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Need help with the second part:

Need help with the second part:-example-1
User Orhun
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1 Answer

3 votes

Answer:

3.27 meters

Explanation:

The given equation gives a y(0) = 0, for the deflection of 3 cm. Then for a deflection of 2.5 cm, we want y(x) = 0.005, half a centimeter above the low point on the beam. Using this value, we can solve for x:

0.005 = 3/6400x^2

32/3 = x^2

4√(2/3) = x ≈ 3.26599 ≈ 3.27 . . . . meters

The deflection of the beam is 2.5 cm at a distance of 3.27 meters from center.

Need help with the second part:-example-1
User Unexpectedvalue
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