Question

Pls answer fast......​

Answer 1

$\displaystyle\large\boxed {\sf {\Sigma\tau=-110\ \hat t\ Nm}}$

Here the anticlockwise rotation is taken as positive and so the clockwise rotation is taken as negative.

Let $\displaystyle\sf {\hat t}$ be the unit vector acting perpendicularly to the plane lamina but along the direction in front of the lamina towards our eye sight.

★ The force acting at the point A is of $\displaystyle\sf {20\ N}$ at a distance of $\displaystyle\sf {1\ m}$ perpendicular to O, which gives the lamina an anticlockwise rotation about O. Hence the torque acting at A is,

• $\displaystyle\sf {\tau_A=1\times20\ \hat t}$

• $\displaystyle\sf {\tau_A=20\ \hat t\ Nm}$

★ The force acting at the point B is of $\displaystyle\sf {30\ N}$ at a distance of $\displaystyle\sf {4\ m}$ perpendicular to O, which gives the lamina a clockwise rotation about O. Hence the torque acting at B is,

• $\displaystyle\sf {\tau_B=4\times30\ (-\hat t)}$

• $\displaystyle\sf {\tau_B=-120\ \hat t\ Nm}$

★ The force acting at the point C is of $\displaystyle\sf {80\ N}$ at a distance of $\displaystyle\sf {3\ m}$ perpendicular to O, which gives the lamina an anticlockwise rotation about O. Hence the torque acting at C is,

• $\displaystyle\sf {\tau_C=3\times80\ \hat t}$

• $\displaystyle\sf {\tau_C=240\ \hat t\ Nm}$

★ The force acting at the point D is of $\displaystyle\sf {200\ N}$ towards O, which gives the lamina neither a clockwise rotation nor anticlockwise about O. Hence the torque acting at D is,

• $\displaystyle\sf {\tau_D=0\times200\ \hat t}$

• $\displaystyle\sf {\tau_D=0\ \hat t\ Nm}$

★ The force acting at the point E is of $\displaystyle\sf {50\ N}$ at a distance of $\displaystyle\sf {5\ m}$ perpendicular to O, which gives the lamina a clockwise rotation about O. Hence the torque acting at E is,

• $\displaystyle\sf {\tau_E=5\times50\ (-\hat t)}$

• $\displaystyle\sf {\tau_E=-250\ \hat t\ Nm}$

Therefore, the net torque acting on the plane lamina about O is,

$\displaystyle\longrightarrow\sf {\Sigma\tau=\tau_A+\tau_B+\tau_C+\tau_D+\tau_E}$

$\displaystyle\longrightarrow\sf {\Sigma\tau=(20-120+240+0-250)\ \hat t\ Nm}$

$\displaystyle\longrightarrow\sf {\underline {\underline {\Sigma\tau=-110\ \hat t\ Nm}}}$

I.e., the net torque is of $\displaystyle\sf {110\ Nm}$ acting in the direction behind the plane lamina, away from our eye.

Answer 2

Answer:

30 × 4

20 x 1

=

80x 3 m

50x 5m