Statics problem

[Mo]

I am having a bit of trouble solving statics problems. its actually only 1 issue. I can do the trig and rest.

The question , like the one below...

http://img25.imageshack.us/img25/5403/q10lq.th.gif (http://img25.imageshack.us/my.php?image=q10lq.gif)

...usually asks to find the tension in one of the strings.In this case they asked for the tension in AC.

My question is, what do i take as the hypotenuse?

Do i take the weight as the hypotenuse or do i take the tension as the hypotenuse?

Below is how i have attempted to solve the question.

http://img133.imageshack.us/img133/2118/a15kz.th.gif (http://img133.imageshack.us/my.php?image=a15kz.gif)

However, taking the weight as the 'opposite', i get T = 3g / sin 30
Where as if i was taking the hypotenuse as the weight, then the answer would be correct. (14.7 N)

Am i drawing the diagram correctly? I am inserting the angles correctly?

I asked my teacher this question, and he said that you usually take the weight as the hypotenuse - however, he said this concerning mass-on-an -inclined-plane type questions, not these types of questions.

This is probably the only thing that will bring me down on my mechanics paper :frown: , so i would be very happy to solve this once and for all.

Thanks.

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Initially i pressed submit instead of preview, hence the reason the link didnt work!
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[Hootenanny]

Can't see the image and the link doesnt work.

[lightgrav]

The only thing to be sure of in Statics is that the Sum Force Vectors is equal to zero (= m a_vector) ! You add Vectors by placing them (sequentially) tail-to-tip, keeping each one parallel to its Free-Body-Diagram arrow .... a polygon will be formed.
In the case of three Force vectors, the polygon will be a triangle, but NOT always a Right triangle, so a lot of the time NONE of the Forces are a hypotenuse.

[civil_dude]

This is a two unknown problem so you need two equations.

I would use the strings as the hypotenuses and T1(AC) & T2(BC) as their respective tensions.

So, T1cos30 has to equal T2cos60 to cancel x forces Eq.1
And T1sin30 + T2sin60 = mg Eq.2

T1 = 0.577 T2 & 0.5T1 + 0.866T2 = 29.4N

So from Eq.1, T1 = 0.577T2, and plug & chug to get

T1 = 13.16N & T2 = 25.45N

I think..

[lightgrav]

If you draw the Force Addition Polygon, you see that In this case,
the two Tensions are the legs of a right triangle, with 29.4 N hypotenuse.
T_CA = 29.4 N sin(30) , and T_CB = 29.4 sin(60) .

In more typical situations, you obtain horizontal and vertical components of each Diagonal Force by treating it as the hypotenuse of a right triangle.

[Mo]

Thanks for your reply.Thing is the answer is given in the question. It says, "show that tension of AC is 14.7N". So it can't be 29.4N, unless i have misunderstood your post :confused:

[lightgrav]

The way it works is that you :
1) draw the Free-Body-Diagram
2) Draw the Force Polygon
. . . if the solution is not obvious yet, choose coordinate axes
3) separate each Force into components along the axes, and
4) sum the Force components separately.

Step 2 was what I drew for you, you should know some trig !

- - - - - - - - - - -

Multiply by sin(30) !

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[Mo]

:blushing: I can see now! In my defence i have been travelling all day :zzz: . Thanks for your help. I will put this into practice with a few other questions and if i have any further queries, i will repost!

Once again, thanks :smile:

ps: damn, i didnt even read your reply correctly, i shouldve noticed the
T_CA = 29.4 N sin(30) !

[behnamphysics]

hi every body i want to exchange my data with someone that like to more give information from statics.

thanks