Rigid Objects In Static Equilibrium: Mass on a Bar

[mickellowery]

Homework Statement

One end of a uniform 4.00m long rod of weight F_g is supported by a cable. The cable is attached to the bar at an angle of 37.0^o. 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 $$\mu$$_s=0.500. Determine the minimum distance x from point A at which an additional object with the same weight F_g can be hung without causing the rod to slip off the wall.

Homework Equations

$$\Sigma$$F_x= F_h-Tcos37=0
$$\Sigma$$F_y= 0.500-F_go-F_gb+Tsin37=0
$$\Sigma$$$$\tau$$_z= x(F_go)-2(F_gb)+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

 

[JaWiB]

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.

 

[inky]

Homework Statement

One end of a uniform 4.00m long rod of weight F_g is supported by a cable. The cable is attached to the bar at an angle of 37.0^o. 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 $$\mu$$_s=0.500. Determine the minimum distance x from point A at which an additional object with the same weight F_g can be hung without causing the rod to slip off the wall.

Homework Equations

$$\Sigma$$F_x= F_h-Tcos37=0
$$\Sigma$$F_y= 0.500-F_go-F_gb+Tsin37=0
$$\Sigma$$$$\tau$$_z= x(F_go)-2(F_gb)+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

 

[mickellowery]

Oh sorry its the minimum distance away from the wall.

 

[inky]

Oh sorry its the minimum distance away from the wall.

According to your Eqns.,first eqn. is correct. F_h=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]\sumF_y[/tex]= 0.500F_h-Fg-Fg+Tsin37=0

3rd eqn.(how do you get Tsin14?)angle is 37 degree
change Tsin37
instead of F_go and F_gb , write F_g
So you can get the answer.

 

[mickellowery]

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.