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Chapter 13: Problem 77

The cylindrical plug has a weight of 2 lb and it is free to move within the confines of the smooth pipe. The spring has a stiffness \(k=14 \mathrm{lb} / \mathrm{ft}\) and when no motion occurs the distance \(d=0.5 \mathrm{ft}\). Determine the force of the spring on the plug when the plug is at rest with respect to the pipe. The plug is traveling with a constant speed of \(15 \mathrm{ft} / \mathrm{s}\), which is caused by the rotation of the pipe about the vertical axis.

Short Answer

Expert verified
The force of the spring on the plug when the plug is at rest with respect to the pipe is 2 lb.

Step by step solution

01

Identify all of the Inputs

The weight of the cylindrical plug is given as 2 lb. The stiffness of the spring \(k\) is given as 14 lb/ft and the initial distance \(d\) is 0.5 ft. The speed at which the plug is traveling is not necessary for this problem as it is mentioned that the plug is at rest with respect to the pipe.
02

Calculate the Spring Force

The force of the spring on the plug can be determined using Hooke's Law, which states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed. The spring force \(F\) can be calculated using the formula \(F = -k \cdot d\), where \(k\) is spring stiffness, and \(d\) is extension or compression distance. By substituting the given values, \(F = -14 \cdot 0.5\). Note that the negative sign indicates the direction opposite to the direction of displacement.
03

Determine the Net Force

In this case the plug is at rest with respect to the pipe. This means the net force acting on the plug is zero. The weight of the plug acts downward and the spring force acts upward. Therefore, we can write the net force equation as \(F_s + W = 0\), where \(F_s\) is the spring force and \(W\) is the weight of the plug. The weight of the plug can be determined using the formula \(W = m \cdot g\), where \(m\) is the mass of the plug and \(g\) is the acceleration due to gravity. Given that \(W = 2 lb\) (equivalent to the force due to gravity on the object in pound-mass unit), the upward force applied by the spring must be equal in magnitude and opposite in direction to maintain equilibrium, i.e., \(F_s = -W\). Thus, the spring force \(F_s\) equals 2 lb.

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Key Concepts

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

Hooke's Law
In Engineering Mechanics, Hooke's Law plays an important role in understanding how forces work in components like springs. Hooke's Law states that the force exerted by a spring is proportional to the distance it is stretched or compressed from its natural length. This can be simply expressed as
  • Formula: \( F = -k \cdot d \)
  • Where \( F \) is the spring force
  • \( k \) is the stiffness of the spring
  • \( d \) represents the amount of stretch or compression
  • The negative sign signifies that the spring force acts in the opposite direction to the deformation.
Understanding Hooke's Law helps in predicting how a spring will react when forces are applied. The concept is crucial in designing systems where springs are used to absorb shock or maintain equilibrium. The proportional relationship is linear, meaning doubling the extension will double the force provided the elastic limit is not surpassed.
Spring Force Calculation
Calculating the spring force involves substituting known values into the formula derived from Hooke's Law. In our exercise, the spring has a stiffness \( k = 14 \text{ lb/ft} \), and the initial distance, which is the compression or stretch, \( d = 0.5 \text{ ft} \). By applying these values,
  • Formula: \( F = -k \cdot d \)
  • Substitution: \( F = -14 \cdot 0.5 \)
  • Calculation: \( F = -7 \text{ lb} \)
The spring force acting on the plug is \(-7 \text{ lb}\). The negative value indicates the force direction is opposite to the direction of compression. This calculation is practical for determining the force a spring can exert to maintain mechanical systems in equilibrium or for holding components in place.
Equilibrium of Forces
Equilibrium of forces occurs when the net force acting on a system or object is zero, meaning the object remains at rest or in constant motion. In the context of our exercise, the plug is at rest relative to the pipe, indicating that there is equilibrium.To achieve equilibrium in the exercise:
  • Net Force: \( F_s + W = 0 \)
  • Where \( F_s \) is the spring force (upward direction)
  • \( W \) is the weight of the plug (downward direction)
  • The weight of the plug is given as \( 2 \text{ lb} \).
Since equilibrium is achieved when the two forces are equal in magnitude but opposite in direction, the spring force must balance the weight of the plug. Thus, \( F_s = -W \). This results in the spring exerting a force of \(-2 \text{ lb}\), confirming that forces are balanced.

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Most popular questions from this chapter

The tractor is used to lift the 150 -kg load \(B\) with the 24-m-long rope, boom, and pulley system. If the tractor travels to the right at a constant speed of \(4 \mathrm{m} / \mathrm{s}\), determine the tension in the rope when \(s_{A}=5 \mathrm{m} .\) When \(s_{A}=0, s_{B}=0\)

At the instant shown the 100 -lb block \(A\) is moving down the plane at \(5 \mathrm{ft} / \mathrm{s}\) while being attached to the \(50-1 \mathrm{b}\) block \(B\). If the coefficient of kinetic friction between the block and the incline is \(\mu_{k}=0.2,\) determine the acceleration of \(A\) and the distance \(A\) slides before it stops. Neglect the mass of the pulleys and cables.

The 10-lb block has a speed of 4 ft/s when the force of \(F=\left(8 t^{2}\right)\) lb is applied. Determine the velocity of the block when it moves \(s=30 \mathrm{ft}\). The coefficient of kinetic friction at the surface is \(\mu_{s}=0.2\).

The forked rod is used to move the smooth 2-lb particle around the horizontal path in the shape of a lima莽on, \(r=(2+\cos \theta)\) ft. If \(\theta=\left(0.5 t^{2}\right)\) rad, where \(t\) is in seconds, determine the force which the rod exerts on the particle at the instant \(t=1\) s. The fork and path contact the particle on only one side.

A \(0.2-\mathrm{kg}\) spool slides down along a smooth rod. If the rod has a constant angular rate of rotation \(\dot{\theta}=2 \mathrm{rad} / \mathrm{s}\) in the vertical plane, show that the equations of motion for the spool are \(\ddot{r}-4 r-9.81 \sin \theta=0\) and \(0.8 r+N_{s}-1.962 \cos \theta=0,\) where \(N_{s}\) is the magnitude of the normal force of the rod on the spool. Using the methods of differential equations, it can be shown that the solution of the first of these equations is \(r=C_{1} e^{-2 t}+C_{2} e^{2 t}-(9.81 / 8) \sin 2 t .\) If \(r, \dot{r},\) and \(\theta\) are zero when \(t=0,\) evaluate the constants \(C_{1}\) and \(C_{2}\) determine \(r\) at the instant \(\theta=\pi / 4\) rad.
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