Question

An elevator cab is pulled upward by a cable. The cab and its single occupant have a combined mass of 1510 kg. When that occupant drops a coin, its acceleration relative to the cab is 7.40 m/s2 downward. What is the tension in the cable?

Answer 1

Answer:

Explanation:

Given

Combined mass of cab and elevator is $M=1510\ kg$

acceleration of coin with respect to elevator $a_{ce}=-7.4\ m/s^2$

$a_{ce}$ can be written as

$a_{ce}=a_{cg}-a_{eg}$

where

$a_{cg}$=acceleration of coin with respect to ground

$a_{eg}$=acceleration of elevator with respect to ground

$a_{ce}=a_{cg}-a_{eg}$

$-7.4=-9.8-a_{eg}$

$a_{eg}=-9.8+7.4=-2.4\ m/s^2$

Forces on elevator

$T-Mg=Ma_{eg}$

$T=M(g+a_{eg})$

$T=1510\times (9.8-2.4)$

$T=11,174\ N$

Here we considered downward direction as negative and upward as positive