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. Assume the rope can provide a tension force of up to 29 KN before breaking, and use a “safety factor” of 10 (that is, the rope should only be required to undergo a tension force of 2.9 kN). Bob’s mass is 68.5 kg. (Figure 1) Determine the distance z that the rope must sag at a point halfway across if it is to be within its recommended safet range. Express your answer to two significant figures and include the appropriate units. li µ? ? I = 1.33 nm Submit Previous Answers Request Answer Figure 1 of 1 X Incorrect; Try Again; 2 attempts remaining
The Correct Answer and Explanation is :
To determine the distance zz that the rope must sag at the halfway point across the chasm, we will approach the problem using the concept of tension in the rope and the safety factor provided. Here is the step-by-step breakdown:
Given:
- Distance between trees (span): L=25 mL = 25 \, \text{m}
- Tension force before breaking: Tmax=29 kNT_{\text{max}} = 29 \, \text{kN}
- Safety factor: Safety factor=10\text{Safety factor} = 10
- Allowed tension force: Tsafe=2910=2.9 kN=2900 NT_{\text{safe}} = \frac{29}{10} = 2.9 \, \text{kN} = 2900 \, \text{N}
- Bob’s mass: m=68.5 kgm = 68.5 \, \text{kg}
- Acceleration due to gravity: g=9.81 m/s2g = 9.81 \, \text{m/s}^2
- Bob’s weight: W=mg=68.5×9.81=672.4 NW = mg = 68.5 \times 9.81 = 672.4 \, \text{N}
Step 1: Understand the Tension in the Rope
When Bob is halfway across, the rope forms a catenary curve. However, for simplicity, we can approximate this as a parabolic shape. The maximum tension will occur at the point of greatest sag, which is the middle of the rope.
Step 2: Use the Vertical Force Balance
At the midpoint, the rope will experience both vertical and horizontal components of the tension force. The tension is split into two components: the vertical component, which supports Bob’s weight, and the horizontal component, which is constant throughout the rope.
At the midpoint of the rope:
- The vertical component of the tension, TyT_y, should balance Bob’s weight.
- The horizontal component of the tension, TxT_x, is constant and must be the same throughout the rope.
We can set up the equation for vertical forces: Ty=WT_y = W
The total tension at the midpoint will be the resultant of the horizontal and vertical components: Tsafe=Tx2+Ty2T_{\text{safe}} = \sqrt{T_x^2 + T_y^2}
Now, using the given values: 2900=Tx2+672.422900 = \sqrt{T_x^2 + 672.4^2}
Solve for TxT_x: 29002=Tx2+672.422900^2 = T_x^2 + 672.4^2 Tx2=29002−672.42=8410000−452728.96=7957271.04T_x^2 = 2900^2 – 672.4^2 = 8410000 – 452728.96 = 7957271.04 Tx=7957271.04=2822.3 NT_x = \sqrt{7957271.04} = 2822.3 \, \text{N}
Step 3: Determine the Sag Distance zz
Now, the sag of the rope zz is determined by the geometry of the rope and the tension forces. For a rope hanging between two points with a small sag (parabolic approximation), the relation between the sag and tension can be approximated using the formula: z=WL28Txz = \frac{W L^2}{8 T_x}
Substitute the known values: z=672.4×2528×2822.3z = \frac{672.4 \times 25^2}{8 \times 2822.3} z=672.4×6258×2822.3=42025022578.4=18.6 mz = \frac{672.4 \times 625}{8 \times 2822.3} = \frac{420250}{22578.4} = 18.6 \, \text{m}
Final Answer:
The sag zz of the rope at the halfway point is 18.6 meters.
This ensures the rope remains within the safe tension range, as the tension force does not exceed the allowed value of 2.9 kN.