Solving an Equilibrium Problem

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SUMMARY

The discussion focuses on solving an equilibrium problem involving a mass attached to a wall via a light string on a trolley with an inclined plane. The key equations derived are Tsin(α) = G and Tcos(α) = F, leading to F = mg/tan(α). A critical insight is that the tension T must equal mg*sin(α) when considering the forces acting on both the trolley and the block separately. This approach clarifies the relationship between the forces and resolves the initial confusion regarding infinite force when α approaches 0°.

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Domestikus
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Equilibrium problem

Homework Statement


http://www.deviantpics.com/image.php?id=5029_4A04401C

Mass m is atached to the wall with a light string and set on a trolley with an inclined plane.
The string is parallel with the plane. Let's say that there is no friction at all.
Determine intensity of the horizontal force which has to act upon the trolley so that the system stays motionless.


2. The attempt at a solution

Tsin(alpha)=G
Tcos(alpha)=F

F=mg/tg(alpha)

But I'm not sure if this is correct. If we let alpha be 0° than force F would have to be infinite. Also this would mean that tension T is greater than G and that seems odd too.Can you please tell we what am I doing wrong?


Thanks
 
Last edited:
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In the picture above you've considered the trolley and the block as one object, but that's not really valid because they are not rigidly attached to each other. Try drawing a separate free-body diagram for each, and taking into account the normal force that the trolley exerts on the block (and vice-versa). I get what seems like a more reasonable answer doing it that way.
 
OK this is what I get:

http://www.deviantpics.com/image.php?id=CDD1_4A071DF0

Is this any good?
In this case T would be mg*sin(a).
 

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