Question

дам 30 балов: Лифт, висящий в лифтовой шахте на тросе, опустился на 2 см, когда в него зашли два человека массой 90 кг и 70 кг. Определите жёсткость троса.​

Answer 1

Ответ:

80 кН/м

Объяснение:

x=0.02 м    m1=90 кг    m2=70 кг    k=?

===

F=k*x

F=(m1+2)*g

k=F/x=(m1+m2)*g/x=(90+70)*10/0.02=80000 Н/м

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Answer 2

Calculation of the Cable Stiffness

To determine the stiffness of the cable, we can use the concept of Hooke's Law, which states that the force applied to a spring or elastic material is directly proportional to the displacement it undergoes. In this case, the cable acts like a spring, and the force applied to it is the weight of the people in the elevator.

The formula for Hooke's Law is:

F = k * x

Where: - F is the force applied to the cable (weight of the people) - k is the stiffness of the cable - x is the displacement of the cable (2 cm in this case)

To find the stiffness of the cable, we need to calculate the force applied to it. The force is equal to the sum of the weights of the two people:

F = m1 * g + m2 * g

Where: - m1 is the mass of the first person (90 kg) - m2 is the mass of the second person (70 kg) - g is the acceleration due to gravity (approximately 9.8 m/s^2)

Substituting the values into the equation, we get:

F = (90 kg * 9.8 m/s^2) + (70 kg * 9.8 m/s^2)

Calculating the force:

F = 882 N + 686 N = 1568 N

Now, we can rearrange Hooke's Law to solve for the stiffness:

k = F / x

Substituting the values into the equation, we get:

k = 1568 N / 0.02 m = 78400 N/m

Therefore, the stiffness of the cable is 78400 N/m.