A weightless rod is acted on by upward parall | vedclass

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(D) For the rod to be in equilibrium,the net force must be zero and the net torque about any point must be zero.$1$. Net Force: The sum of upward forc

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Downwards at a point $D$ such that $BD = 1m$.

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(D) For the rod to be in equilibrium,the net force must be zero and the net torque about any point must be zero.$1$. Net Force: The sum of upward forces is $2N + 4N = 6N$. To balance this,a downward force of $6N$ must be applied.$2$. Net Torque: Let the point of application of the $6N$ downward force be at a distance $x$ from end $A$. Taking the torque about point $A$:$	au_A = (2N 	imes 0) + (4N 	imes 3m) - (6N 	imes x) = 0$$12 = 6x$$x = 2m$ from $A$.Since the total length $AB = 3m$,the distance from $B$ is $3m - 2m = 1m$.Therefore,the force must act downwards at a point $D$ such that $BD = 1m$.

$A$ weightless rod is acted on by upward parallel forces of $2N$ and $4N$ at ends $A$ and $B$ respectively. The total length of the rod $AB = 3m$. To keep the rod in equilibrium,a force of $6N$ should act in the following manner:

Similar Questions

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