12072010, 02:29 PM  #1 
Member
Join Date: Jun 2009
Posts: 1,211

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.
__________________

Last edited by Ingolf; 12072010 at 03:50 PM. 
12072010, 07:21 PM  #2 
Member
Join Date: Jun 2009
Location: India
Posts: 2,119

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. 
Last edited by 3.14159; 12092010 at 11:48 AM. 
12082010, 01:04 AM  #3 
Banned
Join Date: Aug 2009
Location: Deep in the ocean
Posts: 2,286

Re: A Bar Suspended by Two Wires
The problem is lacking data.

12082010, 08:41 PM  #4 
Member
Join Date: Jun 2009
Posts: 1,211

Re: A Bar Suspended by Two Wires
No it wasn't. I finally solved it using equilibrium.
__________________

Last edited by Ingolf; 12132010 at 05:33 AM. 
12092010, 12:18 AM  #5 
Banned
Join Date: Aug 2009
Location: Deep in the ocean
Posts: 2,286

Re: A Bar Suspended by Two Wires
Show me your solution, I insist the problem lacks data.

12092010, 12:58 AM  #7 
Member
Join Date: Jul 2009
Posts: 96

Re: A Bar Suspended by Two Wires
How did you use equilibrium?
The strings are massless so mg=0 
12122010, 07:20 PM  #8 
Member
Join Date: Jun 2009
Posts: 1,211

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 ydirections must also be 0, so we have (2) F_x: T_2*cos(phi_2)T_1*cos(phi_1)=0 (3) F_y: wT_1*sin(phi_1)T_2*sin(phi_2) where T_1 is the tension in wire #1. I have taken the positive ydirection 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)
__________________

12122010, 09:03 PM  #9 
Banned
Join Date: Aug 2009
Location: Deep in the ocean
Posts: 2,286

Re: A Bar Suspended by Two Wires
I guess you dont use flexibility at all.... But meh, that is ok.

Last edited by Shamanics; 12122010 at 09:21 PM. 