A weightless rod is acted upon 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:

Question

A weightless rod is acted upon 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: <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-dy7je4lmpu4fai7g.jpg" /> (A) downward at any point between A and B (B) downward at the midpoint of AB (C) downward at a point C such that AC=1m (D) downward at a point D such that BD=1m

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-dy7je4lmpu4fu3yd.jpg" /> for equilibrium net torque at B end =0 -2(.3.)+6× x=0;x=1 m

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

downward at any point between $A$ and $B$

downward at the midpoint of $A B$

downward at a point $C$ such that $A C = 1 m$

downward at a point $D$ such that $B D = 1 m$

for equilibrium net torque at $B$ end $= 0$
$- 2 (3) + 6 \times x = 0; x = 1 m$

Q. A weightless rod is acted upon 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:

downward at any point between A and B

downward at the midpoint of AB

downward at a point C such that AC=1m

downward at a point D such that BD=1m