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What fraction of the sign's weight is supported by each string?

  • Thread starter decathlonist
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  • #1
decathlonist
I am stuck on one of the AP Physics problems. Could you please help me out and give some pointers on how to solve the following problem:

A hemispherical sign 1 m in diameter and of uniform mass density is supported by two strings. What fraction of the sign's weight is supported by each string?

[see attachment for the diagram]

Thank you
 

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  • #2
russ_watters
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[searches in vain for an attachment...]
 
  • #3
HallsofIvy
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I was able to down load the attachement without any trouble.

I shows a semicircle with two lines (the supporting cables) going up from (first cable) one end of a diameter and (2nd cable) 0.75 meters along (I assume) the diameter and so 0.25 meters from the other end.

Let W be the weight of the sign, F1 and F2 be the forces on the respective cables. Obviously F1+F2= W. The total torque must be 0 (about any point- in particular about the point of attachment of cable 1) so -W(1/2 meter) (the torque due to the weight of the sign)plus F_2(0.75) must be 0.
(0.75)F_2- 0.5 W= 0 or F_2= (0.5/0.75)W= (2/3)W.

From F_1+ F_2= W we get F_1+ (2/3)W= W or F_1= (1/3)W.

The cable at the back is supporting 1/3 of the weight and the one in front is supporting 2/3 of the weight.
 

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