A thin uniform rod of mass m=1kg moves translationally with acceleration a=2 `m s^(-1)` due to two antiparallel forces of arm-length l=20 cm . One forces is of magnitude 5 N and acts at one extreme end. Find the length of the rod.

A thin uniform rod AB of mass 1kg move translationally with acceleration a=2m//s^(2) due to two antiparallel force as shown. If l=20cm then:

A thin uniform rod of mass m moves translationally with acceleration a due to two antiparallel forces of lever arm l. One force is of magnitude F and acts at one extreme end. The length of the rod is

A Thin uniform rod AB of mass m=1.0kg moves translationally with acceleration a=2.0 m//s^(2) due to two antiparallel forces F_(1) and F_(2) . The distance between the points at which these forces are applied is equal to d=20 cm . Besides, it is known that F_(2)=5.0 N . Find the length of the rod.

A thin uniform rod AB of mass m=1kg moves translationally with acceleration a=2m//s^(2) and to two anitiparallel forces F_(1) and F_(2) . The distance between the points at which these forces are applied is equal to l=20cm besides it is known that F_(2)=5N find the length of the rod.

A thin uniform rod AB of mass m=1kg moves translationally with acceleration a=2m//s^(2) and to two anitiparallel forces F_(1) and F_(2) . The distance between the points at which these forces are applied is equal to l=20cm besides it is known that F_(2)=5N find the length of the rod.

A thin uniform rod AB of mass m=1.0kg moves translationally with acceleration w=2.0m//s^2 due to two antiparallel forces F_1 and F_2 (figure). The distance between the points at which these forces are applied is equal to a=20cm . Besides, it is known that F_2=5.0N . Find the length of the rod.