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?
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
To calculate the tension in the rope, we can use the equation Tension = Weight. The weight of your friend can be calculated by multiplying their mass by the acceleration due to gravity (9.8 m/s²).
Weight = mass × acceleration due to gravity
Weight = 75 kg × 9.8 m/s²
Weight = 735 N
So, the tension in the rope would be 735 Newtons.
Now, let's move on to determining the minimum diameter of the rope required. We know that the rope can withstand 1.25 kilonewtons (kN) of stress per square centimeter (cm²) of cross-sectional area.
Stress = Force / Area
To determine the force on the rope, we first need to convert the tension from Newtons to kilonewtons (1 kN = 1000 N).
Tension = 735 N = 0.735 kN
Now, we can rearrange the equation to solve for the area of the rope's cross-section:
Area = Force / Stress
Area = 0.735 kN / 1.25 kN/cm²
Area = 0.588 cm²
We can use the formula for the area of a circle to find the minimum diameter (D) of the rope:
Area = π × (D/2)²
0.588 cm² = π × (D/2)²
Now, let's solve for D:
(D/2)² = 0.588 cm² / π
(D/2)² = 0.18745
D/2 = √(0.18745)
D/2 ≈ 0.433
D ≈ 0.433 × 2
D ≈ 0.866 cm
So, the minimum diameter of the rope needed is approximately 0.866 centimeters.