Solving an Equilibrium Problem

  • Thread starter Thread starter Domestikus
  • Start date Start date
  • Tags Tags
    Equlibrium
AI Thread Summary
The discussion revolves around solving an equilibrium problem involving a mass attached to a trolley on an inclined plane. The user initially attempts to relate tension and force using trigonometric functions but expresses confusion about the implications when the angle approaches zero. Another participant suggests that the user should analyze the system by drawing separate free-body diagrams for both the trolley and the block to account for the normal forces. This approach leads to a revised understanding, indicating that tension should equal mg*sin(a). The conversation emphasizes the importance of correctly analyzing forces in non-rigid systems for accurate solutions.
Domestikus
Messages
3
Reaction score
0
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:
Physics news on Phys.org
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).
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top