A bag of cement of weight Fg hangs from three wires as .....??


Kiddo , Wednesday, 11th of August 2010 05:23:33 PM

A bag of cement of weight Fg hangs from three wires as shown in the 
Kiddo
attached diagram (link). Two of the wires make angles theta1 and theta2 
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with the horizontal. If the system were in equilibrium, show that the 
Joined: Monday, 17th of May 2010, 14:26:37
tension in the left hand wire is:

T1 = (Fg.costheta2) / 
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sin(theta 1 + theta2)

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MiLKy , Thursday, 12th of August 2010 07:00:23 AM

Because the system is in equilibrium & there is no  
MiLKy
acceleration, we can deduce the following:  
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Joined: Wednesday, 9th of June 2010, 01:45:39
T3 is balancing the weight. Therefore, T3 = Fg ----> (1)  
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Horizontal components of T1 & T2 are balancing each other. Therefore,  
T1costheta1 = T2costheta2 ----> (2)  
 
And, finally vertical components of T1 & T2 are balancing T3. Therefore,  
T1sintheta1 + T2sintheta2 = T3 ----> (3)  
 
Now we have three equations, to get what is required, we just have to  
manipulate those equations & eliminate T2 & T3, because these terms don't  
appear in the required equation.  
 
T1costheta1 = T2costheta2  
 
=> T2 = T1costheta1 / costheta2  
 
Putting this value in eq (3),  
 
T1sintheta1 + (T1costheta1 / costheta2)sintheta2 = Fg  
 
=> T1 [ sintheta1costheta2 + costheta1sintheta2 ] = Fgcostheta2  
 
=> T1[sin(theta1 + theta2)] = Fgcostheta2 (Because, sin(a + b) =  
sinacosb + cosasinb)  
 
 
=> T1 = (Fgcostheta2) / sin(theta1 + theta2)  
 
 
QED  
 
 
 
 
 



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