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Chapter 6: Problem 90

End \(A\) of the uniform 5 -kg bar is pinned freely to the collar, which has an acceleration \(a=4 \mathrm{m} / \mathrm{s}^{2}\) along the fixed horizontal shaft. If the bar has a clockwise angular velocity \(\omega=2\) rad/s as it swings past the vertical, determine the components of the force on the bar at \(A\) for this instant.

Short Answer

Expert verified
Forces at A: 20 N (right), 49.05 N (down).

Step by step solution

01

Understand the problem

We have a 5 kg bar pinned at point A, which is moving with an acceleration of 4 m/s虏 horizontally. The bar has an angular velocity of 2 rad/s, moving clockwise. We need to find the force components at point A.
02

Define forces acting on the bar

The forces acting on the bar are gravity, the normal force at the collar (since it's pinned), and the inertial forces due to angular motion and linear acceleration.
03

Analyze linear acceleration

The linear acceleration contributes to a horizontal force on the center of mass of the bar. By Newton's second law, the horizontal force is \( F_{a} = m imes a \) where \( m = 5 \, \text{kg} \) and \( a = 4 \, \text{m/s}^2 \).
04

Calculate horizontal force due to linear acceleration

Substitute: \( F_{a} = 5 \, ext{kg} \times 4 \, ext{m/s}^2 = 20 \, \text{N} \). This force acts horizontally to the right.
05

Analyze angular motion

The angular motion causes a centripetal force at the center of mass. The force is given by \( F_{c} = m imes rac{L}{2} imes heta'' \) where \( L \) is the length of the bar and \( \theta'' \) is the angular acceleration. However, since \( \omega = 2 \, ext{rad/s} \) is constant, \( \theta'' = 0 \).
06

Consider gravitational force

The gravitational force is acting downwards at the center of mass of the bar: \( F_{g} = m imes g = 5 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 49.05 \, \text{N} \).
07

Determine net force at A

Combine the forces vectorially. The force due to linear acceleration is horizontal (20 N to the right), and the gravitational force is vertical (49.05 N down). No vertical component from any other forces means vertical is unchanged.
08

Resolve forces at A

The net force at A combines the horizontal and vertical forces. Therefore, \( F_{Ax} = 20 \, \text{N} \) right, and \( F_{Ay} = 49.05 \, \text{N} \) down.

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Key Concepts

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

Angular Velocity
Angular velocity is a measure of how fast an object rotates or revolves around a point. It is typically denoted by the symbol \( \omega \) and is expressed in radians per second (rad/s).

In our original exercise, the bar is described as having an angular velocity of 2 rad/s clockwise as it swings past the vertical. This means that every second, the bar rotates through an angle of 2 radians. Angular velocity is an essential concept in dynamics because it helps us understand rotational motion.
  • Direction: Angular velocity can be clockwise or counter-clockwise. In problems, the direction is crucial for sign conventions.
  • Relation to Angular Displacement: Angular velocity is the rate of change of angular displacement with respect to time.
  • Equation: For uniform rotation, \( \theta = \omega t \) describes the angular position, where \( \theta \) is the angular displacement.
Understanding angular velocity helps in determining other rotational parameters like angular acceleration and centripetal forces in more complex scenarios.
Newton's Second Law
Newton's second law is a fundamental principle in dynamics and physics. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. In formula terms, it is expressed as \( F = ma \).

This principle plays a crucial role in breaking down the forces in any dynamic system, as demonstrated in the step-by-step solution of the exercise. Here, we calculate the horizontal force at point A using the bar's mass and linear acceleration.
  • Application: It allows us to understand how acceleration is affected by forces, providing insights into the motion dynamics of the bar.
  • Components: Forces can be split into components (e.g., horizontal and vertical), making it simpler to solve multi-directional force problems.
  • Relation to Motion: It ties the motion (acceleration) directly to force, giving a comprehensive view of the object's tendencies.
By applying Newton's second law, we generated the force values that act on the bar, specifically at point A, clarifying its movement.
Linear Acceleration
Linear acceleration refers to the change in velocity of an object along a straight path, and it is crucial in analyzing motion in dynamics. It is denoted by the symbol \( a \) and is measured in meters per second squared (m/s虏).

