An elevator (mass 4025 kg) is to be designed so that the maximum acceleration is 0.0700g.
What is the maximum force the motor should exert on the supporting cable?(N)
What is the minimum force the motor should exert on the supporting cable?(N)
I am so sorry but ellll tothero cannot teach so.
To find the maximum and minimum forces exerted by the motor on the supporting cable, we need to consider the maximum acceleration of the elevator and the force of gravity acting on the elevator.
To find the maximum force, we will use the maximum acceleration and the mass of the elevator. The maximum force is given by Newton's second law, which states that force (F) equals mass (m) multiplied by acceleration (a).
First, we need to convert the acceleration from a percentage of gravity (0.0700g) to a value in meters per second squared (m/s²). The acceleration due to gravity is approximately 9.8 m/s².
Maximum acceleration (a) = (0.0700g) * (9.8 m/s²)
= (0.0700) * (9.8 m/s²)
= 0.686 m/s²
Now, we can find the maximum force by multiplying the maximum acceleration by the mass of the elevator.
Maximum force (F_max) = mass (m) * acceleration (a)
= 4025 kg * 0.686 m/s²
To find the minimum force exerted by the motor on the supporting cable, we consider the force of gravity acting on the elevator. The minimum force would occur when the elevator is moving at a constant speed, and the net force is zero. In this case, the force exerted by the motor should exactly balance the force of gravity.
Minimum force (F_min) = weight of the elevator
= mass (m) * gravity (g)
= 4025 kg * 9.8 m/s²
Therefore, the maximum force the motor should exert on the supporting cable is 2771 N (rounded to the nearest whole number), and the minimum force the motor should exert on the supporting cable is 39445 N (rounded to the nearest whole number).