Rigid Objects In Static Equilibrium: Mass on a Bar

In summary, the problem involves a uniform 4.00m long rod with one end supported by a cable at an angle of 37.0o and the other end resting against a wall with a coefficient of static friction of 0.500. The goal is to determine the minimum distance from the wall at which an additional object with the same weight as the rod can be hung without causing the rod to slip off the wall. The equations used are the sum of forces in the x-direction, the sum of forces in the y-direction, and the sum of torques. The first equation is correct, but the second equation has a mistake with the frictional force. The third equation also has an error with the angle, and it is
  • #1
mickellowery
69
0

Homework Statement


One end of a uniform 4.00m long rod of weight Fg is supported by a cable. The cable is attached to the bar at an angle of 37.0o. The other end of the bar rests against the wall where its held by friction. The coefficient of static friction between the wall and the rod is [tex]\mu[/tex]s=0.500. Determine the minimum distance x from point A at which an additional object with the same weight Fg can be hung without causing the rod to slip off the wall.


Homework Equations


[tex]\Sigma[/tex]Fx= Fh-Tcos37=0
[tex]\Sigma[/tex]Fy= 0.500-Fgo-Fgb+Tsin37=0
[tex]\Sigma[/tex][tex]\tau[/tex]z= x(Fgo)-2(Fgb)+4Tsin14=0

Since the two weights are equal can I add them together in the equations and call it 2F? I was thinking I couldn't do this because the additional object's weight will be in a different area than the center of gravity of the beam. Also my subscripts are go is the object and gb is the beam. Fh is the horizontal force of the wall on the beam.

The Attempt at a Solution

 
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  • #2
For finding the net force you should be able to just add the x and y components of each force together. For the torque, you'll have to take into account where each force is acting.
 
  • #3
mickellowery said:

Homework Statement


One end of a uniform 4.00m long rod of weight Fg is supported by a cable. The cable is attached to the bar at an angle of 37.0o. The other end of the bar rests against the wall where its held by friction. The coefficient of static friction etween the wall and the rod is [tex]\mu[/tex]s=0.500. Determine the minimum distance x from point A at which an additional object with the same weight Fg can be hung without causing the rod to slip off the wall.


Homework Equations


[tex]\Sigma[/tex]Fx= Fh-Tcos37=0
[tex]\Sigma[/tex]Fy= 0.500-Fgo-Fgb+Tsin37=0
[tex]\Sigma[/tex][tex]\tau[/tex]z= x(Fgo)-2(Fgb)+4Tsin14=0

Since the two weights are equal can I add them together in the equations and call it 2F? I was thinking I couldn't do this because the additional object's weight will be in a different area than the center of gravity of the beam. Also my subscripts are go is the object and gb is the beam. Fh is the horizontal force of the wall on the beam.

The Attempt at a Solution


Where is the point A?
0.5 is coefficient of friction. Not frictional force. f=0.5N
 
  • #4
Oh sorry its the minimum distance away from the wall.
 
  • #5
mickellowery said:
Oh sorry its the minimum distance away from the wall.

According to your Eqns.,first eqn. is correct. Fh=normal force=horizontal force
2nd eqn. is in correct.
F(g)=F(go)=F(go)
Don't write a lot of unknown values. Take F(g).
[tex]\sumFy[/tex]= 0.500Fh-Fg-Fg+Tsin37=0

3rd eqn.(how do you get Tsin14?)angle is 37 degree
change Tsin37
instead of Fgo and Fgb , write Fg
So you can get the answer.
 
  • #6
Alright thanks much Inky. I don't know where the 14 came from. I guess I have to make a stupid math mistake so I'll just substitute in nonexistent numbers.
 

FAQ: Rigid Objects In Static Equilibrium: Mass on a Bar

1. What is static equilibrium when referring to rigid objects?

In physics, static equilibrium refers to the state where an object is at rest and all the forces acting on it are balanced. This means that there is no acceleration or movement in any direction.

2. How is the mass of an object on a bar calculated?

The mass of an object on a bar can be calculated by using the equation: mass = weight / acceleration due to gravity. The weight of the object can be determined by multiplying its mass by the acceleration due to gravity, which is approximately 9.8 m/s^2 on Earth.

3. What is the relationship between the center of mass and static equilibrium?

In order for an object to be in static equilibrium, the center of mass must be directly above the point of support. This ensures that the object will not tip over or rotate due to unbalanced forces.

4. How does the length of the bar affect the equilibrium of the object?

The longer the bar, the more stable the object will be in static equilibrium. This is because a longer bar will have a larger moment arm, which means that the forces acting on the object will have a greater distance to act upon, making it more difficult to tip over.

5. Can an object be in static equilibrium if there are no external forces acting on it?

Yes, an object can be in static equilibrium even if there are no external forces acting on it. This is because the forces acting within the object, such as the weight and normal force, can still be balanced. However, if there are no external forces, the object will remain at rest and will not experience any acceleration.

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