The value of M, as shown, for which the rod will be in equilibrium is:
      

In the figure given below, O is the centre of an equilateral triangle ABC and $\overrightarrow{{\mathrm{F}}_1},$ $$ $\overrightarrow{{\mathrm{F}}_2},$ $$ $\overrightarrow{{\mathrm{F}}_3}$ are three forces acting along the sides AB, BC and AC. What should be the magnitude of $\overrightarrow{F_3}$ $$ so that total torque about O is zero?
        
1.  $|$ ${\overrightarrow{F}}_3|$ $=$ $\left|\overrightarrow{F_1}\right|$ $+$ $|\overrightarrow{F_2|}$

2. $|$ $\overrightarrow{F_3}|$ $=$ $\left|\overrightarrow{F_1}\right|$ $-$ $|\overrightarrow{F_2|}$

3. $|$ $\overrightarrow{F_3}|$ $=$ $\overrightarrow{F_1}$ $+$ $2\overrightarrow{F_2}$

4. Not possible

A wheel with a radius of 20 cm has forces applied to it as shown in the figure. The torque produced by the forces of 4 N at A, 8N at B, 6 N at C, and 9N at D, at the angles indicated, is:
                    
1.  5.4 N-m anticlockwise

2.  1.80 N-m clockwise

3.  2.0 N-m clockwise

4.  3.6 N-m clockwise

A rod of weight \(w\) is supported by two parallel knife edges, A and B, and is in equilibrium in a horizontal position. The knives are at a distance \(d\) from each other. The centre of mass of the rod is at a distance \(x \) from A. The normal reaction on A is:

For L = 3.0 m, the total torque about pivot A provided by the forces as shown in the figure is: