Elevator pulled upward by cable - find tension

[shawpeez]

here's the question,

An elevator cab is pulled upward by a cable. The cab and its single occupant have a combined mass of 2000kg. When that occupant drops a coin, its acceleration relative to the cab is 8.00m/s^2 downward. What is the tension in the cable?

could someone explane to me how to relate the accerleration of the coin downward to the acceleration of the cab being pulled up.

 

[Chi Meson]

IF the elevator is going up, and the apparent acceleration in the cab is less than 9.8 m/s/s, then the elevator is at that moment slowing down.

If an object is being pulled up at a constant velocity, then there is zero acceleration because forces are balanced (specifically, the tension in the cable balances the weight). Dropping the coin inside at this moment, the coin would display normal acceleration to the person inside the cab, no matter what speed the elevator was doing.

If the elevator is slowing down, then the upward tension in the cable must be slightly less than the total weight (thereby making net force NOT zero but slightly downward).

So what's the accelration of the cab, if gravitational acceleration "seems to be" 1.8 m/s/s less than normal?

 

[shawpeez]

My first instinct was that the upward acceleration of the cab was slowing down, this is what i did

Fnet= T - mg = ma
T - 2000(9.8) = 2000(-1.8)
T - 19600 = -3600
T= 19600 - 3600
T = 16000 N

Is this correct ?

 

[Chi Meson]

You are correct. Follow that instinct (but don't say "the acceleration was slowing down," it's the speed that is slowing down.)

 

[shawpeez]

thanks for the help