Statics: Internal forces at a point along a beam

[yaro99]

Homework Statement

Homework Equations

ƩF_x=0
ƩF_y=0
ƩM=0

The Attempt at a Solution

FBD of the entire thing:

ƩM_B = 780(0.3) + (5/13)T*(0.72) - (12/13)T*(0.6) = 0
T = 845N

ƩF_x = B_x + (5/13)*845 = 0
B_x = 325N ←
ƩF_y = B_y + (12/13)*845 = 0
B_y = 780N ↓

Now here is where I tried 2 different methods for the forces at J:


Method 1:

FBD of CDJ:

ƩF_x = (5/13)*845 - (12/13)V + (5/13)F = 0
ƩF_y = (12/13)*845 - 780 + (5/13)V + (12/13)F = 0
θ = arctan(5/13) = 22.6°
F = 125N 67.4° down and to the left
V = 300N 22.6° up and to the left

horizontal distance from D to J = 0.24/tan(67.4°) = 0.1m
ƩM_J = (5/13)*845*(0.24) - (12/13)*845*(0.4) + 780*(0.1) + M = 0
M = 156 N*m counterclockwise

This method gives me the correct magnitudes, but the directions are supposed to be opposite of what I got, which I don't understand.

Method 2:

FBD of BJ:

ƩF_x = (12/13)V - (5/13)F - 325 = 0
ƩF_y = -(5/13)V - (12/13)F - 780 = 0

F = 845N 67.4° up and to the right
V = 0N

horizontal distance from J to B = 0.48/tan(67.4°) = 0.2m
ƩM_J = 325*(0.48) -780*(0.2) + M = 0
M = 144 N*m counterclockwise

I am not sure why this method gives me incorrect answers.

 

[TSny]

ƩF_y = B_y + (12/13)*845 = 0
B_y = 780N ↓

Did you leave out one of the vertical forces here?

I agree with your answers for Method 1 including your directions.

 

[Maiq]

There is no By force. The horizontal force in the AC rope is 780. It looks like your forgot to add that in on your Sum of y-direction forces.

 

[yaro99]

There is no By force. The horizontal force in the AC rope is 780. It looks like your forgot to add that in on your Sum of y-direction forces.

Oh yeah, thanks for the correction. I have the right numbers and directions now, using the second method.

Did you leave out one of the vertical forces here?

I agree with your answers for Method 1 including your directions.

Doing method 2 I get the "correct" directions according to the book, which are opposite those in method 1.

 

[Maiq]

Your answers for both methods should have opposite directions. If you were to put those two sections back together the forces at J would cancel out and the whole system would be in static equilibrium. I guess your book just used method two but either way would be correct.

 

[yaro99]

Your answers for both methods should have opposite directions. If you were to put those two sections back together the forces at J would cancel out and the whole system would be in static equilibrium. I guess your book just used method two but either way would be correct.

Ah, ok, this makes sense. Thanks!