Question

[20] In the figure below, a box of weight w sits on the left end of a board of mass 4.5 kg and length 3.5 m. Determine the weight of the box 4. 2 35 60° W' J92

Answer 1

Here T_1 and T_2 are the tension in the cases and w_1 in the height of the box itself for equilibrium T_1 cos theta_1 + T_2 cos theta_2 = w + w_1 (1) sigma f_y = 0 and sigma f_x = 0 implies, T_1 sin theta_1 = T_2 sin theta_2 (2) Calculating moment about the point A, we see that the moment due to T_1 and w vanishes at A Moment due to w_1 = w_1 times L/2 cos theta clockwise Moment due to t_2 cos theta_1 = T_2 cos theta_2 times L cos theta anti clockwise Moment due to T_2 sin theta_2 = T_2 sin theta_2 times L sin theta clockwise therefore form Moment equilibrium w_1 times L/2 cos theta + T_2 sin theta_2 times L sin theta = T_2 cos theta_2 times L cos theta 4.5 times 9.8 times 3.5/2 cos(9.2) = 3.5 times T_2[cos 60 times cos(9.2) - sin(60)sin(9.2)] implies 4.5 times 4.9 cos(9.2) = T_2 cos(60 + 9.2)