Mechanics- tension in a coupling

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Homework Help Overview

The problem involves two trucks being pulled up a slope by a string, with a rigid coupling between them. The first truck has a mass of 5kg and the second truck has a mass of 2kg, with the slope inclined at 10 degrees. The task is to demonstrate that the coupling remains in tension while the trucks are moving up the slope, despite the resistances acting on each truck.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss drawing free body diagrams (FBD) for each truck and writing equations of motion. There is confusion regarding the division of masses in the equations and the forces acting on the trucks. Questions arise about the role of the slope's reaction force and the contributions of tension from both the coupling and the string.

Discussion Status

The discussion is ongoing, with participants exploring the equations of motion and the forces involved. Some guidance has been offered regarding the need to consider the reaction forces and the implications of positive or negative tension in the coupling. However, there is no explicit consensus on the resolution of the problem.

Contextual Notes

Participants note the need to clarify the forces acting on the trucks, particularly the normal reaction forces and how they relate to the overall system. There is an emphasis on understanding the conditions under which the coupling remains in tension.

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Homework Statement


A truck that has a mass of 5kg is being pulled up a slope by a string. The truck is also pulling another truck of mass 2kg up the slope with a rigid coupling inbetween the trucks. Ths slope is inclined at 10 degrees to the horizontal. The 5kg truck has a resistance of 8N and the smaller truck a resistance of 6N. Show that the coupling remains in tension, regardless of the tension in the string, whilst the trucks are moving up the slope.


Homework Equations


I've been told that this is true:
gsin10+6/2 > gsin10+8/5


The Attempt at a Solution


I understand why the smaller truck has got to be greater. But I don't understand why the masses are divided on the equaiton. I thought that it would be 2gsin10 etc. but that doesn't work. Any help at all would be greatly appreciated.
 
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Draw a FBD for each truck and write the equations of motion parallel and perpendicular to the slope. That should get you started.
 
For the 5kg truck I've got:
5gsin10+8=T1 (in the string) parallel to the slope
5gcos10=? (perpendicular) I'm not sure what it would equal because there is no other force that is acting at an angle, and 8cos90 would be 0.

then for the 2kg truck, again similar:
2gsin10+6=T2 (in the coupling) parallel to the slope
2gcos10=? (perpendicular) Again, I'm not sure about the other perpendicular values.

Thanks again, are these equations right?
 
Does the slope push back on these two trucks?

You need to look again at the FBD for the 5kg truck. You have overlooked part of the system there.
 
Dr.D said:
Does the slope push back on these two trucks?

You need to look again at the FBD for the 5kg truck. You have overlooked part of the system there.

So the perpendicular equations will equal the reaction force (R), totally forgot about that. Thanks.

Oh and will the 5kg truck also have the Tension from the coupling and tension from the string?

So it will be:
5gsin10+8-T2=T1

From there will I have to do simultaneous equations? But I'm still not sure how it is proving that the coupling remains in tension. Thanks again.
 
If you get a positive number for the coupling tension, then it remains in tension. If you get a result that could go negative, then it might not remain in tension.

There will be separate normal reaction forces on each truck.
 

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