Arlene is to walk across a “high wire” strung horizontally between two buildings 10.0 m apart. The sag in the rope when she is at the midpoint is 10.0º as shown in the figure below. If her mass is 50.0 kg, what is the tension in the rope at this point?
To find the tension in the rope at the midpoint, we need to consider the forces acting on Arlene.
Since Arlene is at the midpoint of the wire, the wire is symmetrically sagging in the shape of a catenary curve. The angle of sag given in the problem (10.0º) is the angle between the horizontal and the tangent line at the midpoint.
In this scenario, the tension in the wire can be resolved into two components: the vertical component (T_v) and the horizontal component (T_h).
To calculate the tension, we can use the following steps:
1. Analyze the forces acting on Arlene:
- The weight of Arlene acts vertically downwards. The magnitude of the weight can be calculated using the formula: weight = mass × gravitational acceleration.
- The vertical component of the tension acts upwards.
- There is no horizontal force acting on Arlene.
2. Resolve the tension into vertical and horizontal components:
- The tension can be resolved by using trigonometry. The vertical component (T_v) can be found using the formula: T_v = T × sin(angle).
- The horizontal component (T_h) is equal to the tension in the wire at the midpoint since there is no horizontal force acting on Arlene.
3. Use equilibrium conditions:
- At the midpoint, since Arlene is not accelerating, the vertical forces are balanced. Therefore, the weight of Arlene (mg) is equal to the vertical component of the tension (T_v).
4. Calculate the tension in the rope:
- Using the equation from step 3, we can solve for T_v: T_v = mg.
- Since T_v = T × sin(angle), we can substitute T_v with mg in the equation.
- The equation becomes: mg = T × sin(angle).
- Rearranging the equation, we get: T = mg / sin(angle).
Now, we can insert the given values into the equation and find the tension in the rope:
mass (m) = 50.0 kg
angle = 10.0º
T = (50.0 kg × 9.8 m/s^2) / sin(10.0º)
Calculating this will give you the tension in the rope at the midpoint.
Well, Arlene's certainly got some balance! She's defying gravity and sagging ropes. But let's get down to business.
To find the tension in the rope at the midpoint, we need to consider the gravitational force acting on Arlene.
Now, I could give you a serious answer with a bunch of physics equations, but I'm a clown bot, so I'll give you a silly one instead.
The tension in the rope at the midpoint depends on Arlene's ability to keep calm. If she starts freaking out, the tension in the rope will skyrocket! So, let's hope she's a cool customer.
In all seriousness, we can actually calculate the tension in the rope using trigonometry.
The sag in the rope forms a right triangle with the horizontal distance between the buildings. The angle of sag, 10.0º, is half the angle of the triangle.
So, in this case, the angle of the triangle is 20.0º. Now, we can use the sine function to find the height of the triangle (the sag in the rope).
sin(20.0º) = sag / (10.0 m / 2)
sag = sin(20.0º) * (10.0 m / 2)
sag ≈ 1.72 m
Now, we can calculate the gravitational force acting on Arlene:
F = m * g
F = 50.0 kg * 9.8 m/s²
F ≈ 490 N
Since the rope is at an angle with the vertical, we need to resolve this force into two components: one perpendicular to the rope (the tension we're trying to find) and one parallel to the rope.
The vertical component is equal to the gravitational force, and the horizontal component is zero since there's no force pulling the rope horizontally.
Now, using some more trigonometry magic, we can find the tension in the rope at the midpoint.
Tension = √(F² - perpendicular component²)
Tension = √(490 N² - 490 N²)
Tension ≈ √0 N²
So, the tension in the rope at the midpoint is essentially zero since the perpendicular component of the gravitational force is canceled out by the tension in the rope.
In simpler terms, Arlene's got nothing to worry about! The tension in the rope at the midpoint is negligible. Now, let's hope she doesn't have a fear of heights or ridiculous clown bots giving her physics lessons!
2 T sin(Θ) = m g
2 * T * sin(10.0º) = 50.0 * 9.81
answer is in Newtons