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Chapter 4: Problem 48

A 55-kg gymnast hangs vertically from a pair of parallel rings. (a) If the ropes supporting the rings are attached to the ceiling directly above, what is the tension in each rope? (b) If the ropes are supported so that they make an angle of \(45^{\circ}\) with the ceiling, what is the tension in each rope?

Short Answer

Expert verified
(a) 269.5 N (b) 381.03 N in each rope.

Step by step solution

01

Understanding the Forces

The gymnast's weight is distributed equally between the two ropes. Therefore, each rope supports half of the gymnast's weight.
02

Calculate Weight of Gymnast

Calculate the weight of the gymnast using the formula for weight: \[ W = mg \]where \( m = 55 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \). Thus, \( W = 55 \times 9.8 = 539 \text{ N} \).
03

Calculate Tension When Ropes are Vertical

Since the ropes are vertical, the weight is equally distributed between the two ropes. Therefore, the tension in each rope is:\[ T_1 = T_2 = \frac{W}{2} = \frac{539}{2} = 269.5 \text{ N}. \]
04

Analyze Forces with Angled Ropes

When the ropes make a \(45^{\circ}\) angle with the ceiling, each tension \(T\) is a component of the vertical force equal to half the weight of the gymnast.
05

Use Trigonometry for Angled Ropes

The vertical component of the tension in each rope is \( T \times \cos(45^\circ) \). This must equal half the weight:\[ T \cdot \cos(45^\circ) = \frac{539}{2} \].
06

Calculate Tension in Each Rope at Angle

Solving for \(T\), we get:\[ T = \frac{539/2}{\cos(45^\circ)} = \frac{269.5}{0.7071} \approx 381.03 \text{ N}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tension in Ropes
Tension in ropes is a fundamental concept in physics, especially when studying equilibrium problems. When a gymnast hangs from ropes, the tension is the force exerted by the ropes that supports the weight of the gymnast. This happens because the ropes are pulling upwards against gravity, preventing the gymnast from falling.
  • The tension will always adjust to be equal to the forces necessary to keep the system (like the gymnast) in equilibrium.
  • When the ropes are vertical, the entire gravitational force or weight is shared equally between the ropes. This means that if there are two ropes, each rope carries half of the load.
  • When the ropes are angled, the situation becomes a bit more complex, but tension still serves the same purpose鈥攈olding the gymnast in place.
Understanding how tension in ropes works helps us analyze how systems remain stable and how forces are transmitted through structures. It's crucial to remember that tension keeps objects from accelerating by balancing out the forces of gravity.
Trigonometry in Physics
Trigonometry is a key mathematical tool used in physics to resolve forces acting at angles. In our gymnast example, when the ropes are angled, trigonometry comes into play to determine the tension in the ropes.

Using trigonometry allows us to break down forces into components:
  • The vertical component, which opposes gravity.
  • The horizontal component, which is often balanced by an equal and opposite force.
For an angle of \(45^{\circ}\), the cosine function is used to find the vertical component of the tension:
  • The formula \(T \cdot \cos(45^{\circ}) = \frac{539}{2}\) is derived from setting the vertical component of tension equal to the gymnast鈥檚 weight supported by each rope.
  • Knowing \(\cos(45^{\circ}) = 0.7071\), the required tension \(T\) is solved by rearranging the equation to \(T = \frac{269.5}{0.7071}\).
Trigonometry not only provides a practical way of solving these problems but also helps us gain a deeper understanding of how forces interact in real-world scenarios.
Forces in Equilibrium
Forces in equilibrium are essential when evaluating how objects, like our gymnast, remain stationary despite multiple forces acting on them. When discussing forces in equilibrium:
  • A system is said to be in equilibrium if the sum of all forces and the sum of all moments (torques) acting on it are zero.
  • The gymnast hanging at rest implies that all forces are balancing out.
  • The ropes' tension fully supports the gymnast鈥檚 weight, providing a net force of zero.
When applying equilibrium conditions:
  • Vertically, the downward gravitational force is balanced by the upward component of the tension in the ropes.
  • Horizontally, if any, forces are balanced such that there is no lateral movement.
Each situation, whether with vertical or angled ropes, demonstrates the application of equilibrium conditions to solve for unknowns like the tension in the ropes. This understanding is pivotal, as it helps us in designing safe structures and understanding balance in scenarios involving multiple forces.

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Most popular questions from this chapter

A crate sits on a flat-bed truck that is traveling with a speed of \(50 \mathrm{~km} / \mathrm{h}\) on a straight, level road. If the coefficient of static friction between the crate and the truck bed is 0.30 , in how short a distance can the truck stop with a constant acceleration without the crate sliding?

(a) When an object is on an inclined plane, the normal force exerted by the inclined plane on the object is (1) less than, (2) equal to, (3) more than the weight of the object. Why? (b) For a \(10-\mathrm{kg}\) object on a \(30^{\circ}\) inclined plane, what are the object's weight and the normal force exerted on the object by the inclined place?

A 20-kg box sits on a rough horizontal surface. When a horizontal force of \(120 \mathrm{~N}\) is applied, the object accelerates at \(1.0 \mathrm{~m} / \mathrm{s}^{2} .\) (a) If the applied force is doubled, the acceleration will (1) increase, but less than double; (2) also double; (3) increase, but more than double. Why? (b) Calculate the acceleration to prove your answer to part (a).

IE .?? Three horizontal forces (the only horizontal ones) act on a box sitting on a floor. One (call it \(F_{1}\) ) acts due east and has a magnitude of \(150 \mathrm{lb}\). A second force (call it \(F_{2}\) ) has an easterly component of \(30.0 \mathrm{lb}\) and a southerly component of \(40.0 \mathrm{lb}\). The box remains at rest. (Neglect friction.) (a) Sketch the two known forces on the box. In which quadrant is the unknown third force: (1) the first quadrant; (2) the second quadrant; (3) the third quadrant; or (4) the fourth quadrant? (b) Find the unknown third force in newtons and compare your answer to the sketched estimate.

In an Olympic figure-skating event, a 65-kg male skater pushes a \(45-\mathrm{kg}\) female skater, causing her to accelerate at a rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\). At what rate will the male skater accelerate? What is the direction of his acceleration?
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