/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 A Truck on a Drawbridge. A loade... [FREE SOLUTION] | 91Ó°ÊÓ

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A Truck on a Drawbridge. A loaded cement mixer drives onto an old drawbridge, where it stalls with its center of gravity three-quarters of the way across the span. The truck driver radios for help, sets the handbrake, and waits. Meanwhile, a boat approaches, so the drawbridge is raised by means of a cable attached to the end opposite the hinge (Fig. P11.56). The drawbridge is 40.0 \(\mathrm{m}\) long and has a mass of \(18,000 \mathrm{kg} ;\) its center of gravity is at its midpoint. The cement mixer, with driver, has mass \(30,000\) kg. When the drawbridge has been raised to an angle of \(30^{\circ}\) above the horizontal, the cable makes an angle of \(70^{\circ}\) with the surface of the bridge. (a) What is the tension \(T\) in the cable when the drawbridge is held in this position? (b) What are the horizontal and vertical components of the force the hinge exerts on the span?

Short Answer

Expert verified
(a) Tension in the cable is approximately 164,200 N. (b) Horizontal hinge force is about 56,195 N, vertical hinge force is about 316,157 N.

Step by step solution

01

Identify Forces and Calculate Weight Components

Firstly, identify the forces acting on the drawbridge. There are two primary forces:1. The weight of the drawbridge itself, acting downward through its center of gravity, located at the midpoint.2. The weight of the cement mixer, located three-quarters of the way along the span.The weight of the drawbridge, \( W_d = m_d \times g \), where \( m_d = 18000 \ kg \) and \( g = 9.8 \ m/s^2 \), results in \( W_d = 18000 \times 9.8 = 176,400 \ N \).The weight of the cement mixer, \( W_c = m_c \times g \), where \( m_c = 30000 \ kg \), results in \( W_c = 30000 \times 9.8 = 294,000 \ N \).
02

Calculate Torque for Equilibrium

Since the drawbridge is not moving, it is in rotational equilibrium; thus, the sum of the torques around the hinge must be zero. The torque due to a weight is given by \( \tau = F \cdot d \cdot \sin(\theta) \), where \( d \) is the perpendicular distance from the hinge.For the drawbridge's weight, \( \tau_d = W_d \cdot (\frac{40}{2}) \cdot \sin(30^\circ) = 176,400 \cdot 20 \cdot 0.5 = 1,764,000 \ N\cdot m \).For the cement mixer, \( \tau_c = W_c \cdot (\frac{3}{4} \times 40) \cdot \sin(30^\circ) = 294,000 \cdot 30 \cdot 0.5 = 4,410,000 \ N\cdot m \).
03

Solve for Cable Tension Using Torque Balance

The tension in the cable creates torque that opposes the sum of the drawbridge and cement mixer torques. For equilibrium:\[ T \cdot 40 \cdot \sin(70^\circ) = \tau_d + \tau_c \]\[ T \cdot 40 \cdot 0.9397 = 1,764,000 + 4,410,000 \]\[ T \cdot 37.588 = 6,174,000 \]Solving for \(T\) gives:\[ T = \frac{6,174,000}{37.588} = 164,199.5 \ N \approx 164,200 \ N \]
04

Determine the Components of Hinge Force

The hinge supports the bridge, exerting two force components: horizontal (\(F_{hx}\)) and vertical (\(F_{hy}\)).**Horizontal Components:**The horizontal component of the tension is balanced by the horizontal hinge force:\[ T \cdot \cos(70^\circ) = F_{hx} \]\[ F_{hx} = 164,200 \cdot 0.342 \approx 56,194.8 \ N \]**Vertical Components:**The vertical forces should also sum to zero (considering both weights and the vertical component of tension):\[ F_{hy} + T \cdot \sin(70^\circ) = W_d + W_c \]\[ F_{hy} + 164,200 \cdot 0.9397 = 176,400 + 294,000 \]\[ F_{hy} + 154,243.14 = 470,400 \]\[ F_{hy} = 470,400 - 154,243.14 \approx 316,156.86 \ N \]

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Key Concepts

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

Rotational Mechanics
Rotational mechanics is a branch of physics that deals with objects rotating about a pivot point. In this exercise, the drawbridge is a classic example of rotational mechanics in action. The bridge can rotate around its hinge, which serves as the pivot. For such a system to be stable and in equilibrium, the torques around this point must balance out, meaning the sum of the clockwise torques must equal the sum of the counterclockwise torques.

Torque ( \( \tau \) ) is a measure of how much a force causes an object to rotate. It's calculated using the formula: \( \tau = F \cdot d \cdot \sin(\theta) \) , where \( F \) is the force applied, \( d \) is the distance from the pivot, and \( \theta \) is the angle between the force direction and the line from the pivot.

For the bridge scenario, each weight (drawbridge and cement mixer) contributes a torque around the hinge. These must be countered by the torque from the tension in the cable. Solving these equations helps us find the tension required to keep the bridge stable. Understanding this concept is crucial for solving problems involving rotating systems and equilibrium.
Tension in Cables
The tension in cables is a pivotal concept in mechanics, particularly when supporting structures like the drawbridge in this problem. Tension is the force exerted along a cable or rope, aiming to counteract other forces to keep a system in balance.

In our exercise, the drawbridge's stability relies on the tension in the cable, which counterbalances the torque produced by the masses of both the bridge and truck. To determine this tension, we calculate the torque due to the entire system and equate it to the torque from the cable's tension.

The cable's tension ( \( T \) ) is found using the formula: \( T \cdot 40 \cdot \sin(70^\circ) = 6,174,000 \) . This equation is derived from summing the torques and setting them to zero because the system is at rest and in equilibrium.
  • The length of the bridge, angle of rotation, and the position of weights all affect how the tension is distributed through the cable.
  • Understanding tension aids in designing cables for suspension bridges, elevators, and other structures.
Forces in Equilibrium
Forces in equilibrium is a fundamental topic that revolves around the concept that a system remains stable and unchanging when all forces acting on it are balanced. In our exercise, the drawbridge is held steady due to both rotational equilibrium and the balance of forces at the hinge.

The forces at play include:
  • The gravitational force acting downward due to the weights of the drawbridge and the cement mixer.
  • The tension in the cable pulling upwards and slightly sideways.
  • The hinge's forces, both horizontal and vertical, which keep the bridge from collapsing.
Forces can be broken down into components that are parallel and perpendicular to surfaces. In equilibrium, the sum of these forces in any direction should be zero.

The horizontal and vertical force components at the hinge are calculated through equilibrium equations. For example:
  • The horizontal component is given as \( F_{hx} = T \cdot \cos(70^\circ) \) , responsible for balancing lateral tensions.
  • The vertical component balances gravitational forces, where \( F_{hy} + T \cdot \sin(70^\circ) = W_d + W_c \) .
Mastering forces in equilibrium is essential for solving real-world problems where stability and balance are key.

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