Math  /  Geometry

QuestionAfter a mishap, a 76.0kg76.0-\mathrm{kg} circus performer clings to a trapeze, which is being pulled to the side by another circus artist, as shown here. Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution.

Studdy Solution
Solve the system of equations for T1 T_1 and T2 T_2 .
First, solve for T2 T_2 using the horizontal equilibrium equation:
T2=T1sin(15)cos(10) T_2 = \frac{T_1 \sin(15^\circ)}{\cos(10^\circ)}
Substitute T2 T_2 into the vertical equilibrium equation:
T1cos(15)+(T1sin(15)cos(10))sin(10)=mg T_1 \cos(15^\circ) + \left(\frac{T_1 \sin(15^\circ)}{\cos(10^\circ)}\right) \sin(10^\circ) = mg
Simplify and solve for T1 T_1 :
T1(cos(15)+sin(15)sin(10)cos(10))=mg T_1 \left( \cos(15^\circ) + \frac{\sin(15^\circ) \sin(10^\circ)}{\cos(10^\circ)} \right) = mg
T1=mgcos(15)+sin(15)sin(10)cos(10) T_1 = \frac{mg}{\cos(15^\circ) + \frac{\sin(15^\circ) \sin(10^\circ)}{\cos(10^\circ)}}
Calculate T1 T_1 :
T1=76.0×9.8cos(15)+sin(15)sin(10)cos(10) T_1 = \frac{76.0 \times 9.8}{\cos(15^\circ) + \frac{\sin(15^\circ) \sin(10^\circ)}{\cos(10^\circ)}}
T1746.6N T_1 \approx 746.6 \, \text{N}
Now calculate T2 T_2 :
T2=746.6×sin(15)cos(10) T_2 = \frac{746.6 \times \sin(15^\circ)}{\cos(10^\circ)}
T2195.2N T_2 \approx 195.2 \, \text{N}
The tensions in the ropes are approximately T1=746.6N T_1 = 746.6 \, \text{N} and T2=195.2N T_2 = 195.2 \, \text{N} .

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