Post a New Question

engineering

posted by on .

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?

  • engineering - ,

    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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question