An elevator (cabin mass 600 kg) is designed for a maximum load of 2100 kg, and to reach a velocity of 3.1 m/s in 3 s. For this scenario, what is the tension the elevator rope has to withstand?
This is what I've tried and is incorrect.
2100kg + 600 kg= 2700kg (m)
(3.1/3)= 1.033m/s^2 (a)
2700 x 1.033= 2789.1N (F)
Please help.
you seem to have forgotten gravity
add g to your acceleration number and you should be okay
To find the tension in the elevator rope, we need to consider the Net Force acting on the elevator.
First, let's calculate the Net Force by using Newton's second law of motion, which states that the Net Force (F_net) is equal to the mass (m) of the object multiplied by its acceleration (a):
F_net = m * a
The mass of the elevator is given as 600 kg. However, we need to consider the total mass of the system, including the maximum load. So the total mass (m) is the sum of the elevator's mass (600 kg) and the maximum load (2100 kg):
Total mass (m) = 600 kg + 2100 kg = 2700 kg
Next, we need to determine the acceleration (a). Given that the elevator reaches a velocity of 3.1 m/s in 3 seconds, we can calculate the acceleration by dividing the change in velocity (Δv) by the time taken (Δt):
a = Δv / Δt
a = (3.1 m/s) / (3 s)
a = 1.033 m/s²
Now we can substitute the mass (m) and acceleration (a) values into the formula for Net Force:
F_net = m * a
F_net = 2700 kg * 1.033 m/s²
F_net = 2790.1 N
Therefore, the Net Force acting on the elevator is 2790.1 N. Since the elevator is in equilibrium (not accelerating vertically), the tension in the elevator rope should be equal to the Net Force. Consequently, the tension the elevator rope must withstand is 2790.1 N.