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Physics 21 Online
OpenStudy (anonymous):

The 25-kg bar has a center of mass at G. If it is supported by a smooth peg at C, a roller at A, and cord at AB, determine the reactions at these supports.

OpenStudy (anonymous):

OpenStudy (anonymous):

@TuringTest Are you familiar with Equilibrium of a Rigid Body problems?

OpenStudy (anonymous):

This is a statics question isn't it

OpenStudy (anonymous):

Yes it is.

OpenStudy (anonymous):

I can give it a try, but I am taking the course as well so it might not be correct haha

OpenStudy (anonymous):

Haha alright. @agent0smith would you be able to help us with this?

OpenStudy (anonymous):

LOL! You and I have the exact same book, I have it as one of the fundamental problems in chapter 5

OpenStudy (anonymous):

:P Yeah! Question F5-5.

OpenStudy (anonymous):

Doesn't your book have the solutions to the fundamental problems in the back of the book?

OpenStudy (anonymous):

If not I can send you the picture for it....it doesn't explain it fully, but it gives you a better hint to it on what they found for the forces in the x, y directions and the moment about point A

OpenStudy (anonymous):

Looking at it though, I am pretty sure I understand how the free body diagram looks like, however

OpenStudy (anonymous):

Okay, I found it. It was in the middle of my book because mines a combination of Statics and Dynamics. That's the answer, but I could still use some explanations on what they did.

OpenStudy (anonymous):

Alright well let me see if I can explain it (this is perfect by the way because I have an exam on this this week one chapters 1 through 6 so it is good practice)

OpenStudy (anonymous):

Alright I got the sum of all the forces in the x direction correct if you want some explanation on that

OpenStudy (anonymous):

Okay!

OpenStudy (anonymous):

Alright so looking at the picture we can see that we have 2 things that are in the x direction.....the tension and the normal of the peg

OpenStudy (anonymous):

Or in other words the normal of the bar on the peg

OpenStudy (anonymous):

The tension of the rope will be the Tcos(15) The normal of the bar on the peg will be (N_c)cos60 Thus our equation for the x direction is Tcos(15) - (N_c)cos60 = 0

OpenStudy (anonymous):

I see the tension, I don't understand the second part. What does the normal of the bar mean? Where does the 60 come from?

OpenStudy (anonymous):

From what I understand, the bar is being supported by the peg or in other words, the peg is sticking into the opening of the bar kind of like a smooth rod and the bar is resting on it.

OpenStudy (anonymous):

The normal of the bar on the peg just means that.....well let me draw a picture

OpenStudy (anonymous):

|dw:1395871279679:dw|

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