Server Help

[Agurus]

Mmm I cant figure out this ridiculous problem, I hate doing variables only. Can someone take a crack at it for me, I would apreciate if you showed me the steps you did to get there

File b3low



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[Mine GO BOOM]

To help you along without giving answers right out, first list some equations you know that can be used in this example.

[Blocks]

I'll give you some hints with which to get started.

Draw a free-body diagram. Call the horizontal direction x and the vertical direction y. You'll have two y forces from the weights, a force on the line of action of the wire (separate this out into x and y components), and a reaction at point A (defined by Asubx and Asuby).

Write out the equations for equilibrium, which would be the sum of the x forces equals zero, the sum of the y forces equals zero, and the sum of the moments about point A equals zero. You'll be able to write an inequality relating Asubx and Asuby under which equilibrium holds.

Maybe I gave away too much. Oh well, as long as you promise to understand what you're doing. There might be a simpler way to do it too.

[Agurus]

Well here are a few I think we could use:

(SUM OF)F=MA

Fs(max) = Us*Fn

W=MG=-Fn

Net External Torque= I * angular acceleration

or

SUM OF Torques = SUM OF MASSES r^2*alpha

Xcm= (m1(change in X))+m2(change in x2)/ (m1+m2)

I think we are supposed to split it to X and Y components but then have no idea what to do with the torques.

[Cyan~Fire]

Pay attention to blocks, you need to use the three equilibrium equations.
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[Agurus]

O_o didn't see his post, hmmm

[CypherJF]

i dreaded physics... still not sure how i passed the classes? (C in the first semester and B in the following (which the material was a lot harder)). *shrug*
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[Agurus]

So assist me mgb

[Mine GO BOOM]

[Agurus]

Posts: 3001

Gj mgb ;P



Yeah ok I will try that

[Mine GO BOOM]

[Muskrat]

[Cyan~Fire]

[Agurus]

Pardon my handwriting, but here is my work any thing you guy see wrong?

[Cyan~Fire]

Working on a response, I'll edit in a few minutes...

Edit: OK, here's what I have to say. I'm tired right now and I need to get up early tomorrow (... for Physics) so I'll check it in the morning and I'll actually try to solve the problem.

[Agurus]

Fr= Force Reaction, same as Fn basically.

2w= weight of bar
w= weight of ball (go weights from problem)

[Cyan~Fire]

OK, well you're right, it is the normal force. I guess you can call it whatever you want when you're doing your own work, but it would probably be good to call it the normal force when showing your work to others (such as us or your teacher).

So... everything else I said still stands. I realize that w is the weight of the ball (and the bar, reread the problem). So why did you multiply the coefficient of friction by the weight to get the frictional force? Why is the normal force even in (what I'm assuming is) your torque equation?

Finally, you didn't answer my question. What did you pick for your axis?

[Quan Chi2]

I wish I knew how to add.