the physics question confuses me

Question

The cabin of a small freight elevator is secured to a motor by a cable and is moving upward with a constant speed. There is no contact between the cabin and the elevator shaft. Ignore air resistance.

The mas of the elevator cabin is 1225 kg

What is the weight force equal to?

What is the tension force equal to?

The cabin now accelerates downward, speeding up at a rate of 0.5 m/s/s. Which force has changed? The gravitational force? Or the tension force?

Draw the new FBD, for the cabin accelerating downward.

What is the net force, while the cabin is accelerating?

Calculate the new tension force.

Answer 1

Given

Constant Speed

Mass of Elevator: M = 1225 kg

Find

a) Weight force

b) Tension on Cable

c) If cabin accelerates downward, at a rate of 0.5 m/s^2, which force has changed.

d) Draw the new FBD when accelerating downward

e) What is the net force while accelerating?

f) Calculate the tension

Solution

a) W = m*g = 1225*9.81 = 12,017.25 N

b) Tension: T=? Note constant velocity implies no acceleration

Sum force Y = 0

T - W = 0

T = W = 12,017.25 N

c)

Sum force Y = m*a

T - W = -m*a

The tension of the cable changes, it will decrease based on the equation below

T = W-m*a

d) FBD will show the following (All forces in the vertical direction)

T going up

W going down

e)

Fnet = -m*a = -1225 * 0.5 = -612.5 N

f)

T = W-m*a = 1225 - 1225*0.5 = 612.5 N

Answer 2

1. The weight = mg;
2. the tension T1 = W = mg, the direction of T1 is opposite to W;
3. F = ma, F = T1 - mg, now since accelrateion is downward, a = -0.5, F is negtive. Therefore T1' = mg + F < T1, the tension force decreased.