Tension in the coupling between the locomotive and the first car

In summary: Static friction is always opposite the tendency for sliding.In summary, a 97 250kg train is pulling two freight cars with masses of 51 355 kg and 18 400 kg, respectively. The whole train is accelerating at .01750m/s^2. The tension between the locomotive and the first car can be calculated using Ft=ma and is equal to 8987.1 N. The tension between the first and second freight cars can be calculated using (m1-m2)(.01750)=Ft and is equal to 576.7 N. The force of friction on an inclined surface can be either uphill or downhill, depending on the other forces acting on the object. The formula for calculating friction
  • #1
A 97 250kg train pulls two freight cars. The first car has a mass of 51 355 kg and the second has a mass of 18 400 kg. The whole train accelerates at .01750m/s^2. What is the tension in the coupling between the locomotive and the first car?
A. I used Ft=ma and took (51355)(.1750)= 8987.1 N

What is the tension in the coupling between the first and the second freight cars?
a. I took (m1-m2)(.01750)=Ft so (51 355-18400)(.01750)= 576.7 N

I don't know if that is how I was supposed to do it.

also, I wanted to ask if the force of friction is the same on all inclined surfaces (going up), even if a person is trying to go up the hill instead of going down. If a person was going up a 7 degree slope and a force was needed to have constant velocity. Would I still use Ff=umgcosO to find the force of friction and then set it equal to the needed force?
 
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  • #2
ownedbyphysics said:
A 97 250kg train pulls two freight cars. The first car has a mass of 51 355 kg and the second has a mass of 18 400 kg. The whole train accelerates at .01750m/s^2. What is the tension in the coupling between the locomotive and the first car?
A. I used Ft=ma and took (51355)(.1750)= 8987.1 N

What is the tension in the coupling between the first and the second freight cars?
a. I took (m1-m2)(.01750)=Ft so (51 355-18400)(.01750)= 576.7 N

I don't know if that is how I was supposed to do it.

also, I wanted to ask if the force of friction is the same on all inclined surfaces (going up), even if a person is trying to go up the hill instead of going down. If a person was going up a 7 degree slope and a force was needed to have constant velocity. Would I still use Ff=umgcosO to find the force of friction and then set it equal to the needed force?
Your answers are not correct. There are a couple of ways to approach this problem. The best is to draw a free body diagram for each car and consider all the forces acting. If you do that correctly, in the end you will realize that another point of view will work for this problem. Each coupling is providing the force to accelerate a certain amount of mass. If you recognize how much mass a cououpling is pulling, you can use your ma calculations to find the tension.

I'm not sure I fully inderstand your last question, but kinetic friction is always opposite the direction of motion. Static friction is in whichever direction parallel to the surface it needs to be to make ths sum of the forces zero.
 
  • #3
would the train be pulling itself and the coupling? and the coupling pulling itself and the other coupling?
 
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  • #4
ownedbyphysics said:
would the train be pulling itself and the coupling? and the coupling pulling itself and the other coupling?
The train is providing the force that pulls everything. The first coupling pulls both cars. The second coupling pulls only one car.

This is exactly like the problem of 3 blocks connected by two strings. It really is worth your time to set up the free body diagrams (FBD) and equations for each car or block and then show that by adding equations you arrive at the conclusions I just stated. Some problems are more difficult than this one, and learning how to deal with FBDs will pay off down the road.
 
  • #5
My teacher told us that "the force of friction on an inclined plane is always going uphill and is always UmgcosO," but he didn't say if that was just in the case of an object going downhill. When an object is going uphill, is the force of friction against the object or still uphill. If it is against the object, would the formula be Ff=umgsinO?
 
  • #6
ownedbyphysics said:
My teacher told us that "the force of friction on an inclined plane is always going uphill and is always UmgcosO," but he didn't say if that was just in the case of an object going downhill. When an object is going uphill, is the force of friction against the object or still uphill. If it is against the object, would the formula be Ff=umgsinO?
Friction can be in either direction. on a plane. It depends on what else is going on. If there is only gravity and normal force and friction, then static friction is uphill. If you apply a force up the hill that is strong enough, the friction will be downhill. If an object is sliding, friction is always opposite the motion.
 

1. What is "tension in the coupling"?

The tension in the coupling refers to the force that is exerted on the connection between the locomotive and the first car of a train. This force is necessary to transfer power and allow for the movement of the train.

2. How is the tension in the coupling measured?

The tension in the coupling can be measured using a device called a dynamometer. This device is placed between the locomotive and the first car and measures the force being exerted on the coupling.

3. Why is it important to maintain proper tension in the coupling?

Proper tension in the coupling is crucial for the safe and efficient operation of a train. If the tension is too low, there may not be enough force to move the train, and if it is too high, it can cause damage to the coupling or other components of the train.

4. What factors can affect the tension in the coupling?

The tension in the coupling can be affected by a variety of factors, including the weight and length of the train, the speed and acceleration of the train, and the condition of the tracks. Weather conditions such as wind can also impact the tension.

5. How is tension in the coupling adjusted?

Tension in the coupling can be adjusted by changing the position of the locomotive in relation to the first car, using brakes to slow down or speed up the train, or by adjusting the power output of the locomotive. It may also be necessary to make adjustments based on external factors such as changes in track conditions or weather.

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