Ch. 12 - Static Equilibrium; Elasticity and Fracture

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 12 - Static Equilibrium; Elasticity and Fracture

Problem 58

Chapter 12, Problem 58

The subterranean tension ring that surrounds the dome in Fig. 12–39 exerts the balancing horizontal force on the abutments for the dome and is 36-sided, so each segment makes a 10° angle with the adjacent one (Fig. 12–83). Calculate the tension F that must exist in each segment so that the required force of 4.2 x 10⁵ N can be exerted at each corner (Example 12–14).

Verified step by step guidance

Understand the problem: The tension ring is a 36-sided polygon, and each segment of the ring makes a 10° angle with the adjacent one. The goal is to calculate the tension F in each segment such that the horizontal force exerted at each corner is 4.2 × 10⁵ N.

Break down the forces: At each corner of the polygon, the horizontal force is the result of the horizontal components of the tensions in the two adjacent segments. Use symmetry to simplify the problem, as the geometry is regular.

Express the horizontal force: The horizontal component of the tension in one segment is given by F × cos(θ/2), where θ = 10° is the angle between adjacent segments. Since there are two segments contributing to the horizontal force at each corner, the total horizontal force is 2 × F × cos(θ/2).

Set up the equation: Equate the total horizontal force to the required force at each corner. This gives the equation 2 × F × cos(θ/2) = 4.2 × 10⁵ N.

Solve for F: Rearrange the equation to isolate F. This gives F = (4.2 × 10⁵ N) / (2 × cos(θ/2)). Substitute θ = 10° into the equation to find the value of F.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Tension in Structures

Tension refers to the pulling force transmitted through a string, rope, or structural element. In the context of the tension ring surrounding the dome, it is crucial to understand how tension is distributed among the segments. Each segment must exert a force that contributes to the overall stability of the structure, ensuring that the required horizontal force is balanced effectively.

Force Resolution

Force resolution involves breaking down a force into its components, typically along specified axes. In this problem, the horizontal force exerted by the tension segments must be resolved into components that can be analyzed to find the tension in each segment. Understanding how to resolve forces is essential for calculating the necessary tension to achieve the desired force at the corners.

Equilibrium of Forces

The equilibrium of forces states that for a system to be in a state of rest or constant motion, the sum of all forces acting on it must equal zero. In this scenario, the tension forces in the segments must balance the external force of 4.2 x 10⁵ N at each corner. Recognizing the conditions for equilibrium is vital for determining the required tension in the segments of the tension ring.

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Related Practice

Textbook Question

A 50-story building is being planned. It is to be 180.0 m high with a base 46.0 m by 76.0 m. Its total mass will be about 1.8 x 10⁷ kg, and its weight therefore about 1.8 x 10⁸ N. Suppose a 200-km/h wind exerts a force of 950N/m² over the 76.0-m-wide face (Fig. 12–86). Calculate the torque about the potential pivot point, the rear edge of the building (where $\vec{F_{E}}$ acts in Fig. 12–86), and determine whether the building will topple. Assume the total force of the wind acts at the midpoint of the building’s face, and that the building is not anchored in bedrock. [Hint: $\vec{F_{E}}$ in Fig. 12–86 represents the force that the Earth would exert on the building in the case where the building would just begin to tip.]

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Textbook Question

A pole projects horizontally from the front wall of a shop. A 6.1-kg sign hangs from the pole at a point 2.2 m from the wall (Fig. 12–88). If the pole is not to fall off, there must be another torque exerted to balance it. What exerts this torque? Use a diagram to show how this torque must act.

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Textbook Question

Assume the supports of the uniform cantilever shown in Fig. 12–79 (m = 2900 kg) are made of wood. Calculate the minimum cross-sectional area required of each, assuming a safety factor of 9.0.

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Textbook Question

A steel cable is to support an elevator whose total (loaded) mass is not to exceed 3100 kg. If the maximum acceleration of the elevator is 1.8 m/s² , calculate the diameter of cable required. Assume a safety factor of 8.0.

Textbook Question

A heavy load Mg = 62.0 kN hangs at point E of the single cantilever truss shown in Fig. 12–81. Use a torque equation for the truss as a whole to determine the tension F_T in the support cable, and then determine the force $\vec{F_{A}}$ on the truss at pin A. Neglect the weight of the trusses, which is small compared to the load.

Textbook Question

A pole projects horizontally from the front wall of a shop. A 6.1-kg sign hangs from the pole at a point 2.2 m from the wall (Fig. 12–88). What is the torque due to this sign calculated about the point where the pole meets the wall?

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