Answer:
In Fig. 5(HT).1, we have
resolved force of 100 N into two rectangular components: \[100\,\cos
{{30}^{\text{o}}}\]along the horizontal and \[100\,\cos {{30}^{\text{o}}}\]
along the vertical.
The
horizontal component cannot rotate the rod.
Torque
due to \[\text{F}=\text{1}00\text{sin3}0{}^\circ \]about fixed point O It is in
clockwise direction.
Torque
due to \[\text{F}=\text{5}0\text{N}\] about fixed point
It
is also clockwise.
\[\therefore
\] Net torque acting on the rod = \[{{\tau }_{1}}+{{\tau
}_{2}}=25+25=50N\,\,m;\] clockwise.
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