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.
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.
5 answers
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.
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.
so this last answer is correct?
Yes, the last answer is correct. I apologize for the earlier incorrect responses. The tension in the cable when the elevator is at rest is 22050 N, when the elevator is moving up with an acceleration of 0.55 m/s^2 the tension in the cable is 23287.5 N, and when the elevator is moving down with an acceleration of 0.55 m/s^2 the tension in the cable is 20812.5 N.