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# M1 question watch

1. Im struggling with this question (see attached)

For (i) i got the correct answers, 3.36N and 4.48N, but for the wrong strings...
Here's my workings:
5.6 / sin 90 = TA / sin 36.9 = TR / sin 53.1
2. (Original post by Jmun)
Im struggling with this question (see attached)

For (i) i got the correct answers, 3.36N and 4.48N, but for the wrong strings...
Here's my workings:
5.6 / sin 90 = TA / sin 36.9 = TR / sin 53.1
What is the question?
3. (Original post by Jmun)
Im struggling with this question (see attached)

For (i) i got the correct answers, 3.36N and 4.48N, but for the wrong strings...
Here's my workings:
5.6 / sin 90 = TA / sin 36.9 = TR / sin 53.1
???
4. (Original post by Jmun)
Im struggling with this question (see attached)

For (i) i got the correct answers, 3.36N and 4.48N, but for the wrong strings...
Here's my workings:
5.6 / sin 90 = TA / sin 36.9 = TR / sin 53.1
OK, I'll have a guess. Have you drawn a triangle with the forces following each other round? The angles inside the triangle should be the ones that you use, each with the opposite side.
5. oh gosh sorry i forgot to attach the pic... here
Attached Images

6. (Original post by Jmun)
oh gosh sorry i forgot to attach the pic... here
This confirms the theory in my previous posting.
7. Yes bu the answer i get is opposite to theirs... Perhaps they used a trinagle in a different way, but that wouldnt make sense to me. I'm a bit confused about all the necessary forces to be resolved for number (ii)...
8. (Original post by Jmun)
Yes bu the answer i get is opposite to theirs... Perhaps they used a trinagle in a different way, but that wouldnt make sense to me. I'm a bit confused about all the necessary forces to be resolved for number (ii)...
I believe that you are not drawing the correct triangle (or if you are, you've got your angles in the wrong places). The arrows for the forces must follow each other round the triangle. A correct diagram is attached.

For (ii), just consider the forces on the ring: its weight downwards, the tension TR, the normal reaction (which will be horizontal - which way?) and the friction downwards since the ring is on the point of sliding upwards.
Attached Files
9. triangle of forces.doc (29.0 KB, 43 views)
10. TR is 40cm long. And TA is 30cm long. So, angle between TR and the horizontal force of 5.6N is tan^-1 (30/40) = 36.8
11. (Original post by Jmun)
TR is 40cm long. And TA is 30cm long. So, angle between TR and the horizontal force of 5.6N is tan^-1 (30/40) = 36.8
In the original diagram, tan^(-1) (30/40) is the angle between Tr and the vertical.
12. I dont see it like that.. sorry i really cant figure this out
13. (Original post by Jmun)
I dont see it like that.. sorry i really cant figure this out
I've added the original diagram to my attachment and marked on the various angles which equal tan^(-1) (30/40) i.e. 36.9 degrees (I called it 37 degrees on my triangle of forces, for simplicity of drawing).
Attached Files
14. triangle of forces.doc (30.0 KB, 46 views)
15. I understand now... Thanks so so much!!!

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