Question

(a) A man stands a distance d = 0.650 m from the left end of a board as shown in the figure. The board is supported by three lines. Find the tension in each line (In N). Assume the board is uniform, with length L = 2.00 m and mass 28.5 kg, and the weight of the man is 680 N. (Due to the nature of this problem, do not use rounded intermediate values in your calculations-including answers submitted In WebAssign.) 2.00 m 19,1 = 7,1 = T31 = (b) What If? The man now begins to walk to the right. If each of the lines can support a maximum tension of 750 N, which line falls first? line 1 line 2 line 3 At what location does this occur (measured in m from the left end of the board)?

Answer 1

Giver & FK Ji 2m me 2015 d=0.65m. M = 680N. E Fx=0 T3=T, x Cas40° -6 Ta + 7, Sinyo' = 680+ 28.5x9.81 >>> Tot Ti Sinyo = 959.585 - © Considering moment about left and 1 x Sin40 x 2 = 680 x 0.65 + 28.5x981x) T, = 561-293N Tz = 561-293 x cosyo = 429.97N To = 598.79 N

Ti = 561.29 N Ta= 598.79N Ta = 429.97N by If he walks to right, line I will break first ob 750x Sinyoʻx2 = 680xd + 28.5x938)) 964-18 - 2790585 680 d= 1.006 m