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M1 Question Help

I am having trouble with the first part of this (finding tension). You can use resolving and it works but the sine rule doesnt work:

(20/SIN60) = (BC/SIN30). When you then rearrange, you get 11.5 but that is wrong. My M1 teacher reckons it should be (20/SIN30) = (BC/SIN60) and that gives the right answer. But I cant figure out why, can someone explain with a diagram. Surely 60 degrees is opposite 20N? :confused:

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Reply 1

such

The mistake that you are making is that you are confusing the lengths of the strings with the tensions in the strings. You cannot use the sine rule on the triangle because the forces don't form any sort of "triangle" that corresponds to your diagram. (In fact, if you draw the force BC, you will see that it is much longer than the force AC!)
The way to avoid falling into this trap is not to draw forces as lines! Draw them as arrows with only one end connected. (see my post below for a diagram showing where you went wrong)

Although the equation your teacher gives looks like the sine rule, in fact it is not related to the sine rule at all! It is just a misleading way of resolving the forces horizontally. (Multiply both sides by sin30*sin60 to see how)

Resolving forces is the only correct way to do this.

(edited 10 years ago)

Reply 2

TenOfThem

Resolving horizontally

BC Cos60 = AC Cos30

Reply 3

ghostwalker

Original post by momanium

Surely 60 degrees is opposite 20N? :confused:

Nope. The angle opposite 20N is 150 degrees. Which has the same sine value as 180-150 = 30 degrees.

See attached:

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Reply 4

TenOfThem

Original post by ghostwalker

Nope. The angle opposite 20N is 150 degrees. Which has the same sine value as 180-150 = 30 degrees.

See attached:

i think he means opposite in that he is using the sine rule

Reply 5

such

I think the issue is that he is seeing a phantom triangle (highlighted in orange in the picture attached), and is trying to apply the sine rule to that triangle, whereas the reality looks like the force diagram to the right. So all the stuff about sines is bogus and misleading. Just resolve horizontally, as TenOfThem helpfully demonstrated.

(I am aware that the mg force should be larger... Also, I can crappy paint diagrams too! :biggrin:

)

(edited 10 years ago)

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Reply 6

ghostwalker

@all

Yes, the sine rule applied to the triangle ABC doesn't work and that is not the appropriate triangle to use.

I was attempting to support a different methodology, that the OP appeared to have been taught, as far as I could see.

That being the triangle of forces, to which the sine rule could be applied, and is equivalent to using the sine on the angle opposite to the force, at the point of intersection of the three forces.

Sorry if people have misunderstood, or it's caused confusion.

Reply 7

such

Yes, but I thought that it was apparent that OP did not know about this, and the triangle of forces may cause more confusion than it is worth. I was trying to clear up where OP had gone wrong, rather than burden him with another way to approach the problem :tongue:

Reply 8

momanium

I am not really understanding what's going on. You can use the sine rule (just not on the wrong triangle I seem to be making up). However I dont see what angle is opposite of 20N etc.

Reply 9

momanium

Original post by such

I am more than happy to learn where I am going wrong and how to put it right. To all those who are telling me to resolve, you can use the sine rule and I would like to understand how in this question.

Reply 10

momanium

Having done some googling, it seems like I need to learn the "triangle of forces". Any help?

Reply 11

TenOfThem

The only triangle of forces that I can see is a right angled triangle that has mg as the hypotenuse and BC = 20 tan60

Reply 12

TenOfThem

Original post by momanium

Having done some googling, it seems like I need to learn the "triangle of forces". Any help?

I do not understand how you have been taught to use the sine rule without being taught triangle of forces ... You need a triangle for the sine rule

Reply 13

such

Oh, so you are asking about triangle of forces. You form the triangle of forces by joining together the three forces tip-to-tail to form a triangle (see attached). Then, you can apply the sine rule. The opposite angle to 20N is 30 degs, and the opposite angle to T is 60 degs. So 20N/sin30 = T/sin60

ghostwalker is talking about an alternative way which ends up the same. Instead of drawing out a triangle of forces, you just use the angle opposite the force on the force diagram. So 20N is opposite 150degs and T is opposite 120degs. But sin150 = sin30, and sin120 = sin60, so these two methods are actually exactly the same.

HTH

(edited 10 years ago)

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Reply 14

TheFuture001

The method for the sine rule is as the above poster has put it. You basically extend the line of AC to the right (or you can do it the other way if you want) and form the triangle like above. You then label it appropriately and use the sine rule.

The method does work, although you may like to know there's a way easier way to do this question. For the first just resolve horizontally, and for the second resolve perpendicular to AC. Since the tension in AC is at 90 degrees to 'T' you just deal with T and m and it's pretty easy.

Don't bother with the sine rule I say!