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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
Chapter 12, Problem 55a

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 FT in the support cable, and then determine the force FA\(\overrightarrow{F_{A}\)} on the truss at pin A. Neglect the weight of the trusses, which is small compared to the load.
Diagram of a cantilever truss with a 62.0 kN load at point E and dimensions labeled, illustrating equilibrium concepts.

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Identify the forces acting on the truss: The load at point E is given as \( M g = 62.0 \; \text{kN} \), the tension in the support cable is \( F_T \), and the reaction force at pin A is \( \vec{F}_A \). Neglect the weight of the truss itself.
Choose a pivot point for the torque calculation: To simplify the torque equation, select point A as the pivot. This eliminates the reaction force \( \vec{F}_A \) from the torque equation since its lever arm is zero.
Write the torque equation about point A: The torque due to the load \( M g \) at point E is \( \tau_E = M g \cdot d_E \), where \( d_E \) is the horizontal distance from point A to point E. The torque due to the tension \( F_T \) in the cable is \( \tau_T = F_T \cdot d_T \cdot \sin(\theta) \), where \( d_T \) is the length of the cable and \( \theta \) is the angle the cable makes with the horizontal. Set the sum of torques to zero for equilibrium: \( \tau_E - \tau_T = 0 \).
Solve for the tension \( F_T \): Rearrange the torque equation to isolate \( F_T \): \( F_T = \frac{M g \cdot d_E}{d_T \cdot \sin(\theta)} \). Substitute the known values for \( M g \), \( d_E \), \( d_T \), and \( \sin(\theta) \) to calculate \( F_T \).
Determine the reaction force \( \vec{F}_A \): Use the equilibrium conditions for forces. The vertical forces must balance: \( F_{A_y} + F_T \cdot \sin(\theta) = M g \). The horizontal forces must also balance: \( F_{A_x} = F_T \cdot \cos(\theta) \). Solve these equations to find the components of \( \vec{F}_A \), and combine them to find the magnitude and direction of \( \vec{F}_A \).

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Key Concepts

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

Torque

Torque is a measure of the rotational force applied to an object, calculated as the product of the force and the distance from the pivot point (lever arm). In this context, the torque equation is used to analyze the equilibrium of the cantilever truss, where the sum of the torques around a point must equal zero for the system to be in balance.
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Equilibrium

Equilibrium in physics refers to a state where the sum of forces and the sum of torques acting on an object are both zero. For the cantilever truss, this means that the downward force from the load must be balanced by the upward tension in the support cable and any reactions at the pin, ensuring that the structure remains stable and does not rotate.
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Force Analysis

Force analysis involves breaking down the forces acting on a structure to understand how they interact. In this problem, it is essential to determine the tension in the support cable and the reaction force at pin A by applying Newton's laws, ensuring that all forces are accounted for in the equilibrium equations.
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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 FE\(\overrightarrow{F_{E}\)} 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: FE\(\overrightarrow{F_{E}\)} 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

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.

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Textbook Question

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).

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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). What is the torque due to this sign calculated about the point where the pole meets the wall?

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Textbook Question

A 15-cm-long tendon was found to stretch 3.7 mm by a force of 13.4 N. The tendon was approximately round with an average diameter of 8.5 mm. Calculate Young’s modulus of this tendon.

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