To stop an elevator in free fall, brakes are installed between the walls of the elevator car and the walls of the shaft in which the car moves. In addition, springs with compression coils are installed on the floor, at the bottom of the shaft, in order to absorb the residual energy if the brakes fail to completely stop the elevator car.
There is an elevator car, with 3 people on board, whose cables become unhinged. The elevator falls 15 m. The total mass of the car and the people on board is 4000 kg. The sum of the frictional forces applied by the brakes is 5000 N acting over a distance of 15 m. Each spring attached to the floor has a spring constant of 5.0 x 10^6 N/m and may be compressed over a maximum distance of 30 cm.
How many springs are needed on the floor so that the elevator car and its occupants do not compress them by more than 20 cm?
Geepers: a better question for followon is what deacceleration will happen in that 20 cm stopping distance.
OriginalPEnergy-frictionalenergy=FinalPEsprings.
4000*9.8*15-5000*15=n*5E6*1/2*.2^2
solve for n. I get about six.
Now for the better question: what is the deacceleration?
KEfinal=keinitial-force*distance
4000*9.8*15-5000*15 -5000a*.2
a = about 50 g. Wouldn't that crush bones and flesh? check my thinking.
To solve this problem, we need to calculate the maximum compression that each spring can experience and then determine how many springs are required to ensure that the elevator car and its occupants do not compress them by more than 20 cm.
First, let's calculate the maximum compression for each spring. We are given that each spring has a spring constant of 5.0 x 10^6 N/m and can be compressed over a maximum distance of 30 cm.
Using Hooke's Law, we can calculate the force exerted on each spring when it is compressed by its maximum distance:
F = k * x
Where:
F = force exerted on the spring
k = spring constant
x = compression distance
Plugging in the values, we have:
F = (5.0 x 10^6 N/m) * (0.3 m)
F = 1.5 x 10^6 N
So, the maximum force each spring can exert is 1.5 x 10^6 N.
Next, let's calculate the total force exerted by the springs needed to stop the elevator car. We are given that the elevator falls 15 m and its mass, including the occupants, is 4000 kg.
Using the formula for gravitational potential energy:
PE = m * g * h
Where:
PE = potential energy
m = mass
g = acceleration due to gravity
h = height
The potential energy is converted to kinetic energy, equal to the work done by the frictional force of the brakes:
KE = F * d
Where:
KE = kinetic energy
F = frictional force
d = distance
Setting the potential energy equal to the kinetic energy:
m * g * h = F * d
Rearranging the equation for F:
F = (m * g * h) / d
However, we are given that the sum of the frictional forces applied by the brakes is 5000 N acting over a distance of 15 m. Therefore, we simply need to divide the total force by the number of springs to determine how many springs are required to distribute the force evenly.
Total force = 5000 N
Since each spring can exert a maximum force of 1.5 x 10^6 N, the number of springs required is:
Number of springs = Total force / Maximum force per spring = 5000 N / (1.5 x 10^6 N) ≈ 3.33
Since we cannot have a fraction of a spring, we need to round up the number to the nearest whole number.
Therefore, we need at least 4 springs on the floor so that the elevator car and its occupants do not compress them by more than 20 cm.