An elevator and its load have a combined mass of 1962 kg. Find the tension in the supporting cable when the elevator, originally moving downward at 7.5 m/s, is brought to rest with constant acceleration in a distance of 4.2 m.
v²-u²=2aS
v=0
u=-7.5 m/s
S=-4.2 m
Solve for a=acceleration in m/s²
a=7.5²/(-2×4.2)
=6.7 m/s²
To find the tension in the supporting cable, we can use the equation of motion for the elevator. First, let's list down the given information:
Mass of the elevator and load (m) = 1962 kg
Initial velocity (u) = -7.5 m/s (negative because it is moving downward)
Final velocity (v) = 0 m/s (brought to rest)
Displacement (s) = 4.2 m
Acceleration (a) = ?
We can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Plugging in the given values, we have:
0^2 = (-7.5)^2 + 2a(4.2)
Simplifying the equation, we get:
0 = 56.25 + 8.4a
Rearranging the equation:
8.4a = -56.25
Dividing both sides by 8.4:
a = -6.696 m/s^2
Now that we have the acceleration, we can calculate the tension in the supporting cable. The tension force can be found using Newton's second law, F = ma, where F is the force, m is the mass, and a is the acceleration. In this case, the force is equal to the tension in the cable.
Plugging in the values, we have:
F = (1962 kg)(-6.696 m/s^2)
Calculating this, we get:
F = -13135.752 N
However, since tension is a positive value, we take the absolute value of it:
Tension (T) = |F| = 13135.752 N
Therefore, the tension in the supporting cable is approximately 13135.752 N.