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Object Hanging in Railroad Car

  • Thread starter litz057
  • Start date
  • #1
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Homework Statement


A 7.1 kg object hangs at one end of a rope that is attached to a support on a railroad car. When the car accelerates to the right, the rope makes an angle of 7.7 degrees with the vertical, as shown in the figure below. What is the tension in the rope in Newtons?

Homework Equations


Ft/Fg=ma/mg=ag

The Attempt at a Solution


Ft/Fg=ma/mg=ag
tan(7.7)=a/9.81
tan(7.7)*9.81=a
1.326364039 m/s^2 = a
Railroad Car.gif
 

Answers and Replies

  • #2
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You've done the hard part. What's your question?
 
  • #3
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You've done the hard part. What's your question?
How do I find the correct answer? I submitted the answer that I posted above, but it was marked incorrect. Is there a step that I am missing?
 
  • #4
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They asked for the tension in the rope, not the acceleration of the mass.

Chet
 
  • #5
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They asked for the tension in the rope, not the acceleration of the mass.

Chet
How do I use the acceleration to find the tension in the rope? I tried using F=ma
F=ma
F=7.1* 1.3264
F= 9.4172 N
This is not the correct answer. How do I find the tension from the acceleration?
 
  • #6
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You have two accelerations, gravitational, and of the railroad car. Can you do the vector sum of those two?
 
  • #7
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If the tension in the rope is F, how is Ft related to F? How is Fg related to F?

Chet
 
  • #8
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You have two accelerations, gravitational, and of the railroad car. Can you do the vector sum of those two?
I am not completely sure how to do the vector sum. Do I just add 9.81 and 1.3264?
 
  • #9
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If the tension in the rope is F, how is Ft related to F? How is Fg related to F?

Chet
Ft is the force relative to the train and Fg is the force relative to the ground.
 
  • #10
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Ft is the force relative to the train and Fg is the force relative to the ground.
Algebraically, if the tension in the rope is F, and θ is the angle that the rope makes with the vertical, how is Ft related to F? How is Fg related to F?
 
  • #11
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Algebraically, if the tension in the rope is F, and θ is the angle that the rope makes with the vertical, how is Ft related to F? How is Fg related to F?
Does F=tan(Fg/Ft)?
 
  • #12
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Does F=tan(Fg/Ft)?
No. Where did you get the equation Ft/Fg=ma/mg from, and what does it mean to you? Have you had trigonometry yet? Have you learned about vectors yet? If so, if you know two perpendicular components of a vector, how do you find the magnitude of the vector?

Chet
 
  • #13
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No. Where did you get the equation Ft/Fg=ma/mg from, and what does it mean to you? Have you had trigonometry yet? Have you learned about vectors yet? If so, if you know two perpendicular components of a vector, how do you find the magnitude of the vector?

Chet
The equation was given to me by a classmate. It means that the force (acceleration) relative to the train divided by the force relative to the ground equals (the mass of the object times the force relative to the train) divided by (the mass of the object times the gravitational force). I have not had trigonometry for a few years.

Ft/Fg=ma/mg
(1.326364039/9.81)=((7.1)(1.326364039))/((7.1)(9.81))?
 
  • #14
19,919
4,095
The equation was given to me by a classmate. It means that the force (acceleration) relative to the train divided by the force relative to the ground equals (the mass of the object times the force relative to the train) divided by (the mass of the object times the gravitational force). I have not had trigonometry for a few years.

Ft/Fg=ma/mg
(1.326364039/9.81)=((7.1)(1.326364039))/((7.1)(9.81))?
None of this makes any sense to me. It sounds like metaphysics rather than physics. What about you, Bystander?

litz057, what would you have done if your friend wasn't there to help you? Then, you could not even have determined the horizontal acceleration (or as you incorrectly call it, the acceleration relative to the train).

Have you drawn a free body diagram showing the forces acting on the mass? If you don't know the most basic trig and you don't know anything about vectors, you are not going to be able to solve this problem, and we are not going to be able to help you. So, my advise is to learn basic trig, learn vectors, and learn how to apply free body diagrams to write down a force balance on an object.

Chet
 
  • #15
35
1
None of this makes any sense to me. It sounds like metaphysics rather than physics. What about you, Bystander?

litz057, what would you have done if your friend wasn't there to help you? Then, you could not even have determined the horizontal acceleration (or as you incorrectly call it, the acceleration relative to the train).

Have you drawn a free body diagram showing the forces acting on the mass? If you don't know the most basic trig and you don't know anything about vectors, you are not going to be able to solve this problem, and we are not going to be able to help you. So, my advise is to learn basic trig, learn vectors, and learn how to apply free body diagrams to write down a force balance on an object.

Chet
I have similar equations. I had an similar equation T=ma/mg, among a plethora of other equations (relevant or not). In these types of problems, my professor calls the horizontal acceleration the acceleration relative to the train. I know basic trig and basic trig functions. Vectors is an amount that involves a magnitude and direction.
 
  • #16
19,919
4,095
I have similar equations. I had an similar equation T=ma/mg, among a plethora of other equations (relevant or not). In these types of problems, my professor calls the horizontal acceleration the acceleration relative to the train.
I have trouble believing that a professor of physics could use such incorrect terminology.
I know basic trig and basic trig functions. Vectors is an amount that involves a magnitude and direction.
I'm going to have to withdraw from this this thread. I don't feel that I can help you until you have more knowledge of vectors, and are able to draw proper free body diagrams, identify the forces in these diagrams, and write down the relevant force balance equations. Sorry.

Chet
 

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