A-level maths help!

Without looking at the question itself, I think all of them are correct (albeit a bit hard to look at, but anyways). You are just decomposing the forces with respect to different basis (basis here really just means are you decomposing along the slant plane, or the usual x and y's, or whatever).
Though I should emphasize that correct doesn't mean useful. In fact, probably only one out of the three makes your life easier when solving problems. I'd say just try it.

P.S. There is a very legit reason why decomposing forces along the slant plane makes our lives infinitely easier. Do you know why?

(edited 1 year ago)

Reply 2

OP

Original post by tonyiptony

Without looking at the question itself, I think all of them are correct (albeit a bit hard to look at, but anyways). You are just decomposing the forces with respect to different basis (basis here really just means are you decomposing along the slant plane, or the usual x and y's, or whatever).
Though I should emphasize that correct doesn't mean useful. In fact, probably only one out of the three makes your life easier when solving problems. I'd say just try it.

My apologies I was just rushing to make the diagrams but I was just getting at the direction of the vertical and horizontal components of the 40 newtons because you can have them either parallel and perpendicular or horizontal and vertical. But would you say it depends on the question as to which ways you have them?

Reply 3

Original post by skyeforster15

My apologies I was just rushing to make the diagrams but I was just getting at the direction of the vertical and horizontal components of the 40 newtons because you can have them either parallel and perpendicular or horizontal and vertical. But would you say it depends on the question as to which ways you have them?

It mostly depends on the situation.
A parallel is that why we sometimes pick the starting position to be displacement zero, and sometimes the ground be displacement zero - it makes our calculations easier.

Similar in this case, there is a very legit reason why we often choose the decomposition in (A) rather than (B) and (C), even though they are all correct (and will give the right answer in the end). Hint in spoilers:

Spoiler

Reply 4

OP

16801922010142697263160161292294.jpg566.2KB

This was the question I was doing.

Reply 5

OP

Original post by tonyiptony

It mostly depends on the situation.
A parallel is that why we sometimes pick the starting position to be displacement zero, and sometimes the ground be displacement zero - it makes our calculations easier.

Similar in this case, there is a very legit reason why we often choose the decomposition in (A) rather than (B) and (C), even though they are all correct (and will give the right answer in the end). Hint in spoilers:

Spoiler

There is a perpendicular force working on the particle--> The resultant force.

Reply 6

Original post by skyeforster15

There is a perpendicular force working on the particle--> The resultant force.

Hmm... Not sure what you mean.
Let's just look at the forces (after decomposing like in (A)) perpendicular to the plane, and ignore the ones along the plane for now. If the particle never jumps out of the plane, what do you think these forces add up to?

Reply 7

OP

34.4N

(edited 1 year ago)

Reply 8

Hmm... I don't know how you ended up with that number, but that's certainly wrong.
If we have a net force acting on a particle (let's talk in 1D for now, which is the case I posed in #7), then surely the particle will move along that direction, right? But then in our situation in #7, the particle never moves perpendicular to the slant plane...

Reply 9

OP

(Original post by tonyiptony)Hmm... I don't know how you ended up with that number, but that's certainly wrong.
If we have a net force acting on a particle (let's talk in 1D for now, which is the case I posed in #7), then surely the particle will move along that direction, right? But then in our situation in #7, the particle never moves perpendicular to the slant plane...

I don't really understand what you are trying to ask of me. I was relating it to the original question (the one I posted with my working out on and the question).

Reply 10

mqb2766

21

Its probably easier to think of the (simultaneous) equations you produce and how to make them as simple as possible to solve. Here you want to find mu and in order to do that you have to find R. Resolving perpendicular to the plane means the equation is independent of muR and it produces an equation
R = ...
Then as you know R you can resolve in any other direction, but its "easier" to resolve along the plane and get an equation
mu R = ...
as youve already drawn/used that triangle for the first part and that direction removes one of the forces (R) from the calculation

If you resolve in other directions, youll get equations involving both R and muR which youll have to solve simultaneously, which is more work.

(edited 1 year ago)

Reply 11

You.luv.Serenity

3

Original post by skyeforster15

Which diagram is correct because I'm doing this question and I am getting myself confused with the diagram.
--> Ignore the two resultant forces! There is only meant to be one!

i will try it thank you

Reply 12