Question

A 945- kg elevator is suspended by a cable of negligible mass. If the tension in the cable is 8.65 kN, what are the magnitude and direction of the elevator's acceleration?

Answer 1

- Magnitude of the acceleration = 0.65 m/s²

- The acceleration is directed upwards in the direction of the Tension in the suspending cables.

Step-by-step explanation:

The force balance on the elevator consists of the Tension in the cable, acting upwards away from the elevator, the weight of the elevator, mg, acting downwards and the 'ma' force responsible for motion.

The direction of this 'ma' force depends on which side of the force balance is lesser or more.

That is, depending on the one that's higher,

ma = T - mg

OR

ma = mg - T

We need to find the higher force.

T = 8.65 KN = 8650 N

mg = 945 × 9.8 = 9261 N

mg > T, meaning the 'ma' force is on the upwards side of the tension, but motion of the elevator is definitely downwards.

Since mg > T,

ma = mg - T = 9261 - 8650 = 611 N

a = 611/m = 611/945 = 0.65 m/s²

Hope this helps!

Answer 2

Acceleration= -0.6466 m/
`sec^(2)` Downward

Step-by-step explanation:

Given: Mass M= 945 Kg, Tension T = 8.65 kN = 8650 N and g =9.8 m/
`sec^(2)`

Sol: Weight W = mg = 945 Kg 9.8 m/
`sec^(2)` = 9261 N

this show that T < W so the motion is downward so to find acceleration

mass × acceleration = T - W (putting values)

945 Kg × a = 8650 N - 9261 N

a= -0.6466 m/
`sec^(2)` (-ve sign shows the downward direction)