解：$(1)F_{乙}=G-F_{甲}=120\ \text {N}-80\ \text {N}=40\ \text {N}$
$F_{甲}l_{甲}=F_{乙}l_{乙}①$
$l_{甲}+l_{乙}=1.8\ \text {m}②$
联立①②，解得$l_{甲}=0.6\ \text {m}$
$ (2)l_{甲}'=0.6\ \text {m}+0.2\ \text {m}=0.8\ \text {m}$，$l_{乙}'=1.8\ \text {m}-0.8\ \text {m}=1\ \text {m}$
$ F_{甲}'×1.8\ \text {m}=120\ \text {N}×1\ \text {m}$
$ $解得$F_{甲}'≈66.7\ \text {N}$
$ $减小了$∆F_{甲}=80\ \text {N}-66.7\ \text {N}=13.3\ \text {N}$
