/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 The motor pulls on the cable at ... [FREE SOLUTION] | 91Ó°ÊÓ

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The motor pulls on the cable at \(A\) with a force \(F=\left(e^{2 t}\right)\) lb, where \(t\) is in seconds. If the 34 -lb crate is originally at rest on the ground at \(t=0,\) determine the crate's velocity when \(t=2\) s. Neglect the mass of the cable and pulleys. Hint: First find the time needed to begin lifting the crate.

Short Answer

Expert verified
To get the short answer, first calculate when the crate will start moving by equating the given force function with the weight of the crate, and solving for time. Then find the acceleration by applying Newton's second law. Finally, use the kinematic equation to find the required velocity at \(t=2\) s.

Step by step solution

01

Find the Net Force

First we need to find out when the crate begins to move. It begins to lift when the pull force of the motor at A is greater than the weight of the crate. Given force \(F(t)=e^{2t}\) and the weight of the crate \(W = 34 lb\), we can write the inequality \(e^{2t} > 34\). Solve for \(t\) to find the time it begins to lift.
02

Calculate the Acceleration

Once we have the time when the crate starts to move, we can compute the acceleration of the crate using the Newton's second law of motion. This can be written as \(F = ma\), where \(m\) is the mass of the crate and \(a\) is its acceleration. Divide force by the mass to obtain acceleration.
03

Find the Velocity

Since we now have the acceleration, we can find the velocity using the kinematic equation \(V = u + at\), where \(u\) is initial velocity (which is zero as the crate was at rest), \(a\) is acceleration and \(t\) is time. By plugging the values, we can find the crate's velocity at \(t = 2\) seconds.

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Key Concepts

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

Force Analysis
When analyzing forces in a dynamics problem, the goal is to identify and quantify all the forces acting on an object. In this exercise, we examine the force exerted by the motor cable on a crate. The force provided by the motor is time-dependent and given by the expression \( F(t) = e^{2t} \) lb. This changes as time progresses. Additionally, the weight of the crate, a constant force due to gravity, acting downward, is 34 lb.

Force analysis involves comparing these forces to find when the crate begins to move. The crate will start lifting when the upward force from the motor exceeds the downward gravitational force or weight. Establishing this inequality \( e^{2t} > 34 \) allows solving for the specific time \( t \) when the crate starts to lift. This step is crucial in determining the initiation of motion in dynamics problems.
  • Identify all forces involved.
  • Apply conditions for motion initiation.
  • Formulate conclusions based on comparisons of forces.
Kinematics
Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. In this problem, kinematics comes into play once the crate starts moving.

After determining when the force overcomes the weight, we can apply basic kinematic equations to understand how the crate's velocity changes over time. The formula relating velocity to acceleration is \( V = u + at \). Given the initial velocity \( u \) is zero since the crate is at rest, we need to calculate acceleration to proceed.
  • Apply kinematic equations to determine changes in motion.
  • Understand how acceleration affects velocity over time.

These kinematic principles simplify analyzing the movement once motion begins, focusing on changes in speed or velocity.
Newton's Second Law
Newton's Second Law of Motion states that the acceleration \( a \) of an object is directly proportional to the net force \( F \) acting on it and inversely proportional to its mass \( m \). Expressed mathematically: \( F = ma \).

In our problem, once the crate starts moving, we apply this principle to find its acceleration. We already have the force as \( F(t) = e^{2t} - 34 \) once it is greater than the gravitational pull. The mass of the crate is the weight divided by gravity: \( m = 34/32.2 \) slugs. Solving this equation gives the acceleration of the crate.
  • Relate net force to mass and acceleration.
  • Determine acceleration using force and mass.

This law is critical for bridging the gap between force analysis and kinematics, allowing us to calculate how changing forces affect motion.
Acceleration
Acceleration refers to how quickly the velocity of an object changes with time. It is a vector quantity with both magnitude and direction. After determining when the crate starts to move, we calculated its acceleration by using Newton's Second Law.

The net force acting on the crate is \( F(t) = e^{2t} - 34 \), and dividing this by the crate's mass enables computing acceleration. Knowing acceleration is pivotal for the next steps involving the kinematic equation. Acceleration provides the rate at which the crate picks up speed once it begins moving.
  • Compute acceleration from net force and mass.
  • Understand how it influences subsequent motion changes.
Crate Motion
Analyzing crate motion involves synthesizing the elements of dynamics, using expressed equations for force, acceleration, and kinematics. Once acceleration has been determined, the kinematic equation \( V = u + at \) enables solving for the crate's velocity at any given time.

In this specific case, with \( t = 2 \) seconds, and knowing both the initial velocity \( u = 0 \) and calculated acceleration \( a \), it becomes straightforward to calculate the velocity. Our understanding of motion gives a clear picture of how dynamics apply throughout and solves real-world problems efficiently.
  • Determine velocity changes over time using known initial conditions and acceleration.
  • Apply learned concepts to solve for motion features like velocity.

This completes the exercise, showing how theory converts into practical application to predict motion accurately.

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Most popular questions from this chapter

The \(5-\mathrm{Mg}\) bus \(B\) is traveling to the right at \(20 \mathrm{m} / \mathrm{s}\). Meanwhile a \(2-\mathrm{Mg}\) car \(A\) is traveling at \(15 \mathrm{m} / \mathrm{s}\) to the right. If the vehicles crash and become entangled, determine their common velocity just after the collision. Assume that the vehicles are free to roll during collision.

A \(1-\) lb ball \(A\) is traveling horizontally at \(20 \mathrm{ft} / \mathrm{s}\) when it strikes a 10 -lb block \(B\) that is at rest. If the coefficient of restitution between \(A\) and \(B\) is \(e=0.6,\) and the coefficient of kinetic friction between the plane and the block is \(\mu_{k}=0.4,\) determine the time for the block \(B\) to stop sliding.

The boat has a mass of \(180 \mathrm{kg}\) and is traveling forward on a river with a constant velocity of \(70 \mathrm{km} / \mathrm{h}\) measured relative to the river. The river is flowing in the opposite direction at \(5 \mathrm{km} / \mathrm{h}\). If a tube is placed in the water, as shown, and it collects \(40 \mathrm{kg}\) of water in the boat in \(80 \mathrm{s}\) determine the horizontal thrust \(T\) on the tube that is required to overcome the resistance due to the water collection and yet maintain the constant speed of the boat. \(\rho_{w}=1 \mathrm{Mg} / \mathrm{m}^{3}\)

A tankcar has a mass of \(20 \mathrm{Mg}\) and is freely rolling to the right with a speed of \(0.75 \mathrm{m} / \mathrm{s}\). If it strikes the barrier determine the horizontal impulse needed to stop the car if the spring in the bumper \(B\) has a stiffness (a) \(k \rightarrow \infty\) (bumper is rigid), and (b) \(k=15 \mathrm{kN} / \mathrm{m}\).

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.

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