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Old 12-07-2010, 01:29 PM   #1
Ingolf
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Default A Bar Suspended by Two Wires


A nonuniform, horizontal bar of mass m is supported by two massless wires against gravity. The left wire makes an angle phi_1 with the horizontal, and the right wire makes an angle phi_2. The bar has length L.



Find the position of the center of mass of the bar, x, measured from the bar's left end.
Express the center of mass in terms of L, phi_1, and phi_2.
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Last edited by Ingolf; 12-07-2010 at 02:50 PM.
Old 12-07-2010, 06:21 PM   #2
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Default Re: A Bar Suspended by Two Wires

zz
Edit: I wrote ZZ coz i have no idea how to solve this question, not because i think that this problem is lacking data. I wouldnt know if it is.
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Last edited by 3.14159; 12-09-2010 at 10:48 AM.
Old 12-08-2010, 12:04 AM   #3
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Default Re: A Bar Suspended by Two Wires

The problem is lacking data.
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Old 12-08-2010, 07:41 PM   #4
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Default Re: A Bar Suspended by Two Wires

No it wasn't. I finally solved it using equilibrium.
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Last edited by Ingolf; 12-13-2010 at 04:33 AM.
Old 12-08-2010, 11:18 PM   #5
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Default Re: A Bar Suspended by Two Wires

Show me your solution, I insist the problem lacks data.
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Old 12-08-2010, 11:26 PM   #6
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Default Re: A Bar Suspended by Two Wires

How exactly were you able to solve this? And what formula? Torque?
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Old 12-08-2010, 11:58 PM   #7
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Default Re: A Bar Suspended by Two Wires

How did you use equilibrium?

The strings are massless so mg=0
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Old 12-12-2010, 06:20 PM   #8
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Default Re: A Bar Suspended by Two Wires

My solution:

The bar is in equilibrum so the net torque and forces acting on it must be zero. I take the torque of the left end of the bar.

The formula for the net torque looks like this:

(1) L*T_2*sin(phi_2)-w*x=0

where T_2 is the tension in wire #2 and w is the weight of the bar (m*g).

The net forces in x- and y-directions must also be 0, so we have

(2) F_x: T_2*cos(phi_2)-T_1*cos(phi_1)=0
(3) F_y: w-T_1*sin(phi_1)-T_2*sin(phi_2)

where T_1 is the tension in wire #1. I have taken the positive y-direction to be downward so that w (weight) is positive.

By rearranging (2) I get that

(4) T_1/T_2=cos(phi_2)/cos(phi_1)

I solve for x in (1) and get

(5) x=L*T_2*sin(phi_2)/w

Solving for w in (3) and inserting in (5) gives

(6) x=L*T_2*sin(phi_2)/(T_1*sin(phi_1)+T_2*sin(phi_2))
x=L/((T_1/T_2)*(sin(phi_1)/sin(phi_2))+1)

By inserting (4) in (6) I get that

x=L/(cos(phi_2)*sin(phi_1)/(cos(phi_1)*sin(phi_2))+1)
x=L/(tan(phi_1)*tan^(-1)(phi_2)+1)=L/(tan(phi_1)/tan(phi_2)+1)
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Old 12-12-2010, 08:03 PM   #9
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Default Re: A Bar Suspended by Two Wires

I guess you dont use flexibility at all.... But meh, that is ok.
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Last edited by Shamanics; 12-12-2010 at 08:21 PM.
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