Problem 1

T1 and T2: 30*9.8/sin 45 = 415.77 N

Problem 2

T1: 980 N

T2:980 cos 35 = 802.76 N

T3:980/sin 35 = 1708 N

answer One: i anticipate T1 and T2 are rigidity vectors. making a triangle with an prolonged area of M and the different 2 elements as T1 and T2 the perspective opposite T1 could be (ninety ranges – alpha) and the perspective opposite T2 could be (ninety ranges – theta) — i stumble on we’ve a triangle with 2 angles and one area that all of us understand. which could be solved employing the — regulation of Sines the place we are able to symbolize the elements of the triangle as a,b,c. and the Angles of the Triangle as A,B,C. area b = a * sin(B)/sin(A) and area c = a * sin(C)/sin(A) and A = one hundred eighty – B – C — substituting M for a and T2 for b and T3 for c and (ninety-alpha) for C and (ninety-theta) for B then A = one hundred eighty – (ninety-theta) – (ninety-alpha) = theta + alpha we’ve T2 = M * sin(ninety-theta) / sin(theta + alpha) T3 = M * sin(ninety-alpha) / sin(theta + alpha)