Calculate the total torque acting on the body shown in figure about the point O.
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Torque about a point = Total force \times perpendicular distance from the point to that force.
Let anticlockwise torque = +ve And clockwise acting torque = -ve Force acting at the point B is 15N Therefore torque at O due to this force $=15\times6\times10^{-2}\times\sin37^\circ$
$=15\times6\times10^{-2}\times\frac{3}{5}=0.54\text{N-m}$ (anticlock wise) Force acting at the point C is 10N Therefore, torque at O due to this force = $10 \times 4 \times 10^{-2} = 0.4N-m$ (clockwise) Force acting at the point A is 20N Therefore, Torque at O due to this force $=20\times4\times10^{-2}\times\sin30^{\circ}$
$=20\times4\times10^{-2}\times\frac{1}{2}=0.4\text{N-m}$ (anticlockwise) Therefore resultant torque acting at ‘O’ = 0.54 - 0.4 + 0.4 = 0.54N-m.
Let anticlockwise torque = +ve And clockwise acting torque = -ve Force acting at the point B is 15N Therefore torque at O due to this force $=15\times6\times10^{-2}\times\sin37^\circ$
$=15\times6\times10^{-2}\times\frac{3}{5}=0.54\text{N-m}$ (anticlock wise) Force acting at the point C is 10N Therefore, torque at O due to this force = $10 \times 4 \times 10^{-2} = 0.4N-m$ (clockwise) Force acting at the point A is 20N Therefore, Torque at O due to this force $=20\times4\times10^{-2}\times\sin30^{\circ}$
$=20\times4\times10^{-2}\times\frac{1}{2}=0.4\text{N-m}$ (anticlockwise) Therefore resultant torque acting at ‘O’ = 0.54 - 0.4 + 0.4 = 0.54N-m.
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