Bob is lifting a heavy mass in the barn one day. He plans to use a cable system

Question

Bob is lifting a heavy mass in the barn one day. He plans to use a cable system made from a 1.04mm diameter aluminum wire of length 3.44meters.

a) What is the heaviest mass that he can lift using this wire without the wire breaking?

b) Bob's pet is resting directly under the mass in Bob's barn. In order to not drop the mass on top of his pet, Bob decides to use a copper wire of the same diameter and length as the original aluminum wire. By how much does the wire stretch under the load computed in Part A of the problem?

Explanation / Answer

T = m*g

E*A = m*g

1.1*10^8*pi*(0.52*10^-3)^2 = m*9.8

m = 9.535 kg

b)

Y = Fl/Ae

stretch = e = Fl/AY

e = (m*g*l)/(AY)

e = (9.535*9.8*3.44)/((0.52*10^-3)^2*1.2*10^11)

e = 0.009906 m

e = 9.906 mm

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