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Chapter 10: Problem 60

A a uniform drawbridge must be held at a \(37^{\circ}\) angle above the horizontal to allow ships to pass underneath. The drawbridge weighs \(45,000 \mathrm{N},\) is 14.0 \(\mathrm{m}\) long, and pivots about a hinge at its lower end. A cable is connected 3.5 \(\mathrm{m}\) from the hinge, as 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 initial angular acceleration of the bridge?

Short Answer

Expert verified
(a) Tension in cable = 71732.14 N (b) Hinge force = 84732.57 N at 31掳 (c) Angular acceleration = 1.31 rad/s虏

Step by step solution

01

Identify Forces and Set Up Diagram

Draw a diagram showing the drawbridge, hinge, cable, and forces. Label the angle \(\theta = 37^{\circ}\), the weight \(W = 45,000 \, \mathrm{N}\), the length of the bridge \(L = 14.0 \, \mathrm{m}\), and the distance from the hinge to where the cable is attached \(d = 3.5 \, \mathrm{m}\). The cable applies a horizontal tension \(T\) at this point.
02

Calculate Torque Due to Gravity

The torque due to gravity is calculated using the formula \( \tau_g = W \cdot \frac{L}{2} \cdot \cos(\theta) \), where \(L/2\) is the distance from the hinge to the center of mass of the bridge. Substitute: \( \tau_g = 45000 \, \mathrm{N} \cdot 7.0 \, \mathrm{m} \cdot \cos(37^\circ) \approx 251062.5 \, \mathrm{N}\cdot\mathrm{m} \).
03

Calculate Torque Due to Tension

The tension provides torque given by \( \tau_T = T \cdot d \) since it pulls horizontally, and we want equilibrium: \( \tau_T = \tau_g \). Therefore, \( T \cdot 3.5 \, \mathrm{m} = 251062.5 \, \mathrm{N}\cdot\mathrm{m} \). Solve for \( T \): \( T = \frac{251062.5}{3.5} \, \mathrm{N} \approx 71732.14 \, \mathrm{N} \).
04

Determine Hinge Reaction Forces

The hinge exerts forces vertically and horizontally. The sum of horizontal forces must equal zero: \( T = F_{hx} \), so \( F_{hx} = 71732.14 \, \mathrm{N} \). The sum of vertical forces must also equal zero: \( F_{hy} = W = 45000 \, \mathrm{N} \). Use Pythagorean theorem for the resultant: \( F_h = \sqrt{F_{hx}^2 + F_{hy}^2} \approx 84732.57 \, \mathrm{N} \). The direction \( \theta_h \) is given by \( \tan^{-1}\left(\frac{F_{hy}}{F_{hx}}\right) \approx 31^{\circ} \).
05

Calculate Initial Angular Acceleration

Use the equation \( \tau = I\alpha \), where \( \tau = \tau_g\), and \( \alpha \) is angular acceleration. The moment of inertia \( I \) for a bar pivoting at one end is \( \frac{1}{3}mL^2 \). Calculate \( \alpha = \frac{\tau_g}{I} = \frac{251062.5}{\frac{1}{3} \cdot \left(\frac{45000}{9.8}\right) \cdot (14)^2} \approx 1.31 \, \mathrm{rad/s^2} \).

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Key Concepts

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

Static Equilibrium
Static equilibrium is a fundamental concept in physics that describes an object at rest, having no net force or torque acting on it. For an object to be in static equilibrium, there are two primary conditions:

1. The sum of all external forces acting on the object must be zero.
2. The sum of all external torques acting on the object must also be zero.

In the context of the drawbridge problem, the drawbridge remains stable and stationary when the forces and torques are balanced. The tension in the cable provides a horizontal force, while the bridge's weight acts downwards at its center of mass.

To determine if the drawbridge is in static equilibrium, we ensure that the following conditions hold:
  • Horizontal forces: The tension in the cable equals the horizontal component of the hinge force.
  • Vertical forces: The weight of the drawbridge is balanced by the vertical component of the hinge force.
  • Torques: The torque due to gravity (acting at the center of the bridge) is balanced by the torque due to the tension in the cable.
By ensuring these conditions, the drawbridge remains in static equilibrium, allowing us to calculate the tension and forces exerted by the hinge.
Angular Acceleration
Angular acceleration occurs when an object in rotational motion experiences a change in its rate of rotation. In simple terms, it's the rate at which the angular velocity of an object changes with time.

