Three forces F, 2F and 3F act on a rod AB whi | vedclass

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(B) The moment of a force about a point is given by $	au = \pm F \cdot d$, where $d$ is the perpendicular distance from the pivot to the line of acti

Vedclass · Accepted Answer

$-2Fa; +2Fa; +6Fa$

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(B) The moment of a force about a point is given by $	au = \pm F \cdot d$, where $d$ is the perpendicular distance from the pivot to the line of action of the force. By convention, anticlockwise moments are positive $(+)$ and clockwise moments are negative $(-)$.$(i)$ For force $F$: The force $F$ acts downwards at a distance $2a$ from $A$. This creates a clockwise moment.Moment $= -F \cdot (2a) = -2Fa$.$(ii)$ For force $2F$: The force $2F$ acts upwards at a distance $a$ from $A$. This creates an anticlockwise moment.Moment $= +(2F) \cdot a = +2Fa$.$(iii)$ For force $3F$: The force $3F$ acts at $B$ (distance $4a$ from $A$) at an angle of $30^{\circ}$ to the rod. The perpendicular distance from $A$ is $d = 4a \sin 30^{\circ} = 4a \cdot (1/2) = 2a$. This force creates an anticlockwise moment.Moment $= +(3F) \cdot (2a) = +6Fa$.Thus, the moments are $-2Fa, +2Fa, +6Fa$.

Three forces $F, 2F$ and $3F$ act on a rod $AB$ which is pivoted at $A$. The anticlockwise moments of forces $F, 2F$ and $3F$ about the pivot $A$ are respectively:

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