A crane is lifting a wrecking ball supported by a 27 m long, 1.9 cm diameter ste

Question

A crane is lifting a wrecking ball supported by a 27 m long, 1.9 cm diameter steel cable (Ultimate Strength = 540 MN/m2). The mass of the wrecking ball is 870 kg. The modulus of elasticity of the cable is 570x103 MN/m2.

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

(include units with answer)

A Mega-Newton is 1,000,000 Newtons

B.If the wrecking ball is accelerating down at 3.4 m/s2, by how much is the cable stretched?

(include units with answer)

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

A Mega-Newton is 1,000,000 Newtons

Explanation / Answer

given

mass of ball = m = 870 kg

length of supporting cable, l = 27 m

diameter of cable, d = 1.9 cm

Ultimate strength of cable, U = 540*10^6 N/m^2

Modulus of elasticity of cable, E = 570*10^9 N/m^2

1. for stationar ball, let the strech in cable be dl

then stress = 4mg/pid^2

strain = dl/l

so from hookes law

stress/straikn = E

4mg*l/pi*d^2*dl = E

putting in values

4*870*9.81*27/pi*0.019^2*dl = 570*10^9

dl = 1.425 mm

2. when wrecking ball is accelerating down at a = 3.4 m/s/s

stress = 4F/pi*d^2

where mg - F = ma

F = m(g - a)

hence

4*870*(9.81 - 3.4)*27/pi*0.019^2*dl = 570*10^9

dl = 0.9316 mm

3. for upward acceleration of a = 3.4 m/s/s

stress = 4F/pi*d^2

where mg - F = -ma

F = m(g + a)

hence

4*870*(9.81 + 3.4)*27/pi*0.019^2*dl = 570*10^9

dl = 1.92 mm

4. factor of safety in case 1 : stress/ultim atestrength = 4*870*9.81/pi*0.019^2*540*10^6 = 0.055 < 1 so its safe

factor of safety in case 2 : stress/ultim atestrength = 4*870*(9.81-3.4)/pi*0.019^2*540*10^6 = 0.0364 < 1 so its safe

factor of safety in case 3 : stress/ultim atestrength = 4*870*(9.81 + 3.4)/pi*0.019^2*540*10^6 = 0.075 < 1 so its safe

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