## Find the Normal Force of a box in an elevator cab

1. The problem statement, all variables and given/known data

Elevator cabs A and B are connected by a short cable and can be pulled upward or lowered by the cable above cab A. Cab A has mass 1700 kg; cab B has mass 1300 kg. A 12.0 kg box of catnip lies on the floor of cab A. The tension in the cable connecting the cabs is 1.91×10^4 N. What is the magnitude of the normal force on the box from the floor?"

2. Relevant equations

F=ma

3. The attempt at a solution

T-mBg=mBa
T = mB(a+g)
a+g=T/mB

N=m(a+g) = m (T/mB)= (12.0)(1.91×10^4/1300) = 176 N (rounded value)
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Recognitions:
Homework Help
 N=m(a+g) = m (T/mB)
Why did you drop the "g" in this step? Other than that, it looks good to me.

 Quote by ayesha91 T-mBg=mBa T = mB(a+g) a+g=T/mB
I found that a+g=T/mB
So, I replaced a+g by T/mB

Recognitions:
Homework Help

## Find the Normal Force of a box in an elevator cab

You are correct! I bungled the move from Mb to the catnip.