How Is Tension Calculated in an Acrobat's Wire?

[jfri14]

Homework Statement

An acrobat hangs by his hands from the middle of a tightly strestched horizontal wire so that the angle between the wire and the horizontal is 9.15 degrees. If the acrobat's mass is 88.4kg, what is the tension in the wire? Answer in units of N.

Homework Equations

Sum of Fy =0
Sum of Fx =0
F= sqrt(Fx^2 + Fy^2)

The Attempt at a Solution

Fysin(9.15)-(88.4kg)(9.8m/s2)=0
Fy=5447.879N

Fxcos(9.15)-(88.4kg)(9.8m/s2)=0
Fx=877.486N

F= Sqrt(5447.879^2+ 877.486^2)
F=5518.0945N

 

[fliinghier]

draw a free body diagram and add the forces as vectors.

 

[ogb p]

In this scenario, the acrobat is in a state of static equilibrium, meaning that the forces acting on him are balanced and there is no net force causing him to move. To determine the tension in the wire, we can use the equations for sum of forces in the x and y directions. In the y-direction, the force of gravity (mg) is acting downwards while the vertical component of tension in the wire is acting upwards. Since the acrobat is not moving in the y-direction, the sum of forces in this direction must be equal to zero. We can solve for the vertical component of tension (Fy) using the equation Fysin(θ) = mg, where θ is the angle between the wire and the horizontal. Plugging in the given values, we get Fy = 5447.879N.

In the x-direction, the horizontal component of tension in the wire is acting to the left while there is no other force acting in the x-direction. Again, since the acrobat is not moving in the x-direction, the sum of forces in this direction must be equal to zero. We can solve for the horizontal component of tension (Fx) using the equation Fxcos(θ) = 0. Plugging in the given values, we get Fx = 877.486N.

To determine the total tension in the wire (F), we can use the equation F = √(Fx^2 + Fy^2). Plugging in the values for Fx and Fy, we get a tension of 5518.0945N in the wire. This is the force that is keeping the acrobat in static equilibrium, allowing him to hang from the wire without falling.