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Chapter 4: Problem 82

A coil of heavy flexible cable with a total length of \(100 \mathrm{m}\) and a mass of \(1.2 \mathrm{kg} / \mathrm{m}\) is to be laid along a straight horizontal line. The end is secured to a post at \(A,\) and the cable peels off the coil and emerges through the horizontal opening in the cart as shown. The cart and drum together have a mass of \(40 \mathrm{kg} .\) If the cart is moving to the right with a velocity of \(2 \mathrm{m} / \mathrm{s}\) when \(30 \mathrm{m}\) of cable remain in the drum and the tension in the rope at the post is \(2.4 \mathrm{N},\) determine the force \(P\) required to give the cart and drum an acceleration of \(0.3 \mathrm{m} / \mathrm{s}^{2},\) Neglect all friction.

Short Answer

Expert verified
The force \( P \) required is 39.6 N.

Step by step solution

01

Determine the Cable Mass In Motion

Calculate the mass of the cable that is currently being pulled off the coil. Since 30 meters remain in the drum, 70 meters of cable (100 - 30) have been laid out. The mass of these 70 meters is given by:\[\text{Mass of laid cable} = 70\, \text{m} \times 1.2\, \text{kg/m} = 84\, \text{kg}\]
02

Sum the Mass in Motion

Add the mass of the laid cable to the mass of the cart and drum to get the total mass in motion. The cart and drum mass is 40 kg, so:\[\text{Total mass in motion} = 40\, \text{kg} + 84\, \text{kg} = 124\, \text{kg}\]
03

Calculate the Required Force

Use Newton's second law to determine the net force required to accelerate the system. The formula is:\[ F = ma \]where:- \( m = 124\, \text{kg} \) (total mass in motion)- \( a = 0.3\, \text{m/s}^2 \) (desired acceleration)Substitute the values into the equation:\[ F = 124\, \text{kg} \times 0.3\, \text{m/s}^2 = 37.2\, \text{N} \]
04

Include Tension from the Rope

Add the tension in the rope at the post to the net force from Step 3, as tension contributes to the force needed. The tension is 2.4 N, so the total force required, \( P \), is:\[ P = 37.2\, \text{N} + 2.4\, \text{N} = 39.6\, \text{N} \]

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Key Concepts

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

Dynamics
Dynamics is the branch of physics concerned with the study of forces and the changes they cause in the motions of objects. It lays down the basic principles needed to solve the kind of problem presented in the exercise. The fundamental equation that governs dynamics is Newton's Second Law, which can be expressed as:\( F = ma \).
  • \( F \) represents the net force acting on an object.
  • \( m \) is the mass of the object.
  • \( a \) is the acceleration produced due to the applied force.
Understanding dynamics helps us determine how different forces interact to affect the velocity and acceleration of an object. In the exercise, this principle is applied to calculate the force necessary to accelerate a system of a cart and cable. Dynamics serves as the foundation to comprehend more complex systems, making it essential for solving practical mechanics problems.
Cable Mechanics
Cable mechanics involves understanding how forces interact within flexible, lengthy items such as cables or ropes. In the exercise, a cable is gradually being unwound from a coil, which requires consideration of its mass and the forces acting upon it. Cables are often subject to tension, which affects their stability and motion.
  • The cable is pulled until a certain length is uncoiled, exposed to gravity but laid out horizontally.
  • The tension in the cable helps quantify the force required to hold it steady or move it along a path.
  • Any additional forces, like acceleration of a cart, need to account for the mass of the uncoiled part.
This problem also illustrates how mechanics can be calculated assuming no frictional forces are present, simplifying force calculations and focusing purely on mass and tension factors.
Force Calculation
Force calculation is the process of determining the amount of force needed to achieve desired changes in a system's motion. This involves supplementing the principles of dynamics and cable mechanics as illustrated in the exercise.In this scenario, multiple steps are used to calculate the necessary force to accelerate the system:
  • First, determine the total mass in motion, considering both the moving part of the cable and the cart's mass.
  • Apply Newton's Second Law to calculate the force needed for the desired acceleration of 0.3 m/s虏.
  • Account for the tension in the rope by adding it to the calculated force, ensuring all forces are considered.
Thus, combining Newton's law with knowledge about tension and weight of the cable leads to calculating the total force \( P \) required. This holistic approach is crucial in solving mechanical problems accurately and efficiently.

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Most popular questions from this chapter

When only the air of a sand-blasting gun is turned on, the force of the air on a flat surface normal to the stream and close to the nozzle is \(20 \mathrm{N}\). With the nozzle in the same position, the force increases to \(30 \mathrm{N}\) when sand is admitted to the stream. If sand is being consumed at the rate of \(4.5 \mathrm{kg} / \mathrm{min}\), calculate the velocity \(v\) of the sand particles as they strike the surface.

The 300 -kg and 400 -kg mine cars are rolling in opposite directions along the horizontal track with the respective speeds of \(0.6 \mathrm{m} / \mathrm{s}\) and \(0.3 \mathrm{m} / \mathrm{s}\). Upon impact the cars become coupled together, Just prior to impact, a 100 -kg boulder leaves the delivery chute with a velocity of \(1.2 \mathrm{m} / \mathrm{s}\) in the direction shown and lands in the 300 -kg car. Calculate the velocity \(v\) of the system after the boulder has come to rest relative to the car. Would the final velocity be the same if the cars were coupled before the boulder dropped?

The man of mass \(m_{1}\) and the woman of mass \(m_{2}\) are standing on opposite ends of the platform of mass \(m_{0}\) which moves with negligible friction and is initially at rest with \(s=0 .\) The man and woman begin to approach each other. Derive an expression for the displacement \(s\) of the platform when the two meet in terms of the displacement \(x_{1}\) of the man relative to the platform.

The 50,000 -lb flatear supports a 15,000 -lb vehicle on a \(5^{\circ}\) ramp built on the flatear. If the vehicle is released from rest with the flatear also at rest, determine the velocity \(v\) of the flatcar when the vehicle has rolled \(s=40 \mathrm{ft}\) down the ramp just before hitting the stop at \(B\). Neglect all friction and treat the vehicle and the flatcar as particles.

The space shuttle, together with its central fuel \(\operatorname{tank}\) and two booster rockets, has a total mass of \(2.04\left(10^{6}\right) \mathrm{kg}\) at liftoff. Each of the two booster rockets produces a thrust of \(11.80\left(10^{6}\right) \mathrm{N},\) and each of the three main engines of the shuttle produces a thrust of \(2.00\left(10^{8}\right) \mathrm{N}\). The specific impulse (ratio of exhaust velocity to gravitational acceleration) for each of the three main engines of the shuttle is 455 s. Calculate the initial vertical acceleration a of the assembly with all five engines operating and find the rate at which fuel is being consumed by each of the shuttle's three engines.
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