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. 4–65. 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). (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?
The correct answer and explanation is:
To solve this problem, we analyze the forces acting on Bob as he hangs from the midpoint of the rope.
Solution (a): Finding the sag distance xx
We use the following principles:
- Bob’s weight W=mgW = mg, where g=9.8 m/s2g = 9.8 \, \text{m/s}^2.
- The horizontal components of the tension forces are equal (TxT_x).
- The vertical component of the tension force balances Bob’s weight (2Ty=W2T_y = W).
Bob’s weight: W=72.0 kg×9.8 m/s2=705.6 N.W = 72.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 705.6 \, \text{N}.
The total rope length LL is fixed at 25 m. To stay within the safety range, the maximum tension force should not exceed 2.9 kN.
Using trigonometry: tanθ=TyTx,andsinθ=TyT.\tan \theta = \frac{T_y}{T_x}, \quad \text{and} \quad \sin \theta = \frac{T_y}{T}.
At equilibrium: Tx=L2x Ty.T_x = \frac{L}{2x} \, T_y.
Solving for xx and ensuring T≤2.9 kNT \leq 2.9 \, \text{kN}.
Solution (b): Updated tension force.
Solution
Part (a): Sag Distance xx
For the rope to stay within the recommended safety range (tension T≤2.9 kNT \leq 2.9 \, \text{kN}): x=1.53 m.x = 1.53 \, \text{m}.
Part (b): Tension When Sag is Reduced
If the sag is reduced to one-fourth of this value (x=1.53/4=0.383 mx = 1.53 / 4 = 0.383 \, \text{m}), the tension force increases significantly. The calculated tension force in this case is: T=11,519.24 N.T = 11,519.24 \, \text{N}.
This tension exceeds the rope’s breaking point of 29 kN, so the rope will not break.
Explanation
To determine the sag distance xx, we analyze the equilibrium of forces acting on Bob, who is hanging at the midpoint of the rope. Bob’s weight (W=705.6 NW = 705.6 \, \text{N}) creates a vertical force that is balanced by the vertical components (TyT_y) of the tension in the rope. The horizontal components (TxT_x) are equal due to symmetry. Using trigonometric relationships, we calculate the total tension TT, which depends on xx, and ensure it stays within the safety range (T≤2.9 kNT \leq 2.9 \, \text{kN}).
The sag distance x=1.53 mx = 1.53 \, \text{m} ensures the tension remains within the safety range. However, reducing the sag to one-fourth of this value (x=0.383 mx = 0.383 \, \text{m}) increases the horizontal component of tension significantly, as Tx∝1/xT_x \propto 1/x. Consequently, the total tension TT grows to 11,519.24 N11,519.24 \, \text{N}, far exceeding the safety limit. While this is below the rope’s breaking strength of 29 kN29 \, \text{kN}, it is unsafe and violates the safety factor design.
This illustrates the importance of sag in mitigating tension forces in ropes. A larger sag reduces the horizontal forces, keeping the total tension manageable, while a smaller sag increases the horizontal forces exponentially.