Question

QUESTION 5 A force of P=312 N is applied to the 48-kg cart. Determine the reaction at Othe wheel at A. The mass center of the cart is at G. take c = 0.5m and d=0.17m. (See figure (3) c(m) H+ 0.08 m 0.3 m * 0.2 m -

Answer 1

In this problem, we balance the force along the X and Y-axis and moment about the center of gravity of the cart and then find the required reaction at A.

comment if any query.

P apsin3o - - 36 - PCos30 - - OS ก orim - - - - 603-740-2--> apsinzo Proszo 0.5m 48x9.81N. 0.17 7 0.08 L-0.3m 0.2m- we balance Fosice ie & In x direction. PCOs30 = ma PCOs30=48xa. In y direction RA+RB+ Pinzo - 48x9.81=0 RA+RB+312&Sin3o-48x9.81=0 RA+RB = 314.88 -

moment about Centre of Gravity dalanung RAXO.3 - RB X 0.2 + P Sinzox0.38 - PCOs3ox (0.5-0.17) = 0 RAXO.3 - RB XO.2 +312x Sin Box0.38 - 312x cos3oxo.33=0 RAXO.3 - RB XO.2 +59.28-89.1659 zo RAXO:3 - Roxo 2 = 29.8859 0.3RA -0.2RB = 29.8859 - From Ro=314.88-RA OBRA -0.2 (314.88 - RA) = 29.8859 OSRA - 62.976 =29.8859 TRA = 185.1238N)