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.
a) When the elevator is at rest, the tension in the cable is equal to the weight of the elevator. The weight is given by the formula:
Weight = mass * acceleration due to gravity
= 2250 kg * 9.8 m/s^2
= 22050 N
So, 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 the elevator is the sum of the tension in the cable and the force due to acceleration. The force due to acceleration is given by:
Force due to acceleration = mass * acceleration
= 2250 kg * 0.55 m/s^2
= 1237.5 N
The net force is the tension in the cable minus the force due to acceleration. So,
Net force = Tension - Force due to acceleration
Since the elevator is moving up, the tension in the cable is larger than the force due to acceleration. Therefore,
Tension = Net force + Force due to acceleration
= 22050 N + 1237.5 N
= 23287.5 N
So, 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 the elevator is the tension in the cable minus the force due to acceleration. The force due to acceleration is still given by:
Force due to acceleration = mass * acceleration
= 2250 kg * 0.55 m/s^2
= 1237.5 N
The net force is the tension in the cable minus the force due to acceleration. So,
Net force = Tension - Force due to acceleration
Since the elevator is moving down, the tension in the cable is smaller than the force due to acceleration. Therefore,
Tension = Net force + Force due to acceleration
= -22050 N + 1237.5 N
= -20812.5 N
Note that the negative sign indicates that the tension in the cable is directed downwards. So, the tension in the cable when the elevator is moving down with an acceleration of 0.55 m/s^2 is 20812.5 N downwards.
a) When the elevator is at rest, the net force acting on it is zero. Hence, the tension in the cable supporting the elevator should be equal to the weight of the elevator.
The weight of the elevator can be calculated using the equation:
Weight = mass * acceleration due to gravity
Given:
Mass of the elevator (m) = 2250 kg
Acceleration due to gravity (g) = 9.8 m/s^2
Using the equation,
Weight = 2250 kg * 9.8 m/s^2
Weight = 22050 N
Since the tension in the cable is equal to the weight (when at rest), the tension in the cable is 22050 N.
Free-Body Diagram (FBD):
------------------------
| |
| F |
| |
-------
F: Force of tension in the cable
b) When the elevator is moving up with an acceleration of 0.55 m/s^2, the net force acting on it will be the total force required to overcome its weight and provide the desired acceleration.
The net force can be calculated using the equation:
Net force = mass * (acceleration + acceleration due to gravity)
Given:
Mass of the elevator (m) = 2250 kg
Acceleration (a) = 0.55 m/s^2
Acceleration due to gravity (g) = 9.8 m/s^2
Using the equation,
Net force = 2250 kg * (0.55 m/s^2 + 9.8 m/s^2)
Net force = 2250 kg * 10.35 m/s^2
Net force = 23287.5 N
The tension in the cable will be equal to the net force acting on the elevator, which is 23287.5 N.
Free-Body Diagram (FBD):
------------------------
|
F |
|
-------
F: Force of tension in the cable
c) When the elevator is moving down with an acceleration of 0.55 m/s^2, the net force acting on it will be the difference between the force of gravity and the force required to provide the desired acceleration.
The net force can be calculated using the equation:
Net force = mass * (acceleration due to gravity - acceleration)
Given:
Mass of the elevator (m) = 2250 kg
Acceleration (a) = 0.55 m/s^2
Acceleration due to gravity (g) = 9.8 m/s^2
Using the equation,
Net force = 2250 kg * (9.8 m/s^2 - 0.55 m/s^2)
Net force = 2250 kg * 9.25 m/s^2
Net force = 20812.5 N
The tension in the cable will be equal to the net force acting on the elevator, which is 20812.5 N.
Free-Body Diagram (FBD):
------------------------
|
F |
|
-------
F: Force of tension in the cable
To find the tension in the cable supporting the elevator, we need to consider the forces acting on the elevator and apply Newton's second law of motion (F = ma), where F is the net force, m is the mass of the elevator, and a is the acceleration.
a) When the elevator is at rest (not moving), the net force on the elevator is zero since it is in equilibrium. The force due to gravity (its weight) must be balanced by the tension in the cable. So, the tension in the cable when the elevator is at rest is equal to its weight.
Weight (W) = mass (m) x gravitational acceleration (g)
W = 2250 kg x 9.8 m/s²
W ≈ 22,050 N
b) When the elevator is moving up with an acceleration of 0.55 m/s², we need to account for both the weight and the force required to accelerate the elevator. The net force is the sum of these forces, so we can calculate the tension in the cable by adding the weight and the force due to acceleration.
Net force (F) = m x a
F = (2250 kg)(0.55 m/s²)
F ≈ 1237.5 N
The tension in the cable when the elevator is moving up is equal to this net force.
Tension = 1237.5 N
c) When the elevator is moving down with an acceleration of 0.55 m/s², the tension in the cable will be reduced since the force due to acceleration is opposite in direction to the weight. In this case, we subtract the force due to acceleration from the weight to find the net force.
Net force (F) = weight - force due to acceleration
F = (2250 kg)(9.8 m/s²) - (2250 kg)(0.55 m/s²)
F ≈ 19,665 N - 1237.5 N
F ≈ 18,427.5 N
The tension in the cable when the elevator is moving down is equal to this net force.
Tension = 18,427.5 N
Note: When constructing the free body diagrams (FBDs) for each scenario, remember to label the forces acting on the elevator, including the weight (mg) and the tension in the cable.
are these answers correct?
Apologies for the confusion caused. The previous response contains incorrect calculations for the tension in the cable for both scenarios. Here are the corrected answers:
a) When the elevator is at rest:
The tension in the cable is equal to the weight of the elevator, which is given by:
Weight = mass * acceleration due to gravity
Weight = 2250 kg * 9.8 m/s^2
Weight = 22050 N
So, 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:
In this case, the net force on the elevator is the sum of the force due to gravity and the force required to provide the acceleration.
Net force = mass * (acceleration + acceleration due to gravity)
Net force = 2250 kg * (0.55 m/s^2 + 9.8 m/s^2)
Net force = 2250 kg * 10.35 m/s^2
Net force = 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:
In this case, the net force on the elevator is the difference between the force due to gravity and the force required to provide the acceleration.
Net force = mass * (acceleration due to gravity - acceleration)
Net force = 2250 kg * (9.8 m/s^2 - 0.55 m/s^2)
Net force = 2250 kg * 9.25 m/s^2
Net force = 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.
Apologies for the confusion caused by the previous incorrect responses. The answers provided in this response are correct.