A toy freight train consists of an engine and three identical cars. The train is moving to the right at constant speed along a straight, level track. Three spring scales are used to connect the cars as follows: spring scale \(A\) is located between the engine and the first car; scale \(\mathrm{B}\) is between the first and second cars; scale \(C\) is between the second and third cars. (a) If air resistance and friction are negligible, what are the relative readings on the three spring scales \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C} ?\) (b) Repeat part (a), taking air resistance and friction into consideration this time. [Hint: Draw an FBD for the car in the middle. \(]\) (c) If air resistance and friction together cause a force of magnitude \(5.5 \mathrm{N}\) on each car, directed toward the left, find the readings of scales $\mathrm{A}, \mathrm{B},\( and \)\mathrm{C}$

Step by step solution

01

Part (a) - Finding Relative Readings Without Considering Air Resistance and Friction

Since we are not taking air resistance and friction into account, the train is moving at constant speed, and the net force acting on each car is zero. Let's consider the forces acting on the middle car, which has springs B and C on both sides. According to Newton's second law of motion, we can write the following equation: $$F_B - F_C = m \times a$$ But since the train is moving at a constant speed, the acceleration (a) is 0. So, $$F_B - F_C = 0$$ $$F_B = F_C$$ This means that both spring scales B and C will have the same reading. Now let's consider the first car, which has springs A and B on both sides. Using Newton's second law of motion, we can write the equation: $$F_A - F_B = m \times a$$ Since the acceleration (a) is 0, we get: $$F_A - F_B = 0$$ $$F_A = F_B$$ As a result, the readings on spring scales A, B, and C are equal. So, $$F_A = F_B = F_C$$

02

Part (b) - Finding Relative Readings Taking Air Resistance and Friction into Consideration

Now let's consider the effect of air resistance and friction, which act in the opposite direction of the motion of the train. Let \(F_f\) be the force due to air resistance and friction acting on each car. Since all three cars are identical, the air resistance and friction forces on each of them are equal. For the middle car, using Newton's second law, we can write: $$F_B - F_C - F_f = m \times a$$ Since the acceleration (a) is 0, $$F_B - F_C - F_f = 0$$ $$F_B - F_C = F_f$$ For the first car, using Newton's second law, we can write: $$F_A - F_B - F_f = m \times a$$ Since the acceleration (a) is 0, $$F_A - F_B - F_f = 0$$ $$F_A - F_B = F_f$$ As a result, the difference between readings on spring scales A and B, as well as B and C, is equal to the force due to air resistance and friction.

03

Part (c) - Finding the Absolute Readings of the Spring Scales

In this part, we are given the value of the force due to air resistance and friction acting on each car, which is \(5.5\,\mathrm{N}\). From the result we obtained in part (b), the difference in the readings of springs A and B, as well as B and C, is equal to the force due to air resistance and friction. So, if we let \(F_A', F_B'\) and \(F_C'\) be the absolute readings of spring scales A, B, and C, respectively, we can write the following equations: $$F_A' - F_B' = 5.5\,\mathrm{N}$$ $$F_B' - F_C' = 5.5\,\mathrm{N}$$ Since we have two equations with three unknowns, we cannot determine the exact values for the readings of spring scales A, B, and C. However, we can say that the readings on the spring scales are related as: $$F_A' = F_B' + 5.5\,\mathrm{N}$$ $$F_B' = F_C' + 5.5\,\mathrm{N}$$

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