## College Physics (4th Edition)

The tension in the cable that supports the elevator is $16,700~N$
We can find the net force on the elevator: $\sum F = ma = (2010~kg)(1.50~m/s^2) = 3015~N$ We can find the tension $F_T$ in the cable that supports the elevator: $\sum F = mg-F_T$ $F_T = mg - \sum F$ $F_T = (2010~kg)(9.8~m/s^2)-3015~N$ $F_T = 16,700~N$ The tension in the cable that supports the elevator is $16,700~N$