A freight train consists of two 8.00 ×10^4 -kg engines and 45 cars with average masses of 5.50 ×

A freight train consists of two 8.00 ×10^4 -kg engines and...

A freight train consists of two $8.00 \times 10^{4}$ -kg engines and 45 cars with average masses of $5.50 \times 10^{4} \mathrm{kg}$. (a) What force must each engine exert backward on the track to accelerate the train at a rate of $5.00 \times 10^{-2} \mathrm{m} / \mathrm{s}^{2}$ if the force of friction is $7.50 \times 10^{5} \mathrm{N},$ assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems. (b) What is the force in the coupling between the 37 th and 38 th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?

Key Concepts

Internal (Coupling) Forces

These are the forces transmitted between connected cars (or between an engine and a car) within a train. Analysis of internal forces involves considering the train as a series of connected bodies and often requires using free-body diagrams for sections of the train to determine the tension in the couplings, which is essential for understanding how force is propagated through the entire system.

Force Distribution Among Engines

When multiple engines are used to pull a massive system like a train, the total required force (including both accelerating the mass and overcoming friction) must be distributed among them. Assuming identical engines, each engine contributes an equal share of the total force needed to accelerate the train.

Friction Force

Friction, particularly rolling friction in this context, opposes the motion of the train and must be overcome by the engines. Even though rolling friction for trains is relatively small compared to the mass, it still requires careful consideration when computing the net force needed to achieve a specific acceleration.

Total Mass Calculation

For systems composed of multiple objects, such as the two engines and numerous freight cars in a train, calculating the total mass is crucial. The total mass is the sum of the masses of all individual parts. This calculation is necessary to determine the net force required for a given acceleration using Newton's Second Law.

Newton's Second Law

This principle states that the net force acting on an object (or a system of objects) is equal to the product of its mass and acceleration (F = ma). It is fundamental for determining how much force is needed to accelerate a system, and it forms the basis for analyzing both external forces, like the propulsion and friction forces, and internal forces, such as the tension in the couplings.

Transcript

00:02 In letter a, we need to calculate the force exerted by each of the engines to accelerate the entire freight train.

00:17 Our system of interest here is the entire freight train.

00:22 So the freight trains consist of 45 cars and two engines.

00:29 So the freight train experiences, suppose this is our freight train.

00:38 So the freight train experiences an applied force from the engine and of course an opposing friction force so this is along the horizontal and then of course along the vertical since it's on the ground so it experiences a normal force and of course it's acted upon by the gravitational force so this is basically our free body diagram already okay so since there's no acceleration along y so we can just simply write our equation for the forces along y as since this will be zero so this will be be summation f sub y equals zero now along x so the equation so mutton second law equation would be for the forces along x so f sub x equals mass times three acceleration along x so since in this problem we need the force applied by the engine sorry so let's write first the left hand side of the equation which is the summation of the forces so along x.

02:09 So we have here f applied.

02:11 So since it's the one directed to the right minus f friction equals mass times the mass times the acceleration along x.

02:23 So since we're looking for f applied, we just have to transpose the friction force to the other side.

02:31 And we'll have the equation f applied equals mass times.

02:37 The acceleration along x plus the friction force okay so now substitute our given so for the mass we actually have to include the in the computation or consider in the computation the number of cars and the number of engines so since we are given the mass of each car and the mass of each engine so since we have two engines so we have here two times the mass of each engine which is 8 .00 times 10 raise 2 4 kilograms and then we now get the total mass of the 45 cars which make up the fly train so times this will be plus 45 times the mass of each car which is 5 .50 times 10 raised to 4 kilograms then multiply it by the given acceleration which is 5 .00 times 10 raised to negative 2 meter per second squared sorry that didn't fit the line then we have plus 7 .50 times 10 raised to 5 nutons.

04:12 So, for the friction force.

04:15 So input this in your calculator, so you will get, for the applied force, 8.

04:26 So 82 times 10 raised to 5 neutons.

04:34 So this is the force.

04:41 Sorry so since we have two engines so we have and we're asked to determine the force exerted by each engine so we have to divide this by two so the force exerted by each engine would be the half of this so 4 .41 times 10 place to 5 new so that's for letter a...

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