[COLOR=#2D3B45]The diagram shows a model rodOQof 5.72 kg mass. The upper endQof the rod is attached a model string of length 1.34 m, whose other endPis attached to a fixed point on a vertical wall. The endOof the rod is hinged to the vertical wall. The length of the string is equal to the length of the rod. The angleαis 0.83 radians.

Calculate thetension in the string, to 3 decimal places. AssumePQg= 9.81 m/s2.[/COLOR]

I would like to know if im doing it correctly. 1st, i drew a line across the middle OP, connecting it to Q and named the point where Q meets OP C. W=5.72*9.81 and the perpendicular distance from the pivot point to the line of action of the weight is (1.34/2)sin0.83rad. Therefore, clockwise torque=5.72*9.81*(1.34/2)sin0.83rad. OP=2OC. Therefore, OP= 2(1.34)cos0.83rad= 2.68cos0.83rad. The perpendicular distance from the pivot point to the line of action of the tension is 2.68cos0.83rad*sin0.83rad. Therefore, ccw torque= T*2.68cos0.83rad*sin0.83rad. Ccw torque=clockwise torque. Therefore, T= (5.72*9.81*(1.34/2)sin0.83rad)/(2.68cos0.83rad*sin0.83rad)=20.786N. Also, kindly help me check if my final answer is correct .