Three 30mm diameter cables support the car in a cable elevator. Inside the car is a safety notice.
SAFE WORKING LOAD 33 PERSONS 2244 KG
The yield stress of the cables is 200MPa.
What factor of safety has been applied at the safe working load?
Factor of safety=??????????????
To determine the factor of safety applied at the safe working load, we need to compare the ultimate strength of the cables to the actual load they are expected to support.
First, let's calculate the cross-sectional area of one cable. The diameter of each cable is 30mm, so the radius is half of that, which is 15mm or 0.015m.
The formula for calculating the area of a circle is A = πr^2, where A is the area and r is the radius. Plugging in the values, we get:
A = π(0.015)^2 = 0.0007068583 square meters (approximately)
Next, we need to determine the ultimate strength of one cable. The yield stress of the cables is given as 200MPa, which is the stress at which permanent deformation occurs. However, the ultimate strength represents the maximum stress a material can withstand before failure. Typically, the ultimate strength is higher than the yield stress. For simplicity, let's assume the ultimate strength is twice the yield stress: 2 * 200MPa = 400MPa.
The ultimate strength of a material can be expressed in terms of stress (force per unit area) or force. To calculate the ultimate force, we multiply the ultimate strength by the cross-sectional area of one cable:
Ultimate force = Ultimate strength * Area
Ultimate force = 400MPa * 0.0007068583 square meters
Ultimate force = 282.74332 newtons (approximately)
Now we can calculate the safe working load, which is given as 33 persons or 2244 kg. To convert kg to newtons, we use the conversion factor of 9.8 Newtons/kg (acceleration due to gravity). Therefore:
Safe working load = 2244 kg * 9.8 Newtons/kg
Safe working load = 21991.2 newtons (approximately)
Finally, we can determine the factor of safety by dividing the ultimate force by the safe working load:
Factor of safety = Ultimate force / Safe working load
Factor of safety = 282.74332 newtons / 21991.2 newtons
Factor of safety ≈ 0.01285
Therefore, the factor of safety applied at the safe working load is approximately 0.01285.