Your friend, a 75 kg mountain climber is suspended from a cliff by a rope. Calculate the tension in the rope, in Newtons, when you lift your friend to safety. If the rope can withstand 1.25 kN of stress per square centimeter of cross-sectional area, what minimum diameter of rope (cm) is needed so that your friend does not fall into the abyss?

Answer 1

1kg=9.81newton
75kg(9.81)=735.75newton tension

stess= p/a

1.25kn/cm^2=0.73575kn/a

a=1.6989 cm^2

a=3.1415/4(d^2)
1.6989=3.1415/4(d^2)
d=1.471 cm

edsel salariosa
research and development engineer