Finding angle of inclination. Hanging object.

  • Thread starter EVriderDK
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  • #1
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Homework Statement



I have to find the angles from horizontal of AB=30cm and BC=40cm.
http://peecee.dk/upload/view/356001
I don't know how to tackle this one.

I was told that i could see them as lever arms, and find out at which angle they cancel eachother out. I'm not sure how to do that.

Thanks you in advance.
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi EVriderDK! Welcome to PF! :wink:

Draw the forces (the two weights), and take moments about B …

what do you get? :smile:
 
  • #3
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Thank you :)

I could do that, if it was a straight line, but how to do it when it is bend in 90 degrees?
 
  • #4
tiny-tim
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Have you done moments of forces?
 
  • #6
tiny-tim
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But do you know how to calculate the moment of a general force about a general point? :confused:
 
  • #7
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Yes. It is just Force/length ?
 
  • #8
tiny-tim
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Yes. It is just Force/length ?

Noooo :redface:

You need to look it up in your book.

(and i'm off to bed :zzz:)
 
  • #9
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Woops meant F*r :D
 
  • #10
tiny-tim
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  • #11
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Radius or length
 
  • #12
tiny-tim
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of what?

from what to what?
 
  • #13
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From the "rotation point" B to the end of an arm/lever.
 
  • #14
tiny-tim
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From the "rotation point" B to the end of an arm/lever.

ah, no :redface:

it's the perpendicular distance from B to the line of the force

(in this case, the force is the weight, so the line of the force is the vertical line through the centre of the rod)

another formula is F.d (or Fdsinθ)

where d is the vector from B to the centre of mass, and θ is the angle between F and d

see http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html :wink:
 
  • #15
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Hmm.

B is a point. How can something be perpendicular to a point?

Centre of the mass? So i have to make a triangle out of it, and find the center of gravity?
 
  • #17
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I'm looking at the site, but it doesn't make sense. Maybe i'm just not that good at english :) It's confusing.

I don't know what you mean by F.d. What does "." represent ?

If the Force is the force of gravity, and the force of gravity is a vertical force through the point B, how can i make a line perpendicular to the force(Fg) and B?
 
Last edited:
  • #18
tiny-tim
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I don't know what you mean by F.d. What does "." represent ?

That's the dot product (or scalar product), see http://hyperphysics.phy-astr.gsu.edu/hbase/vsca.html

If you don't like the idea of dot product, jsut use the Fdsinθ formula :smile:
If the Force is the force of gravity, and the force of gravity is a vertical force through the point B, how can i make a line perpendicular to the force(Fg) and B?

No, you make it through B (or from B), and perpendicular to that vertical line.

In other words: if d is the distance from B to the centre of the rod,

then dsinθ is the horizontal distance from B to the vertical line through the centre of the rod.​
 
  • #19
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So d is a line from the middle of AB, perpendicular to AB, which goes to F_g, which is the vertical force through B ?

Dot product and vector calculations is long gone. Cannot remember how it works, so another method is preferrable :)
 
  • #20
tiny-tim
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So d is a line from the middle of AB, perpendicular to AB, which goes to F_g, which is the vertical force through B ?

No!!

Read the instructions

it's perpendicular to the force

and it's through B (the pivot point)​
 
  • #21
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I found another solution, where the Fg goes vertically down in the middle of AB(a). From that point you draw a vertical line to the top, and a horizontal line to the pivot point B. The same with BC(c). To calculate the ange, i first have to find out, where the two moments are =0. So that will be F_g_a*Cb*sin(alpha)-F_g_b*Ca*cos(alpha)=0.

Right?
 
  • #22
tiny-tim
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I found another solution …

That is the solution I've been telling you about!
… where the Fg goes vertically down in the middle of AB(a). From that point you draw a vertical line to the top, and a horizontal line to the pivot point B. The same with BC(c). To calculate the ange, i first have to find out, where the two moments are =0. So that will be F_g_a*Cb*sin(alpha)-F_g_b*Ca*cos(alpha)=0.

Right?

You mean F_g_a*Ba*sin(alpha)-F_g_c*Bc*cos(alpha)=0 ?

Right! :smile:

(if alpha is what I think it is)

Now what is the ratio of F_g_a and F_g_c? :wink:
 
  • #23
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60.64deg/29.538=2.053

1:2.053 ?
 
  • #24
tiny-tim
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Now what is the ratio of F_g_a and F_g_c? :wink:
60.64deg/29.538=2.053

1:2.053 ?

uhh? :confused:

F_g_a/F_g_c = weight of BA / weight of BC = length BA / length BC
 

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