T1*sin148 + T2*sin74 = 725 N.
Eq1: 0.53T1 + 0.96T2 = 725.
T1*Cos148 + T2*Cos74 = 0.
Eq2: -0.85T1 + 0.28T2 = 0.
-0.85T1 = -0.28T2.
T1 = 0.324T2.
In Eq1, Replace T1 with 0.324T2:
0.53*0.324T2 + 0.96T2 = 725.
0.172T2 + 0.96T2 = 725.
1.13T2 = 725.
T2 = 640.5 N.
T1 = 0.324*640.5 = 207.5 N. = Ta.
In a rescue, a 74.1 kg police officer is suspended by two cables, as shown in the figure. Cable A makes an angle θA = 32.0° with respect to the horizontal, and cable B, θB = 74.0°. Sketch a free body diagram. Calculate the tension in cable A.
1 answer
1 answer