An elevator (mass 4125 kg) is to be designed so that the maximum acceleration is 0.0300g.
What is the maximum force the motor should exert on the supporting cable?
What is the minimum force the motor should exert on the supporting cable?
F= ma= Tension-Weight
ma+Weight=Tension
(4125kg)(0.0300 * 9.8)+ (4125kg)(9.81)=Tension
Calculate and this will give you the answer for maximum force.
For part (2)
You have to subtract them (ma-weight)
To determine the maximum and minimum forces exerted by the motor on the supporting cable, we need to consider the weight of the elevator and the maximum acceleration.
Let's start by calculating the weight of the elevator using the formula:
Weight = mass * gravity
where the mass is given as 4125 kg and the acceleration due to gravity is approximately 9.8 m/s^2.
Weight = 4125 kg * 9.8 m/s^2 = 40,425 N
Next, let's calculate the maximum force the motor should exert on the supporting cable. This force will be equal to the weight of the elevator plus the force required to accelerate it.
Max Force = Weight + m * a
where m is the mass of the elevator and a is the maximum acceleration, given as 0.0300g.
Max Force = Weight + 4125 kg * 0.0300 * 9.8 m/s^2
Max Force = 40,425 N + 120 N = 40,545 N
Therefore, the maximum force the motor should exert on the supporting cable is 40,545 N.
To find the minimum force, we need to consider the minimum acceleration. Since the elevator is at rest initially, the minimum force required to start the motion is equal to the weight of the elevator.
Therefore, the minimum force the motor should exert on the supporting cable is 40,425 N.
To find the maximum and minimum forces the motor should exert on the supporting cable, we can use Newton's second law of motion:
F = m * a
where F represents the force, m is the mass, and a is the acceleration.
First, let's find the maximum force when the elevator is accelerating upwards:
Given:
Mass of the elevator (m) = 4125 kg
Maximum acceleration (a) = 0.0300g
To find the maximum force (Fmax), we need to convert the acceleration to meters per second squared (m/s^2) since g is the acceleration due to gravity:
a = 0.0300g
where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
a = 0.0300 * 9.8 m/s^2
a ≈ 0.294 m/s^2
Now we can calculate the maximum force (Fmax):
Fmax = m * a
Fmax = 4125 kg * 0.294 m/s^2
Fmax ≈ 1211.55 N
Therefore, the maximum force the motor should exert on the supporting cable is approximately 1211.55 Newtons.
Next, let's find the minimum force when the elevator is accelerating downwards. In this case, the acceleration will be negative, so we can use the same approach.
Given:
Mass of the elevator (m) = 4125 kg
Minimum acceleration (a) = -0.0300g
To find the minimum force (Fmin):
a = -0.0300 * 9.8 m/s^2
a ≈ -0.294 m/s^2
Fmin = m * a
Fmin = 4125 kg * (-0.294 m/s^2)
Fmin ≈ -1211.55 N
Therefore, the minimum force the motor should exert on the supporting cable is approximately -1211.55 Newtons. The negative sign indicates that the force is directed downwards.