mass M- 12.0 kg, hangs from the end of a cylindrical steel rod of diameter 10.0
ID: 1777004 • Letter: M
Question
mass M- 12.0 kg, hangs from the end of a cylindrical steel rod of diameter 10.0 cm and A sign, length 1.60 m and a cable anchored to the wall at a= 45.0°· extending horizontally from a wal. The rod is supported by a pivot at the wall steel = 20.0 * 1010 m2 A) List & draw the 4 forces acting on the rod, and the dire torque they produce around the pivot. Ft SIGNS lv B) Find the magnitude of the tension acting in the cable. Sin Ys Sin 166.3 N .ssnor) C) What is the compression in the steel rod due to the tension? Recall that only forces acting perpendicular to the cross-sectional area create tensile compression.Explanation / Answer
B) balancing torque about the right end,
W*L- T sin 45 degree *L =0
T = (W) / sin 45 degree = 12*9.8/( sin 45 degree)
= 166.3 N
C) stress = T cos 45 degree / area
= 166.3 cos 45 degree/(pi*0.05^2)
= 14972 Pa
By Hookes law, Y = stress / strain
compression = strain = stress/Y
= 14972/20e10
= 7.486*10^-8*1.6
= 1.2*10^-7 m answer
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