A crane is lifting a wrecking ball supported by a 12 m long, 2.2 cm diameter ste

Question

A crane is lifting a wrecking ball supported by a 12 m long, 2.2 cm diameter steel cable (Ultimate Strength = 560 MN/m2). The mass of the wrecking ball is 920 kg. The modulus of elasticity of the cable is 570x103 MN/m2.

When the wrecking ball is stationary, by how much is the cable stretched?

If the wrecking ball is accelerating down at 3.5 m/s2, by how much is the cable stretched?

If the wrecking ball is accelerating up at 3.5 m/s2, by how much is the cable stretched?

For the three cases examined in parts A-C, what is the minimum factor of safety?

Explanation / Answer

using stress = (modulus of elasticity )(strain)

(F /A) = Y ( deltaL / L)

((920 x 9.81)/ ( pi x 0.011^2)) = (570 x 10^3 x 10^6) ( deltaL / 12)

deltaL = 5 x 10^-4 m ....Ans

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accelerating down then using Fnet = ma

mg - F = ma

F = m(g -a)

using the same equation,
((920 x (9.81-3.5))/ ( pi x 0.011^2)) = (570 x 10^3 x 10^6) ( deltaL / 12)

deltaL = 3.21 x 10^-4 m

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accelerating upthen using Fnet = ma

F - mg = ma

F = m(g + a)

using the same equation,
((920 x (9.81+3.5))/ ( pi x 0.011^2)) = (570 x 10^3 x 10^6) ( deltaL / 12)

deltaL = 6.78 x 10^-4 m

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