Math / Geometry

QuestionAfter a mishap, a $76.0-\mathrm{kg}$ 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

STEP 1

1. The system is in static equilibrium, meaning the net force in both the horizontal and vertical directions is zero.
2. The weight of the circus performer acts vertically downward.
3. The angles given are with respect to the vertical and horizontal as described.

STEP 2

1. Draw a free-body diagram.
2. Set up equations for equilibrium conditions.
3. Solve for the tensions in the ropes.

STEP 3

Draw a free-body diagram of the forces acting on the trapeze. Include: - The weight $w = mg$ acting downward, where $m = 76.0 \, \text{kg}$ and $g = 9.8 \, \text{m/s}^2$ . - Tension $T_1$ at an angle $15^\circ$ from the vertical. - Tension $T_2$ at an angle $10^\circ$ from the horizontal.

STEP 4

Set up the equilibrium conditions for the vertical forces. The sum of vertical components of the tensions must equal the weight:
$T_1 \cos(15^\circ) + T_2 \sin(10^\circ) = mg$

STEP 5

Set up the equilibrium conditions for the horizontal forces. The horizontal components of the tensions must cancel each other out:
$T_1 \sin(15^\circ) = T_2 \cos(10^\circ)$

STEP 6

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

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QuestionAfter a mishap, a $76.0-\mathrm{kg}$ 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

STEP 1

1. The system is in static equilibrium, meaning the net force in both the horizontal and vertical directions is zero.
2. The weight of the circus performer acts vertically downward.
3. The angles given are with respect to the vertical and horizontal as described.

STEP 2

1. Draw a free-body diagram.
2. Set up equations for equilibrium conditions.
3. Solve for the tensions in the ropes.

STEP 3

Draw a free-body diagram of the forces acting on the trapeze. Include: - The weight $w = mg$ acting downward, where $m = 76.0 \, \text{kg}$ and $g = 9.8 \, \text{m/s}^2$ . - Tension $T_1$ at an angle $15^\circ$ from the vertical. - Tension $T_2$ at an angle $10^\circ$ from the horizontal.

STEP 4

Set up the equilibrium conditions for the vertical forces. The sum of vertical components of the tensions must equal the weight:
$T_1 \cos(15^\circ) + T_2 \sin(10^\circ) = mg$

STEP 5

Set up the equilibrium conditions for the horizontal forces. The horizontal components of the tensions must cancel each other out:
$T_1 \sin(15^\circ) = T_2 \cos(10^\circ)$

STEP 6

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QuestionAfter a mishap, a 76.0−kg76.0-\mathrm{kg}76.0−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

STEP 1

1. The system is in static equilibrium, meaning the net force in both the horizontal and vertical directions is zero.2. The weight of the circus performer acts vertically downward.3. The angles given are with respect to the vertical and horizontal as described.

STEP 2

1. Draw a free-body diagram.2. Set up equations for equilibrium conditions.3. Solve for the tensions in the ropes.

STEP 3

STEP 4

Set up the equilibrium conditions for the vertical forces. The sum of vertical components of the tensions must equal the weight:T1cos⁡(15∘)+T2sin⁡(10∘)=mg T_1 \cos(15^\circ) + T_2 \sin(10^\circ) = mg T1​cos(15∘)+T2​sin(10∘)=mg

STEP 5

Set up the equilibrium conditions for the horizontal forces. The horizontal components of the tensions must cancel each other out:T1sin⁡(15∘)=T2cos⁡(10∘) T_1 \sin(15^\circ) = T_2 \cos(10^\circ) T1​sin(15∘)=T2​cos(10∘)

STEP 6

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionAfter a mishap, a $76.0-\mathrm{kg}$ 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.

1. The system is in static equilibrium, meaning the net force in both the horizontal and vertical directions is zero.
2. The weight of the circus performer acts vertically downward.
3. The angles given are with respect to the vertical and horizontal as described.

1. Draw a free-body diagram.
2. Set up equations for equilibrium conditions.
3. Solve for the tensions in the ropes.

Set up the equilibrium conditions for the vertical forces. The sum of vertical components of the tensions must equal the weight:
$T_1 \cos(15^\circ) + T_2 \sin(10^\circ) = mg$

Set up the equilibrium conditions for the horizontal forces. The horizontal components of the tensions must cancel each other out:
$T_1 \sin(15^\circ) = T_2 \cos(10^\circ)$