Statics....is this problem easy or hard?

nvn

Dory: Let the vertical distance between point A and C be an unknown. You have three unknowns, and you can write three equilibrium equations. The summation of moment equation can be about any point you prefer. After you create your three equilibrium equations, solve the three equations for the three unknowns.

Femme_physics

I can't get to the point of 3 equations and 3 unknowns because I don't have AC and I don't have the angle...here, look, this is where I keep getting stuck at:

Sum of all moments on A = 0 = T(sin alpha) x 1.41 - mg x 1.41/2 = 0
Sum of all moments on B = 0 = -NA x 1.41 + mg 1.41/2 = 0
Sum of all moments on mg = 0 ; -Na(cos45) x 1 + T(sin alpha) x 1 = 0

I don't know how to include AC is one of the unknowns....regardless, since the angle is also an unknown I don't see how can this be solved...

tiny-tim

Hi Dory!

(btw, it's moments about a point (or axis), not on a point )

There's usually no reason to take moments (of forces on the same body) about two different points (the second equation usually doesn't give any extra information that can't more easily be obtained from a linear equation) …

take moments about one point, and use linear components of force to get the other equations …

in this case, the moments about A (i don't think yours are correct, btw) together with the two equations for x and y components should do it

however, it would be simpler in this case to notice that all three forces must go through the same point (otherwise the moment about any one meeting point would be non-zero) …

so you can re-draw the diagram so that the wire goes through the meeting point of the nomal force and the weight, and finding the angle then become a pure geometry question!

try it both ways

Femme_physics

Last time I've taken this advice it didn't work out well for either me or my boyfriend, but I'll try it again just for you ;)....

[hehe...sorry...had to go there]



Okay, back to the problem-- I've tried taking the easier route by drawing a triangle and solving it in a geometrical fashion. Yes, I know where those points meet, shown in my last uploaded img. I have a right triangle. I know mg, and I know the angle opposite to T (90 degrees). I don't know anything else.



Uploaded with ImageShack.us

With respect to sum of all forces - you're right, I should've used these equations instead, but I still don't see how they let me find the angle....I hope I'm not being lazy here... I'm really trying my best to solve it but there's a limit to how many times a girl can write statics equations and fail! Regardless, if you say it can be solved in a geometrical fashion I rather check that out instead of working myself to death again with equations. I still don't see how I can find the angle.

tiny-tim

Hi Dory!
ohhh … I assumed you were a John …

being a goldfish, I naturally immediately thought of John Dory
Do you mean http://www.physicsforums.com/attachm...4&d=1296732455, in post #27, with the cute picture of you?

But your T there doesn't go through the black dot … the forces should all meet there, which is immediately gives you tanα = 2.
Let's review the options:

You can take moments about that dot, which gives you α geometrically, and then you can find T from the y components.

You can take moments about some other point, such as A, which gives you an equation for T and α, and then again use the y components to get another equation for T and α (or use a vector triangle ).

Femme_physics

And you didn't think of Dory without the John? :)



But then, there is no point where all the forces meet! Presuming my diagram is correct, those forces will go on forever and never meet at 3 places. There are 2 forces that meet at 2 points, though...

If I get there and just there that'll be good enough.

This I've tried, but I'll put it on the backburner for now.. I wanna get the geometry way right first. Sounds easier.... "sounds"...

tiny-tim

Oh, that Dory!!

But … isn't she famous for being pretty … but a bit dumb?
"presuming"?!

Dory, your diagram is wrong!!
It is easier …

but you seem to have such an aversion to drawing forces on diagrams that you can't bring yourself to drawing three together!

Femme_physics

hehe....

Hey um.... *scratches head* I...feel..... strangely stung....

Right, the diagram (where I've drawn myself) is wrong, but the last pic I've uploaded is of a triangle presuming that all 3 points meet somewhere between C and B. I only know for a fact it's a right triangle since NA and MG are straight lines....what of T... how can I find it out if I only have 1 angle (90) and 1 force known.

No no...no aversion! here!

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tiny-tim

Then why is your diagram so wonky?

Dory, I've decided that you need a lesson in drawing …

Draw the wall first.

Draw the beam at 45°, to go exactly 8 squares horizontally (as you've done).

Draw the vertical line through the centre of the beam (yes! it can be drawn vertically!! ).

Draw the horizontal line from the top of the beam, to meet the vertical line.

Then say to yourself … "Since the beam is in equilibrium, the only three forces on it must all go through the same point, so I must draw the string through that point, if I can force myself to do it … yes I can! I can! I can! " …

and then draw the string through that point, to the wall.

You should now have two triangles on the same base, one with twice the height of the other

Femme_physics

Okay, I've redrawn it. Here, do you mean the triangles I filled in in the uploaded pic?

I still don't see how can I find out their angles!

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tiny-tim

No, I meant the two triangles you get if you use the horizontal line from the bottom of the beam to the wall as the base of the triangles.

Femme_physics

Are these the two triangles then?

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tiny-tim

No!

The two triangles both with the same baseline, one with hypotenuse along the beam, the other along the string.

(oh, and make the baseline go from the end of the string, not from the adjacent corner of the beam that's almost a square away! )

Femme_physics

Are these?

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tiny-tim

Hunky-dory!

ok. now you know the base and the heights of both triangles …

sooo … ?

Femme_physics

Okay, fine, so now I got the ratios in meters of the triangle I need. So, I use the ratios in Newtons now to find for T. When I do that T = 109 [N]...the answer book says T = 109 [N]... which means that... wait a second.. I'm getting the same answer as the answer book... YEAAAAAAAAAAAAAAAAA BABYY :DD :D :D

//

;)


The entire solution posted here...thank you so much tiny-tim! I even wrote you a special thanks on the paper :) :) :)

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tiny-tim

awww! you've even given me a smilie-face!

see you around!