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.
I just did this problem for an assignment and used sin not tan. Otherwise I agree.
(.5 * mass * g) / sin(theta)
To find the tension in the wire, we can analyze the forces acting on the acrobat. There are two main forces involved: the tension in the wire pulling upwards and the force of gravity pulling downwards.
First, let's consider the force of gravity. The force of gravity acting on the acrobat can be calculated using the formula:
force of gravity = mass * acceleration due to gravity
Plugging in the given values:
force of gravity = 88.8 kg * 9.8 m/s^2
Next, let's analyze the tension in the wire. Since the acrobat is hanging from the middle of the wire, the tension in the two halves of the wire will be the same. This means that the vertical component of tension is equal in magnitude to the force of gravity:
tension * sin(angle) = force of gravity
Rearranging this equation to solve for tension, we get:
tension = force of gravity / sin(angle)
Plugging in the values we calculated earlier:
tension = (88.8 kg * 9.8 m/s^2) / sin(12.5 degrees)
Now we can calculate the tension using this equation.
Wouldn't tension be 1/2 mg/Tension?
the 1/2 is used because each side of the wire carries 1/2 the weight.
sorry. 1/2 mg/Tension=tanTheta
tension= 1/2 mg/tanTheta