Rigid Objects In Static Equilibrium: Mass on a Bar

AI Thread Summary
The discussion revolves around a physics problem involving a uniform rod supported by a cable and resting against a wall, with a focus on determining the minimum distance from point A to hang an additional weight without causing the rod to slip. Participants clarify the equations of static equilibrium, emphasizing the correct application of forces and torques. There is confusion about whether to combine the weights of the rod and the additional object, with the consensus that their positions affect the torque calculations. The importance of correctly identifying the angles and forces in the equations is highlighted, particularly correcting an error related to the angle used in torque calculations. The conversation underscores the need for precise definitions and careful application of physics principles in solving equilibrium problems.
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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 \mus=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


\SigmaFx= Fh-Tcos37=0
\SigmaFy= 0.500-Fgo-Fgb+Tsin37=0
\Sigma\tauz= 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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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.
 
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 \mus=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


\SigmaFx= Fh-Tcos37=0
\SigmaFy= 0.500-Fgo-Fgb+Tsin37=0
\Sigma\tauz= 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
 
Oh sorry its the minimum distance away from the wall.
 
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).
\sumF<sub>y</sub>= 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.
 
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.
 

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