Two cords support a chandelier in the manner shown in the diagram with the upper wire making an angle of 45 degrees with the ceiling. If the cords can sustain a force of 1300 N without breaking, what is the maximum chandelier weight that can be supported?

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AGRFTr · Accepted Answer

T1sin⁡45=1300  T1=1300sin⁡45  =1838.477 N  maximum chandelier weight = 1838.47N

Jeffrey Jordon · Accepted Answer

We know that the chandelier is not necessarily 200 kg either, but some mass such that the maximum tension in the cords is 1300 N.  Let&#039;s just set up some equations to see which cord is under the most tension, and then, we&#039;ll set that tension to 1300 N&amp;gt;  If we look at the point where the cords come together, you can express the equilibrium of that point as follows:  (The vector sum of F1,F2, and F3 - where  is the weight of the chandelier)  In the x direction: F1x+F2x+F3x=0 −F1cos⁡(45∘)+F2x+0=0  So we know that F2x=F2 (It has no y component) = F1cos⁡(45∘)  In the y direction F1y+F2y+F3y=0 F1sin⁡(45∘)+0−weight=0  So we know that  F1=weightsin⁡(45∘)  And F2=F1cos⁡(45∘)=weightsin⁡45∘cos⁡(45∘)=weight(since sin⁡(45∘)=cos⁡(45∘))  So F1 is always going to be larger than F2 and F3 which are both equal to the weight, since sin⁡(45∘) is less than one,  and F1=weightsin⁡(45∘)  Now let&#039;s solve the problem, making F1 just equal to 1300 N, as this would be the maximum it could be:  In the x direction: F1x+F2x+F3x=0 −(1300N)cos⁡(45∘)+F2x+0=0 −919.24 N+F2x+0=0  So F2x=919.24 N  In the y direction F1y+F2y+F3y=0 (1300N)sin⁡(45∘)+0−weight=0 919.24 - weight = 0  So the weight is 919.24 N = 920 N

Two cords support a chandelier in the manner shown in the diagram with the upper wire making an angle of 45 degrees with the ceiling.

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