Do I need to resolve these forces? A level maths mechanics

leoishush

15

I am struggling with knowing when to resolve forces Screenshot 2023-04-04 11.46.00 AM.png

(edited 1 year ago)

Reply 1

davros

16

Original post by leoishush

I am struggling with knowing when to resolve forces Screenshot 2023-04-04 11.46.00 AM.png

what are your thoughts? How else would you approach the problem without resolving? (that's not a trick question!)

Reply 2

leoishush

OP

15

Original post by davros

what are your thoughts? How else would you approach the problem without resolving? (that's not a trick question!)

Sorry I should of probably asked my question better. Why don't you resolve 2 and mg.
I have already answered the question above but I don't understand why 2 and mg arent resolved

Reply 3

davros

16

Original post by leoishush

Sorry I should of probably asked my question better. Why don't you resolve 2 and mg.
I have already answered the question above but I don't understand why 2 and mg arent resolved

Sorry, I'm not understanding what you're asking! Are you comparing your working to a mark scheme or something? Can you post your working and the actual answers / mark scheme for comparison?

Reply 4

SnowyHornet

4

Original post by leoishush

I am struggling with knowing when to resolve forces Screenshot 2023-04-04 11.46.00 AM.png

so for part (a) as the string is threaded through the smooth bead the tension has to be the same across it, so you can then resolve it horizontally as we do not know mg, so Tsin 30 + Tsin60 = 2 so tension would then equal 1.46N (3sf)

part (b) we would resolve vertically now as we know T (use answer function in calc) Tcos30 = Tcos60 + mg so mg is equal to 0.54N so g is 0.055kg (55g)

Reply 5

leoishush

OP

15

Original post by SnowyHornet

so for part (a) as the string is threaded through the smooth bead the tension has to be the same across it, so you can then resolve it horizontally as we do not know mg, so Tsin 30 + Tsin60 = 2 so tension would then equal 1.46N (3sf)

part (b) we would resolve vertically now as we know T (use answer function in calc) Tcos30 = Tcos60 + mg so mg is equal to 0.54N so g is 0.055kg (55g)

yeahbut can 2N and mgN be resolved?

Reply 6

SnowyHornet

4

Original post by leoishush

yeahbut can 2N and mgN be resolved?

I just explained when you resolve 2N and mgN horizontally with the 2N and vertically for mg

Reply 7

tonyiptony

11

Original post by leoishush

yeahbut can 2N and mgN be resolved?

So a better question might be: how would you reckon to resolve the 2N force and the weight?

EDIT: The thing about resolving forces is that we can pick whichever pair of axes we want to resolve.
You can, if you want to, resolve along the AB direction and the 2N force direction - they don't have to be perpendicular!
But then, good luck with trig.

(edited 1 year ago)

Reply 8

mqb2766

21

Original post by leoishush

yeahbut can 2N and mgN be resolved?

The question strongly hints that you consider (resolve) the horizontal forces for i) then the vertical forces for ii). As mentioned by the other poster, this is because the we want to get an equation involving tension which is independent of m so
T = ....
Then use that to determine m in part ii).

You could resolve in the direction of each string and hence resolve 2N and mg. However this will produce two simultaneous equations in T and m which you'd have to solve by eliminating one of the terms. Not hard, but more work than needed which is why the question pushes you in the previous direction. Try it?

If you wanted to solve for m directly without determing T, you could note that the force in each string is the same so the resultant force due to tension must bisect the right angle and act at an angle 15 degrees above the horizonntal. This means that the resultant of 2 N and mg must act 15 degrees below the horizontal so
m = 2*tan(15)/g = ...

You resolve forces in whichever way gives you the easiest way to determine what you want. Choosing directions which give you independent equations is equivalent to eliminating variables when you solve simultaneously. So think a couple of steps ahead when choosing what to do.

Reply 9

leoishush

OP

15

Original post by mqb2766

The question strongly hints that you consider (resolve) the horizontal forces for i) then the vertical forces for ii). As mentioned by the other poster, this is because the we want to get an equation involving tension which is independent of m so
T = ....
Then use that to determine m in part ii).

You could resolve in the direction of each string and hence resolve 2N and mg. However this will produce two simultaneous equations in T and m which you'd have to solve by eliminating one of the terms. Not hard, but more work than needed which is why the question pushes you in the previous direction. Try it?

If you wanted to solve for m directly without determing T, you could note that the force in each string is the same so the resultant force due to tension must bisect the right angle and act at an angle 15 degrees above the horizonntal. This means that the resultant of 2 N and mg must act 15 degrees below the horizontal so
m = 2*tan(15)/g = ...

You resolve forces in whichever way gives you the easiest way to determine what you want. Choosing directions which give you independent equations is equivalent to eliminating variables when you solve simultaneously. So think a couple of steps ahead when choosing what to do.