questions, why not work though a problem (rather than looking at the ms too soon) two or three different ways and thinking about the pros/cons of the different approaches. That way, its easier to appreciate the decisions you make at the start where youre trying to think a couple of steps ahead about which approach(es) are sensible.

Thank you! This is very helpful, I will try that :smile:
Reply 5
Original post by mqb2766
For the unknowns you dont know T, beta or R. So taking moments about C would be an obvious choice to eliminate both T and beta from the balance equation. If you resolved horizontally, vertically, along the plane, perpendicular to the rope, ... you wouldn't be able to eliminate both at the same time and would need more than one equation to solve simultaneously. Quesiton spotting a bit, you might note that they ask for beta in part b) so its a bit of a hint that its probably not necessary in a). If necessary, roughly write down each force balance equation at the start, and think about which one(s) simply the problem by eliminating unkown terms at the start.

For b) theres not an easy way (see below) to get beta without also considering T, but a bit of foresight would be that resolving vertically and horizontally (for instance) would give Tsin(beta-alpha) = ... and Tcos(beta-alpha) = ... as you know all the rest and then simply divide to get tan and eliminate T. You could do a variety of approaches though (along plane and perpendicular) or another moment equation and one projection as the ms says.

Note that for b) Id be tempted to resolve perpendicular to the string as that would give a single equation which would give beta, though there would be a bit more work setting up the angles/equation in the first place.

Guess the key thing is for forces youre not interested in, try resolving perpendicular to them or taking moments about the point at which theyre applied, rather than setting up "lots" of simultaneous equations and eliminating them. Be clear about which unknowns you want to find and which ones are not necessary.

For a few questions, why not work though a problem (rather than looking at the ms too soon) two or three different ways and thinking about the pros/cons of the different approaches. That way, its easier to appreciate the decisions you make at the start where youre trying to think a couple of steps ahead about which approach(es) are sensible.

Hi. I think I am also a bit confused about forces and components. For the question with the beta angle, you could split this force into components parallel and perpendicular to the plane, but you could also split it into vertical and horizontal components (with a different angle value). Why don't you include the vertical component when resolving vertically, and the same for the horizontal one?
Reply 6
Original post by Har6547
Hi. I think I am also a bit confused about forces and components. For the question with the beta angle, you could split this force into components parallel and perpendicular to the plane, but you could also split it into vertical and horizontal components (with a different angle value). Why don't you include the vertical component when resolving vertically, and the same for the horizontal one?


Whichever direction you choose, the resolved forces must balance in equilibrium. If you consider two directions, they don't have to be perppendicular (but often are). For this part b), when they resolve vertically and horizontally in the ms, they include both Tcos(beta-alpha) and Tsin(beta-alpha) so they do include them?
(edited 10 months ago)
Reply 7
Original post by mqb2766
Whichever direction you choose, the resolved forces must balance in equilibrium. If you consider two directions, they don't have to be perppendicular (but often are). For this part b), when they resolve vertically and horizontally in the ms, they include both Tcos(beta-alpha) and Tsin(beta-alpha) so they do include them?

Oh ok I see now I think. Thanks. :smile:
Reply 8
Original post by Har6547
Oh ok I see now I think. Thanks. :smile:


If youre unsure, just upload what you think. Even if youre wrong, its better to understand where/why.
Reply 9
Original post by mqb2766
If youre unsure, just upload what you think. Even if youre wrong, its better to understand where/why.


I was talking more generally about components and forces. For example, with this one, if I was to resolve perpendicular and parallel to the plane, would I include the green components, and then if I was to resolve vertically and horizontally would I include the yellow? So in this instance it would probably be better to resolve parallel and perpendicular to the plane because you know the angle? Hopefully this makes sense
Reply 10
Original post by Har6547
I was talking more generally about components and forces. For example, with this one, if I was to resolve perpendicular and parallel to the plane, would I include the green components, and then if I was to resolve vertically and horizontally would I include the yellow? So in this instance it would probably be better to resolve parallel and perpendicular to the plane because you know the angle? Hopefully this makes sense

As you say, the green represents the compoents from resolving the force parallel and perpendicular to the plane and yellow represents the components from resolving the force horizontally and vertically. Naively doing perpendicular / parallel would be the obvious thing (when you know theta) for this force, but there would be other information in the question which may change that. As above, the choice about where to take moments, and which directions to resolve forces is usually made based on what makes it easiest to find the unknown.
(edited 10 months ago)
Reply 11
Original post by mqb2766
As you say, the green represents the compoents from resolving the force parallel and perpendicular to the plane and yellow represents the components from resolving the force horizontally and vertically. Naively doing perpendicular / parallel would be the obvious thing (when you know theta) for this force, but there would be other information in the question which may change that. As above, the choice about where to take moments, and which directions to resolve forces is usually made based on what makes it easiest to find the unknown.

Yes, that makes sense now. Thank you :smile:
Reply 12
Original post by Har6547
Yes, that makes sense now. Thank you :smile:


NP. I updated the previous longish post where I discussed the original b). For me the "easiest" way to find beta was to note that R, muR and mg were all horizontal/vertical and known and the unknown beta (and tension) would be easiest to find by drawing the right triangle (resolve horizotal and vertical) with legs muR and mg-R (T is the hypotenuse), then simply doing atan((mg-R)/muR)+alpha using the force triangle. Probably simpler/more natural than simultaneous equations, though its effectively the same thing. You could do pythagoras on the right triangle if you were interested in T, not beta.
(edited 10 months ago)

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