Tension in toy locomotive

  • #1
sheepcountme
80
1

Homework Statement



A child pulls on a toy locomotive of mass 0.979 kg with a force of 3.25 N to the right. The locomotive is connected to two train cars by cables. Friction in the axles results in an effective coefficient of kinetic friction between the floor and the train which is 0.110. One car has a mass of 0.952 kg and the other has a mass of 0.419 kg. (a) What is the acceleration of the train? (b) What is the tension in the cable between the locomotive and the car connected to the locomotive?

Homework Equations



T-mg=ma

The Attempt at a Solution



I found the acceleration just fine but I don't get how to find the tension between the two cars.
 

Answers and Replies

  • #2
PhanthomJay
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Homework Equations



T-mg=ma

The Attempt at a Solution



I found the acceleration just fine but I don't get how to find the tension between the two cars.
Your relevant equation is not correct. How did you find the acceleration using that equation? Assuming you have determined the acceleration correctly, you must now draw a free body diagram of the engine alone. What forces act on the engine? What is the acceleration of the engine? Then solve for the cable tension force betweeen the engine and first car, using Newton's laws. Please also show your work for part a.
 
  • #3
sheepcountme
80
1
Well, I already had a free body diagram, with mg down, Fnormal up, 3.25N right and friction left.

For a, I found mg first for the whole train (23.03N), which would have to be the same as the normal force. So then using the given coefficient for friction and the normal force I found the friction force (23.03 x .110). Since the train is accelerating I subtracted this from applied force and set it equal to ma:
3.25-(23.03 x .110)=ma and plugging everything in solving for a I got .306m/s^2 for the acceleration.

I don't understand how I am just supposed to get the tension between the locomotive and the first car though.
 
  • #4
aim1732
430
2
You missed the tension in the cable in your free body diagram.
 
  • #5
PhanthomJay
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Well, I already had a free body diagram, with mg down, Fnormal up, 3.25N right and friction left.

For a, I found mg first for the whole train (23.03N), which would have to be the same as the normal force. So then using the given coefficient for friction and the normal force I found the friction force (23.03 x .110). Since the train is accelerating I subtracted this from applied force and set it equal to ma:
3.25-(23.03 x .110)=ma and plugging everything in solving for a I got .306m/s^2 for the acceleration.

I don't understand how I am just supposed to get the tension between the locomotive and the first car though.
That looks good for part a. If you call the pulling force P, you used the equation P - friction = ma. Now are you at all familiar with free body diagrams? Draw a crude sketch of the entire train, then isolate the engine alone by drawing a circle around it. Enlarge that circled section in another sketch, and now look at the forces acting on it. You have P acting to the right. There are 2 more forces acting to the left, and 2 more forces acting vertically. What are they?
 
  • #6
sheepcountme
80
1
So it's the normal Force up vertically, mg down vertically, Push to the right, and friction+Tension to the left.

I just tried a few variations of P-(f+T)=ma but they didn't work out...am I on the right track? And should I be using the mass of the entire train or just the first engine?
 
  • #7
aim1732
430
2
You only use the mass of the system on which the law is applied. Here you draw f.b.d of the engine so use it's mass only.
 
  • #8
sheepcountme
80
1
I'm sorry, what are you referring to with f.b.d? Friction and something else?

If I use the mass of the one locomotive in the equation above P-(f+T)=ma for either all masses or the single mass I get the wrong answer so this equation can't be right..can you tell me where it's wrong?
 
  • #9
aim1732
430
2
f.b.d is free body diagram.
What value of friction are you using? Could you show your eqn.?
 
  • #10
sheepcountme
80
1
P-((mu * Fn) +T)=ma
 
  • #11
PhanthomJay
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P-((mu * Fn) +T)=ma
Yes, but now you have to show us what value you are using for F_n.
 
  • #12
sheepcountme
80
1
For Fn I am using mg
 
  • #13
sheepcountme
80
1
Oh! Nevermind it works!

Lesson: never fully trust your calculator
 

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