Math / Trigonometry

Question2. (II) 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 $10.0^{\circ}$ , as shown in Fig. 4-39. If her mass is 50.0 kg , what is the tension in the rope at this point?

Studdy Solution

STEP 1

1. The high wire is strung horizontally between two buildings.
2. The distance between the buildings is $10.0$ meters.
3. The sag in the rope when Arlene is at the midpoint is $10.0^{\circ}$ .
4. Arlene's mass is $50.0$ kg.
5. The gravitational acceleration $g$ is $9.8 \, \text{m/s}^2$ .

STEP 2

1. Calculate the weight of Arlene.
2. Analyze the forces acting on the rope.
3. Use trigonometry to find the tension in the rope.

STEP 3

Calculate the weight of Arlene using the formula:
$W = m \times g$
where $m = 50.0 \, \text{kg}$ and $g = 9.8 \, \text{m/s}^2$ .
$W = 50.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 490 \, \text{N}$

STEP 4

At the midpoint, the forces acting on the rope include the tension $T$ and the weight of Arlene. The rope forms an angle of $10.0^{\circ}$ with the horizontal.
The vertical component of the tension must balance the weight of Arlene:
$2T \sin(10.0^{\circ}) = W$

STEP 5

Solve for the tension $T$ :
$2T \sin(10.0^{\circ}) = 490 \, \text{N}$
$T = \frac{490 \, \text{N}}{2 \sin(10.0^{\circ})}$
Calculate $\sin(10.0^{\circ})$ and then find $T$ :
$\sin(10.0^{\circ}) \approx 0.1736$
$T = \frac{490 \, \text{N}}{2 \times 0.1736}$
$T \approx \frac{490 \, \text{N}}{0.3472}$
$T \approx 1411 \, \text{N}$
The tension in the rope at the midpoint is:
$\boxed{1411 \, \text{N}}$

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