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Mechanics equilibrium of rigid bodies

I'm stuck on part a) of the question. I think you are supposed to take moments about A. I not sure how to do this as you don't get given an angle therefore I don't know what the value of cos theta will be. Thanks

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Reply 1

ghostwalker

Original post by Anonymous_4657

Well you've got a right-angled triangle and two of the sides: That's enough to work out any angles you need.

Reply 2

username4506446

Original post by ghostwalker

Well you've got a right-angled triangle and two of the sides: That's enough to work out any angles you need.

I'm still a bit confused on what to do. Please may you help. In the answers it says cos theta is 2/ square root 13.

Reply 3

ghostwalker

Original post by Anonymous_4657

I'm still a bit confused on what to do. Please may you help. In the answers it says cos theta is 2/ square root 13.

But which angle is theta?

Reply 4

ghostwalker

Original post by Anonymous_4657

I'm still a bit confused on what to do. Please may you help. In the answers it says cos theta is 2/ square root 13.

If you're confused about taking moments, go back to the definition: The perpendicular distance of the line of action of the force multiplied by the magnitude of the force.

So, what's the line of action of the force, and what's its perpendicular distance?

Reply 5

username4506446

I just don't understand how they got the value of Cos theta. @ghostwalker

(edited 4 years ago)

Reply 6

ghostwalker

Original post by Anonymous_4657

I just don't understand how they got the value of Cos theta. @ghostwalker

See my post #5 to start. I don't know where you're upto, what you've been able to understand so far. What angle theta is. You've not posted the markscheme you're looking at, and I'm not a ...... mind reader.

Reply 7

username4506446

Original post by ghostwalker

Here is the mark scheme. Sorry if I have caused you any inconvenience or hassle.

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Reply 8

username4506446

Original post by ghostwalker

Can you help me understand how they get the value of cos theta?

Reply 9

Alex-6781

They get the value of cos theta by just using trigonometry as you know the ratio of two if the sides so you can find the third by using pythagoras then from there you can find the exact value of cos theta

Reply 10

ghostwalker

Original post by Anonymous_4657

Can you help me understand how they get the value of cos theta?

So, theta is in the top left.

Cos theta = adjacent / hypotenuse.

Adjecant is 2a

Hypotenuse, via pythagoras, is

\sqrt{(2a)^2+(3a)^2}=\sqrt{13}a

Hence

\cos\theta = \dfrac{2a}{\sqrt{13}a}=\dfrac{2}{\sqrt{13}}

Reply 11

username4506446

Original post by ghostwalker

So, theta is in the top left.

Cos theta = adjacent / hypotenuse.

Adjecant is 2a

Hypotenuse, via pythagoras, is

\sqrt{(2a)^2+(3a)^2}=\sqrt{13}a

Hence

\cos\theta = \dfrac{2a}{\sqrt{13}a}=\dfrac{2}{\sqrt{13}}

Thanks. I was getting a bit confused where theta was. I didn’t realise it was in the top left. It never specified in the question.

Reply 12

ghostwalker

Original post by Anonymous_4657

Thanks. I was getting a bit confused where theta was. I didn’t realise it was in the top left. It never specified in the question.

Theta is top left in the markscheme (diagram at top), but if you were doing the question without having seen the markscheme you could have chosen a different angle to be theta - e.g. the bottom right. Either would be fine, just give slightly different working, but getting the same results.