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Chapter 15: Problem 137

If the chain is lowered at a constant speed \(v=\) \(4 \mathrm{ft} / \mathrm{s},\) determine the normal reaction exerted on the floor as a function of time. The chain has a weight of \(5 \mathrm{lb} / \mathrm{ft}\) and a total length of \(20 \mathrm{ft}\)

Short Answer

Expert verified
The normal reaction exerted on the floor as a function of time is \(N = 100lb - 5lb/ft \times (4ft/s \times t)\).

Step by step solution

01

Calculate Total Weight of the Chain

The total weight (\(W\)) of the chain can be calculated by multiplying the weight per unit length (\(w\)) by the total length (\(L\)). So, \(W = w \times L = 5lb/ft \times 20ft = 100lb\).
02

Calculate the Weight Lowered After Time t

The amount of chain (\(LT\)) that has been lowered onto the floor after time \(t\) can be calculated by multiplying the velocity (\(v\)) and the time (\(t\)). So, \(LT = v \times t = 4ft/s \times t\).
03

Calculate the Weight Still in the Air After Time t

The weight of the chain still in the air (\(WA\)) after time \(t\) can be calculated by subtracting the weight of the chain that has already been lowered from the total weight. So, \(WA = W - w \times LT = 100lb - 5lb/ft \times (4ft/s \times t)\).
04

Calculate the Normal Reaction Exerted on the Floor

Finally, the normal reaction (\(N\)) is equal to the weight of the chain still in the air. So, \(N = WA = 100lb - 5lb/ft \times (4ft/s \times t)\).

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Key Concepts

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

Chain Weight Distribution
Understanding the chain weight distribution involves grasping the concept that a chain or any object, for that matter, is composed of smaller, discrete units, each contributing to the total weight of the object. In our scenario, the chain has a uniform weight distribution with a weight of 5 lb/ft.

When the chain is being lowered onto the floor at a steady rate, its weight distribution over the floor changes over time. Initially, the entire weight of the chain acts on the section hanging from the lowering point. As time progresses, more of the chain's weight comes to rest on the floor, reducing the weight experienced by the air-bound portion of the chain. This weight transfer continues until the entire chain lies on the floor, at which point the floor supports all the chain's weight. To calculate the weight on different sections of the chain at a given time, you can use the steps provided in the solution to determine the weight suspended in the air and the corresponding normal reaction on the floor.
Constant Velocity
The term 'constant velocity' signifies that an object is moving at a uniform speed and in a straight line. In the context of our problem, the chain is lowered at a constant velocity of 4 ft/s. This uniform motion is key to simplifying the calculation of the chain's position and the weight distribution over time.

Because the velocity is constant, the length of the chain lowered after a given time, denoted as 'LT' in the solution, can be easily computed by multiplying the velocity by the elapsed time. As more time elapses, a larger portion of the chain gets lowered onto the floor. It is this property of constant velocity that allows for the linear relationship between time and the length of the chain lowered, enabling us to calculate the normal reaction at any time during the chain's descent.
Dynamics
Dynamics in physics refers to the study of forces and their effects on objects' motion. In this exercise, dynamics help us understand how the forces due to gravity (weight of the chain) influence the normal reaction on the floor over time as the chain lowers.

When analyzing the scenario dynamically, the only force acting vertically is the weight of the chain. As sections of the chain come to rest on the floor, the floor must provide an upward force to support the chain's weight 鈥� this force is the normal reaction. By calculating the weight of the portion of the chain still airborne, we've effectively measured the force that section exerts downward due to gravity. According to Newton's third law ('for every action, there's an equal and opposite reaction'), the floor must exert an equal force upward, which is precisely the normal reaction we're looking for. The problem's dynamics emerge from how the vertical forces change as the chain's position changes due to its constant lowering velocity.

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Most popular questions from this chapter

The gauge pressure of water at \(C\) is \(40 \mathrm{lb} / \mathrm{in}^{2}\). If water flows out of the pipe at \(A\) and \(B\) with velocities \(v_{A}=12 \mathrm{ft} / \mathrm{s}\) and \(v_{B}=25 \mathrm{ft} / \mathrm{s},\) determine the horizontal and vertical components of force exerted on the elbow necessary to hold the pipe assembly in equilibrium. Neglect the weight of water within the pipe and the weight of the pipe. The pipe has a diameter of 0.75 in. at \(C,\) and at \(A\) and \(B\) the diameter is 0.5 in. \(\gamma_{w}=62.4 \mathrm{lb} / \mathrm{ft}^{3}\)

The 20-kg crate is lifted by a force of \(F=\left(100+5 t^{2}\right) \mathrm{N}\) where \(t\) is in seconds. Determine the speed of the crate when \(t=3 \mathrm{s},\) starting from rest.

The automobile has a weight of 2700 lb and is traveling forward at \(4 \mathrm{ft} / \mathrm{s}\) when it crashes into the wall. If the impact occurs in 0.06 s, determine the average impulsive force acting on the car. Assume the brakes are not applied. If the coefficient of kinetic friction between the wheels and the pavement is \(\mu_{k}=0.3,\) calculate the impulsive force on the wall if the brakes were applied during the crash.The brakes are applied to all four wheels so that all the wheels slip.

The 20 -lb box slides on the surface for which \(\mu_{k}=0.3 .\) The box has a velocity \(v=15 \mathrm{ft} / \mathrm{s}\) when it is \(2 \mathrm{ft}\) from the plate. If it strikes the smooth plate, which has a weight of 10 lb and is held in position by an unstretched spring of stiffness \(k=400 \mathrm{lb} / \mathrm{ft},\) determine the maximum compression imparted to the spring. Take \(e=0.8\) between the box and the plate. Assume that the plate slides smoothly.

Two smooth billiard balls \(A\) and \(B\) each have a mass of 200 g. If \(A\) strikes \(B\) with a velocity \(\left(v_{A}\right)_{1}=1.5 \mathrm{m} / \mathrm{s}\) as shown, determine their final velocities just after collision. Ball \(B\) is originally at rest and the coefficient of restitution is \(e=0.85 .\) Neglect the size of each ball.
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