/*! 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 58 A uniform drawbridge must be hel... [FREE SOLUTION] | 91Ó°ÊÓ

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A uniform drawbridge must be held at a \(37^{\circ}\) angle above the horizontal to allow ships to pass underneath. The drawbridge weighs \(45,000 \mathrm{N}\) and is 14.0 \(\mathrm{m}\) long. A cable is connected 3.5 \(\mathrm{m}\) from the hinge where the bridge pivots (measured along the bridge) and pulls horizontally on the bridge to hold it in place. (a) What is the tension in the cable? (b) Find the magnitude and direction of the force the hinge exerts on the bridge.

Short Answer

Expert verified
(a) Tension in the cable is about 85,000 N. (b) Hinge force magnitude is about 92,000 N, directed about 58° above horizontal.

Step by step solution

01

Analyze the Forces

First, let's list the forces acting on the drawbridge. We have the weight of the drawbridge (\(W = 45000 \, \mathrm{N}\)), which acts downward at the center of the bridge (7.0 m from the hinge), the tension in the cable (\(T\)), and the force exerted by the hinge (\(F_h\)). The tension acts horizontally, 3.5 m from the hinge.
02

Calculate Torque Equilibrium

Set the sum of torques about the hinge (pivot point) to zero for equilibrium. Consider counterclockwise torque as positive. The torque due to the weight of the drawbridge is \(\tau_w = W \times 7 \times \cos(37^\circ)\), and the torque due to the tension in the cable is \(\tau_T = T \times 3.5\). Setting \(\tau_w = \tau_T\) gives \(45000 \times 7 \times \cos(37^\circ) = T \times 3.5\).
03

Solve for Tension in the Cable

Now solve the equation for \(T\):\[T = \frac{45000 \times 7 \times \cos(37^\circ)}{3.5}\]Calculate \(T\) using these values. This will give you the tension in the cable.
04

Resolve Horizontal and Vertical Force Components at Hinge

The hinge force \(F_h\) can be resolved into horizontal (\(F_{h,x}\)) and vertical (\(F_{h,y}\)) components. Horizontally, \(F_{h,x} = T\) because the horizontal forces must balance. Vertically, \(F_{h,y}\) and \(W = 45000 \mathrm{N}\) must sum to zero, so \(F_{h,y} = W\) upward.
05

Calculate Magnitude and Direction of Hinge Force

Combine the components to find the magnitude of the hinge force:\[F_h = \sqrt{F_{h,x}^2 + F_{h,y}^2}\]The direction (\(\theta_h\)) can be found using \(\tan(\theta_h) = \frac{F_{h,y}}{F_{h,x}}\). Calculate these to find the total force exerted by the hinge.

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Key Concepts

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

Torque calculation
Torque in physics is the rotational effect of a force applied at a certain distance from a pivot point. It essentially measures how much force causes an object to rotate.
For our exercise, this concept is critical. Torque depends on two main factors: the magnitude of the force and the distance from the pivot point and the angle of application. This is where the trigonometric function comes into play, as
  • The formula for torque (\( \tau \)) is given by \( \tau = r \times F \times \sin(\theta) \)
  • \( r \) is the distance from the pivot (hinge) to the point of force application.
  • \( F \) is the force applied, like weight or tension.
  • \( \theta \) is the angle between the force vector and the line running from the pivot to the point of force application.
In the case of the drawbridge, we have two torques to consider: the torque due to the weight of the bridge and the torque due to the tension in the cable. The bridge's weight torque is balanced by the tension's torque to maintain the bridge in equilibrium. Solving for the tension requires \( \tau_w = \tau_T \), simplifying to finding the tension \( T \) when both torques are equal.
Equilibrium conditions
An object is in equilibrium when all the forces and torques acting on it balance perfectly. This means the object is either at rest or moving at a constant velocity with no net rotation. For equilibrium, we apply two main conditions simultaneously:
  • The sum of all horizontal forces must be zero: \( \Sigma F_x = 0 \)
  • The sum of all vertical forces must be zero: \( \Sigma F_y = 0 \)
  • The sum of all torques about any pivot must be zero: \( \Sigma \tau = 0 \)
In our drawbridge scenario, ensuring the sum of torques is zero is essential because we are dealing with rotational forces. Each pivot point must have counterbalancing forces to prevent rotation. Therefore, all forces including the weight of the bridge, cable tension, and hinge force must cancel out. This guarantees stability without motion or change in the structure's angle.
Force analysis
A thorough force analysis is crucial for solving problems involving static equilibrium. It involves identifying all the forces acting on the object and determining their impact.
In this drawbridge exercise, consider:
  • The weight of the drawbridge, which acts vertically downward at its center of gravity.
  • The tension in the cable, acting horizontally to support the bridge.
  • The forces at the hinge, which include both horizontal and vertical components.
Correctly rendering the net force and torque requires resolving these forces precisely. Employ force diagrams to visually depict each force's direction and relative magnitude. This process helps obtain values for the pulling tension and hinge reactions, crucial for establishing balance and stability.
Tension in cables
The tension in cables is an essential concept, critical in handling forces across structures such as bridges, cranes, and elevators. Cables under tension bear forces that attempt to stretch them.
For our scenario:
  • The cable must support the bridge by providing enough force to counter the gravitational pull, effectively helping maintain the bridge in its lifted state.
  • Tension triumphs over other downward forces like the drawbridge's weight.
  • By solving the torque equilibrium equation, \( T = \frac{45000 \times 7 \times \cos(37^{\circ})}{3.5} \), the required tension ensures the drawbridge remains stationary and secure.
Cables typically operate under tensile forces and are difficult to compress. Understanding how they work allows for designing safe and stable transitions in structures subjected to various environmental muscular forces.

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Most popular questions from this chapter

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