High School

After a mishap, a 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.

Answer :

Final answer:

To calculate the tension in the ropes, we need to consider the forces acting on the circus performer. The net force acting on the performer is zero, so the sum of the forces in the horizontal direction must be equal to the sum of the forces in the vertical direction. Using trigonometry, we can set up equations to find the tensions in the ropes.

Explanation:

In order to calculate the tension in the ropes, we need to consider the forces acting on the circus performer.

  1. The weight of the performer, given by the formula F = m * g, where m is the mass of the performer and g is the acceleration due to gravity.
  2. The tension in the left rope, T1.
  3. The tension in the right rope, TR.

Since the performer is motionless, the net force acting on him is zero. This means that the sum of the forces in the horizontal direction (T1 and TR) must be equal to the sum of the forces in the vertical direction (the weight of the performer).

To find the tensions in the ropes, we need to use trigonometry. Since the performer is motionless, the vertical component of the tension in each rope must be equal to the weight of the performer. Using the property of similar triangles, we can set up the following equation:

T1*sin(theta) = m * g

where theta is the angle between the left rope and the horizontal direction. Solving for T1, we get:

T1 = (m * g) / sin(theta)

Similarly, we can set up the equation for TR:

TR*sin(theta) = m * g

T1 = (m * g) / sin(theta)

Therefore, the tension in the left rope is (m * g) / sin(theta) and the tension in the right rope is (m * g) / sin(theta).