Quick help please - M1 pulleys
I cant get the answer to part C. The mark schemes answer is half that of mine, and I fail to understand why. The component of the tension for both sides of the string act on the pulley, so why dont thry multiply their answer by two?
Ill post my working too.
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Ill post my working too.
Posted from TSR Mobile
The tension in the pulley is a result of the two forces.. After realising this it seems nonsensical to multiply by two anywhere!
A flag should be raised when you realise your answer is greater than the combination of the two 'Tn's you have labelled! This means you are getting a force greater than the one we would have got if we'd just put them both in the same direction! It's always a good idea to have a think about how realistic your answers are when it comes to mechanics!
(And, in future, remember to keep things to 3 sig figs. The marks schemes sometimes have AWRT written in the marks margin, which means 'anything which rounds to', but this one doesn't. Don't give the examiner an excuse to knock off cheap marks!)
A flag should be raised when you realise your answer is greater than the combination of the two 'Tn's you have labelled! This means you are getting a force greater than the one we would have got if we'd just put them both in the same direction! It's always a good idea to have a think about how realistic your answers are when it comes to mechanics!
(And, in future, remember to keep things to 3 sig figs. The marks schemes sometimes have AWRT written in the marks margin, which means 'anything which rounds to', but this one doesn't. Don't give the examiner an excuse to knock off cheap marks!)
Original post by Jooooshy
The tension in the pulley is a result of the two forces.. After realising this it seems nonsensical to multiply by two anywhere!
A flag should be raised when you realise your answer is greater than the combination of the two 'Tn's you have labelled! This means you are getting a force greater than the one we would have got if we'd just put them both in the same direction! It's always a good idea to have a think about how realistic your answers are when it comes to mechanics!
(And, in future, remember to keep things to 3 sig figs. The marks schemes sometimes have AWRT written in the marks margin, which means 'anything which rounds to', but this one doesn't. Don't give the examiner an excuse to knock off cheap marks!)
A flag should be raised when you realise your answer is greater than the combination of the two 'Tn's you have labelled! This means you are getting a force greater than the one we would have got if we'd just put them both in the same direction! It's always a good idea to have a think about how realistic your answers are when it comes to mechanics!
(And, in future, remember to keep things to 3 sig figs. The marks schemes sometimes have AWRT written in the marks margin, which means 'anything which rounds to', but this one doesn't. Don't give the examiner an excuse to knock off cheap marks!)
Thanks for your answer, but Im sorry.. I tried to read your answer a couple of times, linking it to the diagram but i still cant make sense of what youre saying
if the individual tension components arethat much, dont i need to find the "hypotenuse" for both, and multiply by two because there are two sides?Okay, will do. All my working out has got to use values of 3sfs or should i use 100% accurate fingers right until the end point?
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Original post by jtbteddy
Thanks for your answer, but Im sorry.. I tried to read your answer a couple of times, linking it to the diagram but i still cant make sense of what youre saying if the individual tension components arethat much, dont i need to find the "hypotenuse" for both, and multiply by two because there are two sides?
Okay, will do. All my working out has got to use values of 3sfs or should i use 100% accurate fingers right until the end point?
Posted from TSR Mobile
if the individual tension components arethat much, dont i need to find the "hypotenuse" for both, and multiply by two because there are two sides?Okay, will do. All my working out has got to use values of 3sfs or should i use 100% accurate fingers right until the end point?
Posted from TSR Mobile
Don't apologise! Hopefully somebody will come along with a better explanation than the one I gave!
You don't need to multiply by two.. You are thinking of both triangles, but as if they are laid atop each other, giving the impression that there are almost two hypotenuse. It's not like that! It's like there are two matchsticks at 90deg to each other, and you add another one 45deg to them both. Two triangles, one matchstick. (Sorry, that was crude, but I couldn't think of any other way of explaining it!)
With regards to the 3s.f., they'll probably want you to label your diagram with 3s.f. and use your higher accuracy answer for further calcs. So I'd be writing Tn = 11.8 rather than 11.76, but using 11.76(..) in my calculations!
It's just a force triangle, your two tensions are at right angles to each other and make up the two short sides, your resultant force is the other side (hypotenuse). Since you know T, you can use Pythagoras to work out the resultant force. If it helps you visualise the triangle, draw a line connecting the two particles, this line represents your resultant force.
In m1, if you're using acceleration due to gravity in your calculations, it's acceptable to go to 2sf, as g=9.8 is to 2sf and you can't be more accurate than the information you're using. Otherwise, I'd suggest going to 3sf.
In m1, if you're using acceleration due to gravity in your calculations, it's acceptable to go to 2sf, as g=9.8 is to 2sf and you can't be more accurate than the information you're using. Otherwise, I'd suggest going to 3sf.
(edited 10 years ago)
Makes sense now! Thank you very much 
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