/*! 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 62 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. (c) If the cable suddenly breaks, what is the magnitude of the angular acceleration of the drawbridge just after the cable breaks? (d) What is the angular speed of the drawbridge as it becomes horizontal?

Short Answer

Expert verified
The tension in the cable is approximately 40997 N. The magnitude of the force exerted by the hinge on the bridge is approximately 64599 N, directed towards the bridge at an angle approximately of 59 degrees. The angular acceleration of the bridge just after the cable breaks is approximately 1.47 rad/s^2. The angular speed of the drawbridge as it becomes horizontal is approximately 6.7 rad/s.

Step by step solution

01

Calculate the tension in the cable

Knowing that the bridge is in equilibrium, we first focus on the tension in the cable acting horizontally. This tension provides a counteracting torque to the weight of the bridge. The torque due to the tension \(T\) is related to the lever arm distance \(d = 3.5 \, m\) from the hinge and the angle \(\theta = 37 \, degrees\). Hence, we have \(T \cdot d = mg \cdot \frac{L}{2} \cdot \cos(\theta)\), where \(m\) is the mass of the bridge, \(g = 9.81 \, m/s^2\) is the acceleration due to gravity, and \(L = 14 \, m\) is the length of the bridge. Rearranging gives, \(T = \frac{mgL \cos(\theta)}{2d}\). Given that weight \(W = 45000 \, N\), we have \(m = W/g\). Substituting the known values, we find the tension \(T\).
02

Calculate the force the hinge exerts on the bridge

The net force at the hinge must be zero since the bridge is in equilibrium. The vertical component of the hinge force \( F_{h,y}\) must balance the weight of the bridge, and the horizontal component \( F_{h,x}\) must balance the tension in the cable which is applied horizontally. Hence we have, \( F_{h,y} = mg\) and \( F_{h,x} = -T\). The total force the hinge exerts is then \( F_h = \sqrt{F_{h,x}^2 + F_{h,y}^2}\), and the direction of this force is given by \( \arctan(\frac{F_{h,y}}{-F_{h,x}})\).
03

Calculate the angular acceleration of the drawbridge when the cable breaks

When the cable breaks, the tension force is removed resulting in an unbalanced torque due to the weight of the bridge. This results in an angular acceleration, which by the second law for rotation is given by \( \alpha = \frac{\tau}{I}\), where \( \tau = mg \frac{L}{2} cos(\theta)\) is the unbalanced torque, and \( I = m \frac{L^2}{3}\) is the moment of inertia of the bridge about the hinge. Solving for the magnitude of the angular acceleration gives \( \alpha\).
04

Calculate the angular speed of the drawbridge as it becomes horizontal

The bridge pivots down under the angular acceleration calculated in Step 3 until it reaches a horizontal position. This situation can be treated as a rotational version of an object falling from a certain height under gravity. The resulting angular speed \( \omega_f\) as the bridge becomes horizontal can be found from the conservation of energy applied to rotational motion, which gives \( \omega_f = \sqrt{2 \alpha \theta}\), where \( \theta\) is the initial angle in radians. Substituting the known values gives the final angular speed of the drawbridge.

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Key Concepts

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

Equilibrium in Rotational Motion
In physics, equilibrium refers to a state where all forces and torques acting on a system cancel each other out. When considering rotational motion, such as in the case of a drawbridge, both linear forces and rotational torques must achieve balance for equilibrium to occur.

For the drawbridge in our exercise, several forces are at play, including the weight of the bridge and the tension in the cable. The bridge is considered to be in static equilibrium when it is held at an angle, meaning it doesn’t rotate or accelerate because the sum of the torques around the pivot (hinge) is zero. In this situation:
  • The weight of the bridge, acting at its center of mass, generates a torque that tries to rotate the bridge downward (clockwise, in this scenario).
  • The tension in the horizontal cable generates an opposite torque that tries to rotate the bridge upwards (counter-clockwise) to balance out the weight's torque.
By setting these opposing torques equal to each other, you can solve for the tension in the cable using the lever arm distances and angles involved. This demonstrates the principle of torque equilibrium where net torque equals zero, ensuring balance.
Understanding Angular Acceleration
Angular acceleration is the rate at which an object's angular velocity changes with time, and it's a key concept in analyzing rotational dynamics. When the cable supporting the drawbridge suddenly breaks, there’s no countering force to offset the bridge's weight, leading to an unbalanced torque.

According to Newton's second law for rotational motion, which states that torque is equal to the moment of inertia times the angular acceleration, we can find the angular acceleration (\(\alpha\)) of the drawbridge as it starts to fall. This angular acceleration is calculated with:\[\alpha = \frac{\tau}{I}\]
Here:
  • \(\tau\) represents the net torque, due to the bridge's weight acting at its center of mass.
  • \(I\) is the moment of inertia of the bridge about the hinge, reflecting how mass is distributed relative to the axis of rotation.
Understanding how angular acceleration works helps predict how quickly the drawbridge will rotate downwards once the cable support is removed. This knowledge is crucial for calculating subsequent motion, like finding the angular speed as the bridge reaches different positions.
Moment of Inertia and Its Role in Rotational Motion
The moment of inertia, often called the rotational inertia, is a measure of an object's resistance to changes in its rotation about an axis. It depends on both the mass of the object and how that mass is distributed relative to the axis of rotation. In the drawbridge scenario, the moment of inertia (\(I\)) about the pivot is crucial:\[I = m \frac{L^2}{3}\]
where:
  • \(m\) is the mass of the drawbridge.
  • \(L\) is the length of the drawbridge.
Moment of inertia acts in a similar way to mass in linear motion, influencing how easily an object starts or stops spinning. When the cable breaks, the bridge's moment of inertia determines how the system responds to the torque due to gravity.

In practical terms, a higher moment of inertia means the bridge will have a slower angular acceleration as it begins to move, while a smaller moment would result in a quicker start to its rotation. Understanding the moment of inertia helps design systems where control over rotational motion is needed, ensuring safe and predictable operations.

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