Solution.
Sum the moments about the base of the crane; Σ M = 0.
T2cos35Lcos40+T1cos45Lcos40−T2sin35Lsin40−T1sin45Lsin40−W(L/2)sin40−T1Lsin40=0;T_2cos35Lcos40 + T_1cos45Lcos40 - T_2sin35Lsin40 - T_1sin45Lsin40 - W(L/2)sin40 - T_1Lsin40 =0;T2cos35Lcos40+T1cos45Lcos40−T2sin35Lsin40−T1sin45Lsin40−W(L/2)sin40−T1Lsin40=0;
0.259T2=43kN; ⟹ 0.259T_2=43kN;\implies0.259T2=43kN;⟹ T2=166kN;T_2=166kN;T2=166kN;
Answer: T2=166kN;T_2=166kN;T2=166kN;
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