Finding Tension in a Beam Supported by a Pulley: Solving for Equilibrium

[taxidriverhk]

Homework Statement

[Ans: F_A = -206i + 188j lb, T = 131 lb]

Homework Equations

Since it's assumed that the system is in equilibrium, so I assume that ΣF=0 and ΣM=0

The Attempt at a Solution

I tried to draw a free-body diagram for the system, and set up three equations below:

(I don't know the direction of F_A yet, so I assume that both components are in positive direction)
ΣF_x = F_Ax + Tcos45° + Tcos30° = 0
ΣF_y = F_Ay - 150 - W + Tsin45° + Tsin30° = 0
ΣM_z [Moment centered at A] = (16)(-150 - W) + (40)(Tsin30° + Tsin45°) + (6)(Tcos30° + Tcos45°) = 0

When I solved the third equation to find tension T, it didn't give me the right answer, I am not sure which part I did wrong, maybe I missed some forces on the free-body diagram, so I made a wrong equation of ΣM_z?
Any hints or approaches are much appreciated.

 

[NTW]

In the FBD, and when calculating the moments, you don't put W = 20 kg at mid-point along the beam...

 

[SteamKing]

To clarify what NTW was trying to tell you, you somehow neglected to include the weight of the beam in your equations of static equilibrium.

 

[taxidriverhk]

To clarify what NTW was trying to tell you, you somehow neglected to include the weight of the beam in your equations of static equilibrium.

I think I had included the weight of the beam in FBD and equations, but maybe NTW is trying to tell me that the weight should be in another point along the beam?

 

[SteamKing]

I think I had included the weight of the beam in FBD and equations, but maybe NTW is trying to tell me that the weight should be in another point along the beam?

Homework Statement

[Ans: F_A = -206i + 188j lb, T = 131 lb]

Homework Equations

Since it's assumed that the system is in equilibrium, so I assume that ΣF=0 and ΣM=0

The Attempt at a Solution

I tried to draw a free-body diagram for the system, and set up three equations below:

(I don't know the direction of F_A yet, so I assume that both components are in positive direction)
ΣF_x = F_Ax + Tcos45° + Tcos30° = 0
ΣF_y = F_Ay - 150 - W + Tsin45° + Tsin30° = 0
ΣM_z [Moment centered at A] = (16)(-150 - W) + (40)(Tsin30° + Tsin45°) + (6)(Tcos30° + Tcos45°) = 0

When I solved the third equation to find tension T, it didn't give me the right answer, I am not sure which part I did wrong, maybe I missed some forces on the free-body diagram, so I made a wrong equation of ΣM_z?
Any hints or approaches are much appreciated.

If the beam has a mass of 20 kg, then the weight in lbs is easy to calculate. (Hint: using 2 lbs/kg is not really acceptable. 2.2 lbs/kg is more accurate)

If the beam has a uniform cross-section, then using a c.g. of 16 in. from A is not correct either. The c.g. of a uniform bar is going to be half its length, as measured from one end.

 

[NTW]

I think I had included the weight of the beam in FBD and equations, but maybe NTW is trying to tell me that the weight should be in another point along the beam?

Exactly. Mid-point, as it's a uniform beam...