An acrobat hangs by his hands from the mid-
dle of a tightly stretched horizontal wire so
that the angle between the wire and the hori-
zontal is 12.5◦.
If the acrobat’s mass is 88.8 kg, what is
the tension in the wire? The acceleration of
gravity is 9.8 m/s2.
Answer in units of N.
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To find the tension in the wire, we can use the concept of equilibrium.
In equilibrium, the net force on an object is zero. Since the acrobat is hanging from the wire, the only two forces acting on him are the tension force in the wire and the force due to gravity.
First, we need to resolve the weight of the acrobat into its components parallel and perpendicular to the wire. The perpendicular component will balance out the tension force, and the parallel component will add to the tension force.
Given:
Mass of the acrobat (m) = 88.8 kg
Angle between the wire and horizontal (θ) = 12.5°
Acceleration due to gravity (g) = 9.8 m/s^2
Let's calculate the perpendicular component:
Perpendicular component of weight = m * g * sin(θ)
= 88.8 kg * 9.8 m/s^2 * sin(12.5°)
Next, let's calculate the parallel component:
Parallel component of weight = m * g * cos(θ)
= 88.8 kg * 9.8 m/s^2 * cos(12.5°)
Since the acrobat is in equilibrium, the tension in the wire will be equal to the perpendicular component of weight.
Tension in the wire = perpendicular component of weight
Now, we can plug in the given values to calculate the tension:
Tension in the wire = 88.8 kg * 9.8 m/s^2 * sin(12.5°)
Calculating this value will give us the tension in the wire.