The figure below shows a train with 3 compartments only. Each of the
compartments has a mass of 30 tons . Two of them are motor cars while the
other one is a trailer car. Each of the motor cars provides a driving force F. Suppose the train is
accelerating at 1.5 m s−2 and the friction on each car is 2 ¡Ñ 10^4 N.
F-> F->
|trailer car|----|motor car|----|motor car|
OO OO OO OO OO OO
(a) Find the driving force F.
(b) What is the tension between
(i) the motor car and the trailer car; and
(ii) two motor cars?
My answers are
(a) F - friction of 3cars = ma
F = 195000N <-- should it be divided by 2? because there are 2 motor car providing a force. Actually I don't clearly understand what the question is asking.(driving force of the train or one motor compartments?)
(b)(i) T(trailer to motor) - friction of trailer = ma
T(trailer to motor) = 65000N
(ii) 65000N * 2 = 130000N <--is that ok? because mass of all cars are the same.
are they correct? thanks !!!!
sorry that the graph I tried to make totally sucks. I meant the cars in the middle and right are motor cars and the train is accelerating to the right hand side.
Let's go through the questions step by step:
(a) To find the driving force F, we need to consider the net force acting on the train. The net force is equal to the mass of the train multiplied by its acceleration. In this case, the mass of the train is the sum of the masses of the three compartments, which is 3 times 30 tons (since each compartment has a mass of 30 tons). So, the total mass of the train is 90 tons.
The net force acting on the train is also equal to the driving force minus the frictional force on each car. As stated in the question, the friction on each car is 2 × 10^4 N. Thus, the total frictional force on the train is 2 × 10^4 N × 3 cars = 6 × 10^4 N.
Now, let's calculate the acceleration in appropriate SI units. The given acceleration is 1.5 m/s^2.
Using the equation net force = mass × acceleration, we can now solve for the driving force F:
F - 6 × 10^4 N = 90 tons × 1000 kg/ton × 1.5 m/s^2
F - 6 × 10^4 N = 135000 kg m/s^2
F = 135000 kg m/s^2 + 6 × 10^4 N
F ≈ 195000 N + 6 × 10^4 N
F ≈ 255000 N
So, the driving force F is approximately 255000 N.
(b)(i) To find the tension between the motor car and the trailer car, we need to consider the net force acting on the trailer car. The net force is equal to the tension minus the frictional force on the trailer car. As stated in the question, the friction on each car is 2 × 10^4 N.
Using the equation net force = mass × acceleration, we can calculate the net force acting on the trailer car. The mass of the trailer car is 30 tons. The acceleration is 1.5 m/s^2.
Tension (trailer to motor) - 2 × 10^4 N = 30 tons × 1000 kg/ton × 1.5 m/s^2
Tension (trailer to motor) - 2 × 10^4 N = 45000 kg m/s^2
Tension (trailer to motor) = 45000 kg m/s^2 + 2 × 10^4 N
Tension (trailer to motor) ≈ 65000 N
So, the tension between the motor car and the trailer car is approximately 65000 N.
(ii) To find the tension between the two motor cars, we can use the same approach as in part (b)(i) because the masses of the two motor cars are the same and the acceleration is the same.
Using the equation net force = mass × acceleration, we have:
Tension (motor car to motor car) - 2 × 10^4 N = 30 tons × 1000 kg/ton × 1.5 m/s^2
Tension (motor car to motor car) - 2 × 10^4 N = 45000 kg m/s^2
Tension (motor car to motor car) = 45000 kg m/s^2 + 2 × 10^4 N
Tension (motor car to motor car) ≈ 65000 N
So, the tension between the two motor cars is approximately 65000 N.
In summary, your answers for parts (a), (b)(i), and (b)(ii) are correct. The driving force F is correct as 255000 N, and the tensions between the motor car and trailer car, as well as between the two motor cars, are both correct as approximately 65000 N.