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If it takes \(35 \mathrm{~s}\) for the 50 -Mg tugboat to increase its speed uniformly to \(25 \mathrm{~km} / \mathrm{h}\), starting from rest, determine the force of the rope on the tugboat. The propeller provides the propulsion force \(\mathbf{F}\) which gives the tugboat forward motion, whereas the barge moves freely. Also, determine \(F\) acting on the tugboat. The barge has a mass of \(75 \mathrm{Mg}\).

Short Answer

Expert verified
The force of the rope on the tugboat is \(14.85 \cdot 10^6 \mathrm{N}\) and the propulsion force acting on the tugboat is \(9.9 \cdot 10^6 \mathrm{N}\).

Step by step solution

01

Convert velocity to standard SI units

The tugboat's velocity is given in kilometers per hour (\(25 \mathrm{~km/h}\)). Convert it to meters per second (\(m/s\)) by using the conversion factor \(1 \mathrm{~km/h} = 0.27778 \mathrm{m/s}\). Therefore, the tugboat's final velocity \(v = 25 \cdot 0.27778 = 6.944 \mathrm{~m/s}\).
02

Calculate the acceleration of the tugboat

According to the first kinematic equation, the acceleration \(a\) of an object that starts from rest and achieves velocity \(v\) in time \(t\) is \(a = v / t\). Therefore, the acceleration of the tugboat is \(a = 6.944 \mathrm{~m/s} / 35 \mathrm{s} = 0.198 \mathrm{m/s^2}\).
03

Calculate the propulsion force

According to Newton's second law of motion, the force \(F\) needed to accelerate an object of mass \(m\) with acceleration \(a\) is \(F = ma\). Therefore, the propulsion force \(F\) of the tugboat is \(F = 50 \cdot 10^6 \mathrm{~kg} \cdot 0.198 \mathrm{~m/s^2} = 9.9 \cdot 10^6 \mathrm{N}\).
04

Calculate the force of the rope on the tugboat

The tugboat pulls the barge, which also experiences an acceleration of \(0.198 \mathrm{m/s^2}\). Therefore, the force \(F_r\) of the rope on the tugboat is the force required to accelerate the barge, which is \(F_r = 75 \cdot 10^6 \mathrm{~kg} \cdot 0.198 \mathrm{~m/s^2} = 14.85 \cdot 10^6 \mathrm{N}\).

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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 crucial in understanding how forces act on objects. It tells us that the force acting on an object is the product of its mass and acceleration: \( F = ma \). This means:- If you know the mass and acceleration of an object, you can calculate the force.- Conversely, if you know the force and mass, the acceleration can be deduced.In the context of the tugboat exercise, the mass of the tugboat is large \((50 \mathrm{Mg} \equiv 50,000,000 \mathrm{~kg})\), and it accelerates to a speed of 6.944 meters per second over 35 seconds. By applying Newton's Second Law, you multiply the mass by the calculated acceleration (0.198 \( \mathrm{m/s^2} \)) to determine the propulsion force required for this motion, shown as \( F = 9.9 \times 10^6 \mathrm{~N} \) (Newtons).Understanding this principle explains why engines in large ships must exert enormous forces to achieve seemingly modest changes in speed.
Kinematic Equations
Kinematic equations help us describe the motion of objects without considering the forces that cause the motion. They are derived under the assumption that acceleration is constant.A key kinematic equation used in this exercise is: \[v = u + at\]where:
  • \( v \) is the final velocity,
  • \( u \) is the initial velocity (which is 0 in this exercise),
  • \( a \) is the acceleration,
  • \( t \) is the time taken.
Starting from rest (\( u = 0 \)), the tugboat reaches a velocity of 6.944 \( \mathrm{m/s} \) in 35 seconds, as calculated from the given velocity in the problem. The acceleration \( a \) is then found using this formula: \( a = \frac{v}{t} \). This allows us to break down complex motion into manageable steps, where just by knowing two factors, such as time and final velocity, we can calculate the third, the acceleration.
Propulsion Force Calculation
In this situation, propulsion force is the force exerted by the tugboat's engines to generate movement. It's the primary force required to accelerate the tugboat from rest.Here's a step-by-step breakdown:1. **Determine Mass**: The tugboat's mass is given as 50 Megagrams (Mg), which translates to \( 50,000,000 \mathrm{~kg}\).2. **Find Acceleration**: Using the kinematic equations, the acceleration was found to be 0.198 \( \mathrm{m/s^2} \).3. **Calculate Force**: Applying Newton's Second Law again, \( F = ma \), the propulsion force is:\[F = 50,000,000 \times 0.198 = 9.9 \times 10^6 \mathrm{~N}\]Understanding how to effectively compute propulsion force is crucial in maritime engineering. It ensures that engines and machinery are adequately rated for their required duties. This means matching operational needs to engineering capabilities to ensure smooth, safe, and efficient transport across water.

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Most popular questions from this chapter

The earthmover initially carries \(10 \mathrm{~m}^{3}\) of sand having a density of \(1520 \mathrm{~kg} / \mathrm{m}^{3}\). The sand is unloaded horizontally through a \(2.5 \mathrm{~m}^{3}\) dumping port \(P\) at a rate of \(900 \mathrm{~kg} / \mathrm{s}\) measured relative to the port. Determine the resultant tractive force \(\mathbf{F}\) at its front wheels if the acceleration of the earthmover is \(0.1 \mathrm{~m} / \mathrm{s}^{2}\) when half the sand is dumped. When empty, the earthmover has a mass of \(30 \mathrm{Mg}\). Neglect any resistance to forward motion and the mass of the wheels. The rear wheels are free to roll.

A commercial jet aircraft has a mass of \(150 \mathrm{Mg}\) and is cruising at a constant speed of \(850 \mathrm{~km} / \mathrm{h}\) in level flight \(\left(\theta=0^{\circ}\right) .\) If each of the two engines draws in air at a rate of \(1000 \mathrm{~kg} / \mathrm{s}\) and ejects it with a velocity of \(900 \mathrm{~m} / \mathrm{s}\), relative to the aircraft, determine the maximum angle of inclination \(\theta\) at which the aircraft can fly with a constant speed of \(750 \mathrm{~km} / \mathrm{h}\). Assume that air resistance (drag) is proportional to the square of the speed, that is, \(F_{D}=c v^{2}\), where \(c\) is a constant to be determined. The engines are operating with the same power in both cases. Neglect the amount of fuel consumed.

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) \(k=15 \mathrm{kN} / \mathrm{m}\).

A power lawn mower hovers very close over the ground. This is done by drawing air in at a speed of \(6 \mathrm{~m} / \mathrm{s}\) through an intake unit \(A,\) which has a cross-sectional area of \(A_{A}=0.25 \mathrm{~m}^{2},\) and then discharging it at the ground, \(B\) where the cross-sectional area is \(A_{B}=0.35 \mathrm{~m}^{2}\). If air at \(A\) is subjected only to atmospheric pressure, determine the air pressure which the lawn mower exerts on the ground when the weight of the mower is freely supported and no load is placed on the handle. The mower has a mass of \(15 \mathrm{~kg}\) with center of mass at \(G\). Assume that air has a constant density of \(\rho_{a}=1.22 \mathrm{~kg} / \mathrm{m}^{3}\).

A ball of mass \(m\) is dropped vertically from a height \(h_{0}\) above the ground. If it rebounds to a height of \(h_{1}\), determine the coefficient of restitution between the ball and the ground.

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