(33%) Problem 3: Suppose that we have two hanging rods of the same length 2.8 m.

Question

(33%) Problem 3: Suppose that we have two hanging rods of the same length 2.8 m. However, their cross sections are a circle with radius 2.5 mm and a square with sides (2 2.5) mm. Both rods have the same Young's Modulus of 2.5 x 1011 N/m2 50% Part (a) If we attach a heavy ball of mass 24 kg at the end of the first rod (the one with the circular cross section how much would it in millimeters? 50% Part (b) If we attach the same ball ofmass 24 kg at the end of the second rod (the one with the square cross section), how much would it

Explanation / Answer

part A)

L = 2.8 m

Y = 2.5 *10^11 N/m^2

increase in length = Stress/Y * L

increase in length = (24 * 9.8/(pi * (0.0025)^2))/(2.5 *10^11) * 2.8

increase in length = 0.000134 m

the increase in length is 0.000134 m

part B)

L = 2.8 m

Y = 2.5 *10^11 N/m^2

increase in length = Stress/Y * L

increase in length = (24 * 9.8/(pi * (0.002 * 0.0025)))/(2.5 *10^11) * 2.8

increase in length = 0.000168 m

the increase in length is 0.000168 m

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