A rod of mass 1 kg is kept on a frictionless | vedclass

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(C) The net force acting on the rod is $F_{net} = 6 \ N + 3 \ N = 9 \ N$.To find the position of the equivalent single force,we calculate the net torq

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(C) The net force acting on the rod is $F_{net} = 6 \ N + 3 \ N = 9 \ N$.To find the position of the equivalent single force,we calculate the net torque about the center $C$ of the rod.Taking clockwise torque as positive,the torque due to the $6 \ N$ force is $	au_1 = 6 \ N 	imes 2 \ m = 12 \ N \cdot m$.The torque due to the $3 \ N$ force is $	au_2 = -3 \ N 	imes 1 \ m = -3 \ N \cdot m$ (counter-clockwise).The net torque is $	au_{net} = 12 - 3 = 9 \ N \cdot m$.For a single force $F_{net}$ to have the same effect,it must produce the same net torque about the center $C$ at a distance $x$:$	au_{net} = F_{net} 	imes x$$9 \ N \cdot m = 9 \ N 	imes x$$x = 1 \ m$.

$A$ rod of mass $1 \ kg$ is kept on a frictionless table and is acted upon by $2$ forces as shown. We want to replace these $2$ forces by a single force so that the effect on the rod is the same. At what distance $x$ (in $m$) from the center $C$ of the rod should we apply this single force?

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