Math / Trigonometry

Question49. (II) Bob traverses a chasm by stringing a rope between a tree on one side of the chasm and a tree on the opposite side, 25 m away, Fig. 4-47. Assume the rope can provide a tension force of up to 31 kN before breaking, and use a "safety factor" of 10 (that is, the rope should only be required to undergo a tension force of 3.1 kN ). (a) If Bob's mass is 72.0 kg , determine the distance $x$ that the rope must sag at a point halfway across if it is to be within its recommended safety range. (b) If the rope sags by only one-fourth the distance found in ( $a$ ), determine the tension force in the rope. Will the rope break?
FIGURE 4-47 Problem 49.

Studdy Solution

STEP 1

1. The rope is strung horizontally between two trees 25 meters apart.
2. The maximum safe tension in the rope is 3.1 kN, considering the safety factor.
3. Bob's mass is 72.0 kg.
4. The rope sags symmetrically at the midpoint.

STEP 2

1. Calculate the force due to Bob's weight.
2. Use the geometry of the sagging rope to find the sag distance $x$ for the safe tension.
3. Calculate the tension in the rope if the sag is one-fourth of the distance found in part (a).
4. Determine if the rope will break with the reduced sag.

STEP 3

Calculate the force due to Bob's weight:
$F_{\text{weight}} = m \cdot g$ $= 72.0 \, \text{kg} \times 9.81 \, \text{m/s}^2$ $= 706.32 \, \text{N}$

STEP 4

Use the geometry of the sagging rope to find the sag distance $x$ :
- Assume the rope forms two equal angles with the horizontal at the midpoint. - The horizontal component of the tension $T$ is the same on both sides of the midpoint. - The vertical component of the tension must balance Bob's weight.
Using trigonometry:
$T \cdot \sin(\theta) = \frac{F_{\text{weight}}}{2}$ $T \cdot \cos(\theta) = \frac{3.1 \, \text{kN}}{2}$
Using the small angle approximation, where $\sin(\theta) \approx \frac{x}{L/2}$ and $\cos(\theta) \approx 1$ :
$T \cdot \frac{x}{12.5} = 353.16 \, \text{N}$ $T = 3100 \, \text{N}$
Substitute $T$ into the equation:
$3100 \cdot \frac{x}{12.5} = 353.16$ $x = \frac{353.16 \times 12.5}{3100}$ $x \approx 1.42 \, \text{m}$

STEP 5

Calculate the tension if the sag is one-fourth of the distance found in part (a):
$x = \frac{1.42}{4} \approx 0.355 \, \text{m}$
Using the same trigonometric relationships:
$T \cdot \frac{0.355}{12.5} = 353.16$ $T = \frac{353.16 \times 12.5}{0.355}$ $T \approx 12420 \, \text{N}$

STEP 6

Determine if the rope will break:
- The calculated tension $12420 \, \text{N}$ is greater than the maximum safe tension $3100 \, \text{N}$ . - The rope will break under this tension.
The distance $x$ that the rope must sag for safety is approximately $\boxed{1.42 \, \text{m}}$ . If the rope sags by only one-fourth of this distance, the tension will be $\boxed{12420 \, \text{N}}$ , and the rope will break.

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms