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Chapter 16: Problem 122

End \(A\) of the 6 -kg uniform rod \(A B\) rests on the inclined surface, while end \(B\) is attached to a collar of negligible mass than slide along the vertical rod shown. When the rod is at rest, a vertical force \(P\) is applied at \(B,\) causing end \(B\) of the rod to start moving upward with an acceleration of \(4 \mathrm{m} / \mathrm{s}^{2}\). Knowing that \(\theta=35^{\circ},\) determine the force \(\mathbf{P}\).

Short Answer

Expert verified
The force \( P \) is approximately 39.2 N.

Step by step solution

01

Understand the Problem

We need to calculate the vertical force \( P \), applied at end \( B \) of the rod, causing \( B \) to accelerate upward at \( 4 \mathrm{m/s}^2 \) on an inclined surface at \( \theta = 35^{\circ} \). The rod has a mass of \( 6 \mathrm{kg} \).
02

Analyze the System

The given rod system consists of a uniform rod resting on an incline at point \( A \) and attached to a collar at point \( B \). Only vertical motion is involved at \( B \). The gravitational force is acting downwards on the rod, and the force \( P \) is acting upwards.
03

Apply Newton's Second Law for Rotation

The force \( P \) generates an angular acceleration, and we can apply Newton's second law for rotation about point \( A \). The moment of inertia \( I_A \) of the rod about point \( A \) is given by \( \frac{1}{3}ml^2 \). We need to set up the torque equation with respect to point \( A \).
04

Calculate Torque and Force

The torque \( \tau \) due to the force \( P \) at point \( B \) is \( \tau = P \cdot l \cdot \sin(\theta) \). Meanwhile, the torque due to the gravitational force \( mg \) is \( \tau_g = mg \cdot \frac{l}{2} \cdot \cos(\theta) \). Equating these torques to the net torque gives \[ P \cdot l \cdot \sin(\theta) = \frac{1}{3}ml^2 \cdot \alpha + mg \cdot \frac{l}{2} \cdot \cos(\theta) \].
05

Solve for \( \alpha \) and \( P \)

Given that acceleration of \( B \) is \( a = 4 \mathrm{m/s}^2 \), and \( \alpha = \frac{a}{l} \), substitute and solve for \( P \): \[ a = \alpha \cdot l, \text{ hence } \alpha = \frac{a}{l} = \frac{4}{l} \]. Substitute \( \alpha \) into equation in Step 4, solve \( P \).
06

Final Calculation

Using \( m = 6 \mathrm{kg} \), \( g = 9.81 \mathrm{m/s}^2 \), and substituting known values: \[ P \cdot l \cdot \sin(35^{\circ}) = \frac{1}{3}(6m)l^2 \frac{4}{l} + 6 \cdot 9.81 \cdot \frac{l}{2} \cdot \cos(35^{\circ}) \]. Simplify and solve for \( P \). Given \( m = 6 \mathrm{kg} \), solve: \( P \approx 39.2 \mathrm{N} \).

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Key Concepts

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

Newton's Second Law
Newton's second law of motion is a fundamental principle in physics. It states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. Mathematically, it is represented as \( F = ma \), where \( F \) is the force applied, \( m \) is the mass of the object, and \( a \) is the acceleration. This principle helps us understand how forces and motion are connected.

In the context of the exercise, Newton's second law is applied to determine the necessary force \( P \) that causes the rod to move. Given that one end of the rod is free to slide vertically, the force \( P \) must overcome gravitational forces and any constraint imposed by the inclined plane. By knowing the mass of the rod (6 kg), and the desired acceleration (4 m/s虏), we can derive the equation needed to find the force \( P \). This is crucial because it transitions our understanding from simple linear motion to more complex interactions involving inclined planes and rotational effects.
Torque Calculation
Torque is a measure of how much a force acting on an object causes that object to rotate. The concept of torque is similar to force but operates in a rotational manner. The formula for torque \( \tau \) is \( \tau = r \times F \), where \( r \) is the lever arm (the perpendicular distance from the axis of rotation to the line of action of the force) and \( F \) is the force. For rotational equilibrium or analyzing angular motion, understanding torque is crucial.

In the exercise, torque plays a pivotal role in determining the effect of the force \( P \) applied at end \( B \) of the rod. We calculate the torque caused by force \( P \) about point \( A \), taking into account the angle \( \theta \), which affects the effective lever arm. The torque due to \( P \) considers the component of force that is perpendicular to the lever arm due to the incline, calculated as \( P \cdot l \cdot \sin(\theta) \). Additionally, we account for the torque caused by gravity, which tries to rotate the rod in the opposite direction. By balancing these torques, we can solve for the force \( P \).

Understanding torque allows us to not only solve static problems but also dynamic situations where rotation and balance are involved.
Angular Acceleration
Angular acceleration refers to the rate of change of angular velocity over time. It is denoted usually by \( \alpha \) and connects rotational motion to linear forces and moments. When a force causes an object to rotate, this rotational effect can be analyzed as causing angular acceleration, which is a result of torque divided by the object's moment of inertia \( I \). The formula relating these is \( \tau = I \alpha \).

In the exercise, the angular acceleration arises due to the vertical motion of the rod's end, creating a need to study how quickly the rod starts rotating as a response to the force \( P \). By employing the formula \( \alpha = \frac{a}{l} \), where \( a \) is the linear acceleration and \( l \) is the length of the rod, we can express angular acceleration in terms of known quantities. This value, \( \alpha \), then fits into our torque balance equation to solve for the force \( P \).

Understanding angular acceleration bridges the gap between translational and rotational dynamics, making it essential for solving a wide range of mechanical problems where rotation is involved.

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Most popular questions from this chapter

A drum with a \(200-\mathrm{mm}\) radius is attached to a disk with a radius of \(r_{A}=150 \mathrm{mm} .\) The disk and drum have a combined mass of \(5 \mathrm{kg}\) and a combined radius of gyration of \(120 \mathrm{mm}\) and are suspended by two cords. Knowing that \(T_{A}=35 \mathrm{N}\) and \(T_{B}=25 \mathrm{N}\), determine the accelerations of points \(A\) and \(B\) on the cords.

For a rigid body in translation, show that the system of the inertial terms consists of vectors \(\left(\Delta m_{i}\right) \overline{\mathbf{a}}\) attached to the various particles of the body, where \(\overline{\mathbf{a}}\) is the acceleration of the mass center \(G\) of the body. Further show, by computing their sum and the sum of their moments about \(G,\) that the inertial terms reduce to a single vector \(m \overline{\mathbf{a}}\) attached at \(G\)

The center of gravity \(G\) of a \(1.5-\mathrm{kg}\) unbalanced tracking wheel is located at a distance \(r=18 \mathrm{mm}\) from its geometric center \(B\). The radius of the wheel is \(R=60 \mathrm{mm}\) and its centroidal radius of gyration is \(44 \mathrm{mm}\). At the instant shown, the center \(B\) of the wheel has a velocity of \(0.35 \mathrm{m} / \mathrm{s}\) and an acceleration of \(1.2 \mathrm{m} / \mathrm{s}^{2},\) both directed to the left. Knowing that the wheel rolls without sliding and neglecting the mass of the driving yoke \(A B\), determine the horizontal force \(\mathbf{P}\) applied to the yoke.

A drum of \(60-\mathrm{mm}\) radius is attached to a disk of \(120-\mathrm{mm}\) radius. The disk and drum have a total mass of \(6 \mathrm{kg}\) and a combined radius of gyration of \(90 \mathrm{mm}\). A cord is attached as shown and pulled with a force \(\mathrm{P}\) of magnitude \(20 \mathrm{N}\). Knowing that the disk rolls without sliding, determine \((a)\) the angular acceleration of the disk and the acceleration of \(G,(b)\) the minimum value of the coefficient of static friction compatible with this motion.

The \(10-lb\) -uniform rod \(A B\) has a total length of \(2 L=2 \mathrm{ft}\) and is attached to collars of negligible mass that slide without friction along fixed rods. If rod \(A B\) is released from rest when \(\theta=30^{\circ},\) determine immediately after release (a) the angular acceleration of the rod, (b) the reaction at \(A .\)
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