In our exercise, the bar experiences a linear acceleration of 4 m/s虏, which is responsible for exerting a horizontal force at its center of mass. This force can be calculated by multiplying the bar's mass by its acceleration, following Newton's second law.
  • Key Role: Linear acceleration directly affects the motion and interaction of forces on the bar.
  • Formula: \( F_{a} = ma \) is used to determine the force due to linear acceleration.
  • Vectors: It can act in any direction, but in this exercise, it's horizontal.
By understanding linear acceleration, you can analyze how an object like a bar responds to forces, predicting its motion within a system.
Gravitational Force
Gravitational force is the force that attracts any two bodies with mass toward each other. On Earth, this force is what gives objects weight and causes them to fall toward the ground.

In the context of the exercise, gravitational force acts vertically downward on the bar with a magnitude that depends on the bar's mass and the gravitational acceleration, which is approximately 9.81 m/s虏 on Earth.
  • Formula: Gravitational force \( F_{g} = mg \), where \( g \) is the acceleration due to gravity.
  • Relevance: This force must be considered when calculating the total forces at a certain point, like point A in the bar.
  • Characteristics: It acts uniformly on the center of gravity of objects, providing a vertical force component.
In dynamics, paying attention to gravitational force is vital to solving problems involving mass and motion, as seen with the net forces calculated on the bar in the exercise.

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Most popular questions from this chapter

The 20 -kg wheel has an eccentric mass which places the center of mass \(G\) a distance \(\bar{r}=70 \mathrm{mm}\) away from the geometric center \(0 .\) A constant couple \(M=6 \mathrm{N} \cdot \mathrm{m}\) is applied to the initially stationary wheel, which rolls without slipping along the horizontal surface and enters the curve of radius \(R=600 \mathrm{mm} .\) Determine the normal force under the wheel just before it exits the curve at \(C .\) The wheel has a rolling radius \(r=100 \mathrm{mm}\) and a radius of gyration \(k_{O}=65 \mathrm{mm}\)

The motor \(M\) is used to hoist the 12,000 -lb stadium panel (centroidal radius of gyration \(\bar{k}=6.5 \mathrm{ft}\) ) into position by pivoting the panel about its corner \(A\). If the motor is capable of producing 5000 lb-ft of torque, what pulley diameter \(d\) will give the panel an initial counterclockwise angular acceleration of \(1.5 \mathrm{deg} / \mathrm{sec}^{2} ?\) Neglect all friction.

The 6 -lb pendulum with mass center at \(G\) is pivoted at \(A\) to the fixed support \(C A .\) It has a radius of gyration of 17 in. about \(O-O\) and swings through an amplitude \(\theta=60^{\circ} .\) For the instant when the pendulum is in the extreme position, calculate the moments \(M_{x}, M_{y},\) and \(M_{z}\) applied by the base support to the column at \(C\)

A reel of flexible power cable is mounted on the dolly, which is fixed in position. There are 200 ft of cable weighing 0.436 lb per foot of length wound on the reel at a radius of 15 in. The empty spool weighs 62 lb and has a radius of gyration about its axis of 12 in. A tension \(T\) of 20 lb is required to overcome frictional resistance to turning. Calculate the angular acceleration \(\alpha\) of the reel if a tension of 40 lb is applied to the free end of the cable.

The gear train shown starts from rest and reaches an output speed of \(\omega_{C}=240\) rev/min in 2.25 s. Rotation of the train is resisted by a constant \(150 \mathrm{N} \cdot \mathrm{m}\) moment at the output gear \(C .\) Determine the required input power to the \(86 \%\) efficient motor at \(A\) just before the final speed is reached. The gears have masses \(m_{A}=6 \mathrm{kg}, m_{B}=10 \mathrm{kg},\) and \(m_{C}=24 \mathrm{kg}\) pitch diameters \(d_{A}=120 \mathrm{mm}, d_{B}=160 \mathrm{mm},\) and \(d_{C}=240 \mathrm{mm},\) and centroidal radii of gyration \(k_{A}=\) \(48 \mathrm{mm}, k_{B}=64 \mathrm{mm},\) and \(k_{C}=96 \mathrm{mm}\)
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