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# Mechanics - Static Forces watch

1. Hey, I'm in need of some help with a question on static forces (Mechanics).

"In the diagram above, AOC = 90degrees and BOC = theta degrees. A particle at O is in equilibrium under the action of three coplanar forces. The three forces have magnitudes 8N. 12N and XN and act along OA, OB and OC respectively. Calculate

(a) the value, to one decimal place, of theta,

(b) the value, to 2 decimal places, of X"

Thank you very much in advance, I'm useless at this topic so far
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2. (Original post by AJC98)
Hey, I'm in need of some help with a question on static forces (Mechanics).

"In the diagram above, AOC = 90degrees and BOC = theta degrees. A particle at O is in equilibrium under the action of three coplanar forces. The three forces have magnitudes 8N. 12N and XN and act along OA, OB and OC respectively. Calculate

(a) the value, to one decimal place, of theta,

(b) the value, to 2 decimal places, of X"

Thank you very much in advance, I'm useless at this topic so far
resolve "verically and horizontally"
3. (Original post by TeeEm)
resolve "verically and horizontally"
Well.. Horizontally there is an XN and a 12N at an angle of 180-theta, and vertically there is an 8N and a 12N at an unknown angle
4. (Original post by AJC98)
Well.. Horizontally there is an XN and a 12N at an angle of 180-theta, and vertically there is an 8N and a 12N at an unknown angle
I can see what your problem is now being in 12 I pressume

12sin(180-θ)=8
12cos(180-θ)= X

the first one will find θ
then X

Alternative draw a triangle of forces and use trigonometry
5. (Original post by TeeEm)
I can see what your problem is now being in 12 I pressume

12sin(180-θ)=8
12cos(180-θ)= X

the first one will find θ
then X

Alternative draw a triangle of forces and use trigonometry
That's great, thanks for that, but would you mind explaining how you came up with those equations?
6. (Original post by AJC98)
That's great, thanks for that, but would you mind explaining how you came up with those equations?
if you extend OC to the "left" the angle between the extension and 12 force will be (180-θ)
then I resolved vertically and horizontally
7. (Original post by TeeEm)
if you extend OC to the "left" the angle between the extension and 12 force will be (180-θ)
then I resolved vertically and horizontally
And how did you resolve horizontally and vertically?
8. (Original post by AJC98)
And how did you resolve horizontally and vertically?
try this link for some worked examples.

look at Q1 to 4
9. (Original post by TeeEm)
try this link for some worked examples.

look at Q1 to 4
Ok I understand how you got those equations now, but what do I do now?

I have

x = 12cos(180-θ)
8 = 12sin(180-
θ)

But I don't understand what to do now..
10. (Original post by AJC98)
Ok I understand how you got those equations now, but what do I do now?

I have

x = 12cos(180-θ)
8 = 12sin(180-
θ)

But I don't understand what to do now..
you can solve the 2nd equation to get θ

than you get X
11. (Original post by TeeEm)
you can solve the 2nd equation to get θ

than you get X
Ah, I can rearrange the 2nd equation as normal can I? I didn't know that..

So I'd end up with

8 = 12sin(180-
θ)
8 = 12sin180-θ
8+θ = 12sin180
θ = 12sin180-8
θ = 1.67

Would I?

12. (Original post by AJC98)
Ah, I can rearrange the 2nd equation as normal can I? I didn't know that..

So I'd end up with

8 = 12sin(180-
θ)
8 = 12sin180-θ
8+θ = 12sin180
θ = 12sin180-8
θ = 1.67

Would I?

8=12sin(180-θ)

8=12sin(θ) since sin(180-θ) = sin(θ)

8/12 = sin(θ)

2/3 = sin(θ)

θ = 41.8 degrees

I think it will be more helpful if you sit down with your teacher and ask him to explain this question
13. (Original post by TeeEm)
8=12sin(180-θ)

8=12sin(θ) since sin(180-θ) = sin(θ)

8/12 = sin(θ)

2/3 = sin(θ)

θ = 41.8 degrees

I think it will be more helpful if you sit down with your teacher and ask him to explain this question
Ah yeah of course, I get it now Thanks!

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