Step 2 of 7 CH Free body diagram Step 4 of 7 Sum of moments

Step 2 of 7 CH Free body diagram Step 4 of 7 Sum of moments about C is zero 5%fe=071, eos6x400+TuS.n@xg00-981x1200-0 981 N zur .or..cos T.a cos 400.,sin..S00-981.1200. TAB Tfs cos 26,60 x 400 + Tsa sin 26.6\" x 800-981x 1200 = 0 .. (3) From equation (3), we get 1650N Substituting the value T -1650N in equation(2), we get , 2450 N Substituting the value T-1650N in equation(1), we get C 736 N Comment(0) C. Subsitutingthe.al\"1650N nequation (2, we get R-2450N/ Substituting the value TAB-1650N inequation(1), we getC-736M Comment(o)

Solution

In order to find moment caused by tension AB, we need to calculate perpendicular distance from point C to the line action of tension AB since it is some what laborious to calculate this perpendicular distance, Instead of doing this,it would better and easy process to find moments caused by components of tension AB because it is very easy to calculate perpendicular distances from point C to corresponding tensionAB components

To calculate the components which are equal to product of magnitude of tension multiplied by( cosine of the inclination (or) sine of the complentary of the inclination)