Solving an Equilibrium Problem

  • Thread starter Domestikus
  • Start date
  • Tags
    Equlibrium
In summary: To keep the system in equilibrium, the horizontal force acting on the trolley would have to be equal to mg*tan(a). In summary, the problem involves a mass attached to a wall with a string and set on a trolley on an inclined plane. The goal is to determine the horizontal force needed to keep the system motionless, assuming no friction. The attempted solution involved using trigonometric functions, but the correct approach is to draw separate free-body diagrams for the trolley and the block and take into account the normal force between them. The final solution involves a horizontal force equal to mg*tan(a) to maintain equilibrium.
  • #1
Domestikus
3
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
  • #2
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.
 
  • #3
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).
 

1. How do you determine the equilibrium constant for a chemical reaction?

The equilibrium constant (Keq) can be determined by using the concentrations of the reactants and products at equilibrium. It is calculated by dividing the product of the concentrations of the products raised to their respective coefficients by the product of the concentrations of the reactants raised to their respective coefficients.

2. What is the difference between Keq and Qeq?

Keq is the equilibrium constant for a reaction at a specific temperature, while Qeq is the reaction quotient which is calculated using the initial concentrations of the reactants and products. When the reaction is at equilibrium, the values of Keq and Qeq will be equal.

3. How do you know if a reaction has reached equilibrium?

A reaction has reached equilibrium when the concentrations of the reactants and products remain constant over time. This can be observed by measuring the concentrations at different time intervals and seeing if they change significantly or not.

4. How do you calculate the concentrations of reactants and products at equilibrium?

The concentrations at equilibrium can be calculated using the equilibrium constant (Keq) and the initial concentrations of the reactants and products. By rearranging the equilibrium expression, the concentrations of the reactants and products can be solved for.

5. How does changing the temperature affect the equilibrium constant?

The equilibrium constant (Keq) is temperature-dependent. Changing the temperature of a reaction will also change the value of Keq. In general, an increase in temperature will favor the endothermic reaction, while a decrease in temperature will favor the exothermic reaction.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
225
  • Introductory Physics Homework Help
Replies
2
Views
627
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
2K
Back
Top