The formula to calculate angular acceleration is given by:
\[ \alpha = \frac{\tau}{I} \]
where:
  • \( \alpha \) is the angular acceleration.
  • \( \tau \) is the net torque acting on the object.
  • \( I \) is the moment of inertia of the object.


In the situation with the drawbridge, if the cable suddenly breaks, there is no more tension force to balance the torque due to the bridge's weight. Thus, the drawbridge begins to rotate under the influence of gravity alone. Here, the torque is substantial since it is entirely due to the weight of the bridge. As a result, this torque will cause the drawbridge to accelerate rotationally, initiating its angular acceleration.

Understanding angular acceleration is crucial in predicting how quickly the drawbridge will begin to rotate once it's no longer supported by the cable.
Moment of Inertia
Moment of inertia is a term used to describe how difficult it is to change the rotational motion of an object. It depends not only on the mass of an object but also on how that mass is distributed concerning the axis of rotation.

The moment of inertia \( I \) for an object of mass \( m \) and length \( L \), pivoted about its end, is calculated using the formula:
\[ I = \frac{1}{3}mL^2 \]

In the drawbridge problem, the bridge acts like a long bar, and its moment of inertia plays a crucial role in determining how it reacts when external torques are applied or when it rotates freely. When computing angular acceleration, knowing the moment of inertia allows us to relate the applied torque (such as gravity) to the resulting angular acceleration of the bridge.

Key descriptive points about the moment of inertia include:
  • Greater moment of inertia means more resistance to changes in rotational speed.
  • An object's shape and mass distribution significantly affect its moment of inertia.
  • It's a pivotal factor in understanding rotational dynamics for structures like beams and bridges.
By understanding the moment of inertia, we can better predict how systems will behave under rotational influences, like the motion of the drawbridge once the cable breaks.

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Most popular questions from this chapter

A A large turntable rotates about a fixed vertical axis, making one revolution in 6.00 s. The moment of inertia of the turntable about this axis is 1200 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) A child of mass \(40.0 \mathrm{kg},\) initially standing at the center of the turntable, runs out along a radius. What is the angular speed of the turntable when the child is 2.00 \(\mathrm{m}\) from the center, assuming that you can treat the child as a particle?

A large wooden turntable in the shape of a flat disk has a radius of 2.00 \(\mathrm{m}\) and a total mass of 120 \(\mathrm{kg}\) . The turntable is initially rotating at 3.00 \(\mathrm{rad} / \mathrm{s}\) about a vertical axis through its center. Suddenly, a 70.0 kg parachutist makes a soft landing on the turntable at a point on its outer edge. Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.)

You are trying to raise a bicycle wheel of mass \(m\) and radius \(R\) up over a curb of height \(h .\) To do this, you apply a horizontal force \(\vec{F}\) (Fig. \(10.81 ) .\) What is the smallest magnitude of the force \(\vec{F}\) that will succeed in raising the wheel onto the curb when the force is applied (a) at the center of the wheel, and (b) at the top of the wheel? (c) In which case is less force required?

Disks \(A\) and \(B\) are mounted on shaft \(S S\) and may be connected or dis- connected by clutch \(C .\) (See Figure \(10.82 . )\) Disk \(A\) is made of a lighter material than disk \(B,\) so the moment of inertia of disk \(A\) about the shaft is one-third that of disk \(B\) . The moments of inertia of the shaft and clutch are negligible. With the clutch disconnected, \(A\) is brought up to an angular speed \(\omega_{0 . \text { . The }}\) accelerating torque is then removed from \(A,\) and \(A\) , and \(A\) is coupled to disk \(B\) by the clutch. (You can ignore bearing friction.) It is found that 2400 \(\mathrm{J}\) of thermal energy is developed in the clutch when the connection is made. What was the original kinetic energy of disk \(A\) ?

Back pains during pregnancy. Women often suffer from back pains during pregnancy. Let us investigate the cause of these pains, assuming that the woman's mass is 60 kg before pregnancy. Typically, women gain about 10 kg during pregnancy, due to the weight of the fetus, placenta, amniotic fluid, etc. To make the calculations easy, but still realistic, we shall model the unpregnant woman as a uniform cylinder of diameter 30 \(\mathrm{cm} .\) We can model the added mass due to the fetus as a 10 kg sphere 25 \(\mathrm{cm}\) in diameter and centered about 5 \(\mathrm{cm}\) outside the woman's original front surface. (a) By how much does her pregnancy change the horizontal location of the woman's center of mass? (b) How does the change in part (a) affect the way the pregnant woman must stand and walk? In other words, what must she do to her posture to make up for her shifted center of mass? (c) Can you now explain why she might have backaches?
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