An elevator and its load have a combined mass of 1700 kg. Find the tension in the supporting cable when the elevator, originally moving downward at 13 m/s, is brought to rest with constant acceleration in a distance of 43 m.
The cable tension will increase to
T = M(g+a)
when the elevator is slowing down at deceleration rate a.
You can get a from the relation
V^2/2 = a X
where X is the stopping distance, and V is the initial velocity.
6089
169/2=a*43
84.5=a*43
a=84.5/43
a=1.965
T=M(g+a)
T=1700(-9.8+1.965)
T=1700*(-7.835)
T=-13,319.5
To find the tension in the supporting cable, we can use Newton's second law of motion. The force exerted on the elevator is equal to the product of its mass and acceleration.
First, let's determine the acceleration of the elevator. We are given the initial velocity (13 m/s) and the distance over which it comes to rest (43 m). We can use the following kinematic equation to solve for the acceleration:
(vf)^2 = (vi)^2 + 2as
Where vf is the final velocity (which is 0 m/s as the elevator comes to rest), vi is the initial velocity, a is the acceleration, and s is the distance.
Rearranging the equation, we have:
a = (vf^2 - vi^2) / (2s)
Plugging in the given values:
a = (0^2 - 13^2) / (2 * 43)
= (-169) / 86
≈ -1.97 m/s^2
Note: The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, which makes sense since the elevator is coming to rest from a downward motion.
Now, we can calculate the force exerted on the elevator using Newton's second law:
F = m * a
Given that the combined mass of the elevator and its load is 1700 kg, we can substitute the values:
F = 1700 kg * (-1.97 m/s^2)
≈ -3349 N
Since force is a vector quantity, the negative sign indicates that the tension in the cable acts in the opposite direction of the gravitational force. To get the magnitude of the tension, we can remove the negative sign:
|T| = |-3349 N|
= 3349 N
Therefore, the tension in the supporting cable is approximately 3349 N.