a) When the elevator is at rest, the net force acting on it is equal to zero. Therefore, the tension in the cable should be equal to the weight of the elevator.
The weight of the elevator can be calculated using the formula:
Weight = mass * gravity
where mass is the mass of the elevator and gravity is the acceleration due to gravity. Assuming gravity is approximately 9.8 m/s^2, we have:
Weight = 2250 kg * 9.8 m/s^2 = 22050 N
Therefore, the tension in the cable when the elevator is at rest is 22050 N.
b) When the elevator is moving up with an acceleration of 0.55 m/s^2, the net force acting on it is the sum of the tension in the cable and the force due to the elevator's acceleration. We can represent this on a Free Body Diagram (FBD) as follows:
- Tension in the cable (upward)
- Weight of the elevator (downward)
- Force due to acceleration (upward)
The net force can be calculated using the formula:
Net force = mass * acceleration
where mass is the mass of the elevator and acceleration is the acceleration of the elevator. We have:
Net force = 2250 kg * 0.55 m/s^2 = 1237.5 N
Since the elevator is moving up, the tension in the cable is greater than the weight of the elevator. Therefore, we can write:
Tension in the cable - Weight of the elevator = Net force
Tension in the cable - 22050 N = 1237.5 N
Tension in the cable = 1237.5 N + 22050 N
Tension in the cable = 23287.5 N
Therefore, the tension in the cable when the elevator is moving up with an acceleration of 0.55 m/s^2 is 23287.5 N.
c) When the elevator is moving down with an acceleration of 0.55 m/s^2, the net force acting on it is the difference between the tension in the cable and the force due to the elevator's acceleration. We can represent this on a Free Body Diagram (FBD) as follows:
- Tension in the cable (downward)
- Weight of the elevator (downward)
- Force due to acceleration (upward)
The net force can be calculated using the same formula as in part b:
Net force = mass * acceleration
Net force = 2250 kg * 0.55 m/s^2 = 1237.5 N
Since the elevator is moving down, the tension in the cable is less than the weight of the elevator. Therefore, we can write:
Weight of the elevator - Tension in the cable = Net force
22050 N - Tension in the cable = 1237.5 N
Tension in the cable = 22050 N - 1237.5 N
Tension in the cable = 20812.5 N
Therefore, the tension in the cable when the elevator is moving down with an acceleration of 0.55 m/s^2 is 20812.5 N.
Grade 11 physics
Assignment 5 Newton’s Laws
An elevator has a mass of 2250 kg.
a) What is the tension in the cable that supports the elevator when it is at rest? Complete the FBD
b) What is the tension in the cable when the elevator is moving up with an acceleration of 0.55 m/s2? Complete the FBD.
c) What is the tension in the cable when the elevator is moving down with an acceleration of 0.55 m/s2? Complete the FBD.
1 answer
1 answer