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The legs of a tripod make equal anges of 90 degrees with each other at the apex.What are the compressional forces in three legs?

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The legs of a tripod make equal anges of 90 degrees with each other at the apex.What are the compressional forces in three legs?

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lightgrav

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The compression forces depend on the weight that they support.

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If m is the mass of the object kept,

]each has compressional force=

[tex]\frac{mg}{\sqrt{3}}[/tex]

]each has compressional force=

[tex]\frac{mg}{\sqrt{3}}[/tex]

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Balakrishnan v can u clarify how do you get to [tex]\frac{mg}{\sqrt{3}}[/tex]

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HallsofIvy

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First draw a picture. If the three legs of a tripod are 90 degrees to each other, then it's like one corner of a cube. Taking the length of each leg to be 1, the line segment connecting two legs has length [itex]\sqrt{2}[/tex] (Pythagorean theorem). That means the base is an equilateral triangle with sides of length [itex]\sqrt{2}[/itex]. It's then not too difficult to calculate that the altitude is [itex]\sqrt{\frac{3}{2}}[/itex] and that each leg of the tripod is 2/3 of that, [itex]\sqrt{\frac{2}{3}}[/itex], from the center. By the Pythagorean theorem, the height of the tripod above the ground is [itex]\sqrt{\frac{3}{3}}[/itex]. (I'm sure an obvious answer like that could have been worked out more easily!)

Now, by symmetry, each leg must support 1/3 of the weight: mg/3**vertically**. That means that, letting T be the compression in one leg, we must have

(ratio of forces equal to ratio of lengths)

[tex]\frac{T}{\frac{mg}{3}}= \frac{1}{\frac{\sqrt{3}}{3}}[/tex]

[tex]T= \(\frac{mg}{3}\)\(\frac{3}{\sqrt{3}}\)= \frac{mg}{\sqrt{3}}[/tex]

Notice that that is larger than just [itex]\frac{mg}{3}[/itex]. The extra compression comes from the force of the legs pressing together and pressing into the ground to keep the legs from spreading apart, the horizontal component of force on the legs.

(Nicely done, Balakrishnan v. I was just going to answer "it depends on the weight on the tripod!)

Now, by symmetry, each leg must support 1/3 of the weight: mg/3

(ratio of forces equal to ratio of lengths)

[tex]\frac{T}{\frac{mg}{3}}= \frac{1}{\frac{\sqrt{3}}{3}}[/tex]

[tex]T= \(\frac{mg}{3}\)\(\frac{3}{\sqrt{3}}\)= \frac{mg}{\sqrt{3}}[/tex]

Notice that that is larger than just [itex]\frac{mg}{3}[/itex]. The extra compression comes from the force of the legs pressing together and pressing into the ground to keep the legs from spreading apart, the horizontal component of force on the legs.

(Nicely done, Balakrishnan v. I was just going to answer "it depends on the weight on the tripod!)

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Thanx all.

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