Calculating Tension and Reaction Forces in a Ball and Socket Joint

[goldfish9776]

Homework Statement

Determine the tension cable A and B and the x , y , z reaction of the ball and socket joint at C .

Homework Equations

The Attempt at a Solution

I let the reaction of ball and socket joint at C = Fx , Fy and Fz
Sum of force in z direction :
TA +TB +FZ= 400

Sum of force in x direction :
Fx = 0

Sum of force in y direction :
Fy = 0

Sum of moment about X :
3TB -1.3(400) = 0
3TB+1.6FZ= 520
Sum of moment about Y :
-2TA-2TB = -1.5(400)[/B]
2TA+2TB= -600

Sum of moment about z :
-1.6Fx = 0 , Fx = 0

Since TA +TB= -300 , so i sub this eq into TA +TB +FZ= 400 ,
i gt -300+Fz = 400 , Fz = 700N

3TB +1.6(700) =520 , TB =-200N

TA -200+700=400 , TA = -100N

since the question mention the A, B is in tension , but i gt my TB and TA in negative value , i am wondering my ans is correct or not


-

 

[SteamKing]

Homework Statement

Determine the tension cable A and B and the x , y , z reaction of the ball and socket joint at C .

Homework Equations

The Attempt at a Solution

I let the reaction of ball and socket joint at C = Fx , Fy and Fz
Sum of force in z direction :
TA +TB +FZ= 400

Sum of force in x direction :
Fx = 0

Sum of force in y direction :
Fy = 0

Sum of moment about X :
3TB -1.3(400) = 0
3TB+1.6FZ= 520
Sum of moment about Y :
-2TA-2TB = -1.5(400)[/B]
2TA+2TB= -600

Sum of moment about z :
-1.6Fx = 0 , Fx = 0

Since TA +TB= -300 , so i sub this eq into TA +TB +FZ= 400 ,
i gt -300+Fz = 400 , Fz = 700N

3TB +1.6(700) =520 , TB =-200N

TA -200+700=400 , TA = -100N

since the question mention the A, B is in tension , but i gt my TB and TA in negative value , i am wondering my ans is correct or not


-

No, it's not correct.

From the sum of the moments about the y-axis, you get TA + TB = 300 N, which seems reasonable.

Where you messed up was substituting back into the force equation: TA + TB + FZ = 600

If TA + TB = 300, how can FZ = 700 N?

 

[goldfish9776]

No, it's not correct.

From the sum of the moments about the y-axis, you get TA + TB = 300 N, which seems reasonable.

Where you messed up was substituting back into the force equation: TA + TB + FZ = 600

If TA + TB = 300, how can FZ = 700 N?

so , after changing TB +TA = 300 , i have TA = 480 N , TB=120N and FZ= 100N , are they correct now ?

 

[SteamKing]

so , after changing TB +TA = 300 , i have TA = 480 N , TB=120N and FZ= 100N , are they correct now ?

Why did you do this? It doesn't make any sense.

If TA = 480 and TB = 120, how can TA + TB = 300?

You're not even doing mathematics anymore. You're guessing or something, which I haven't figured out (and don't want to.)

Look, if
TA + TB + FZ = 600

and

TA + TB = 300,

then

300 + FZ = 600

What's FZ? This is a simple equation which you should be able to solve.

 

[goldfish9776]

Why did you do this? It doesn't make any sense.

If TA = 480 and TB = 120, how can TA + TB = 300?

You're not even doing mathematics anymore. You're guessing or something, which I haven't figured out (and don't want to.)

Look, if
TA + TB + FZ = 600

and

TA + TB = 300,

then

300 + FZ = 600

What's FZ? This is a simple equation which you should be able to solve.

so , i have my TA = 180N , TB=120n , FZ= 100N , are they correct now ?

 

[SteamKing]

so , i have my TA = 180N , TB=120n , FZ= 100N , are they correct now ?

These would normally be OK, but you seem to have overlooked that couple of magnitude 200 N-m which wants to rotate the plate around the z-axis.

The presence of this couple complicates things, and changes the tensions in cables A and B.

 

[goldfish9776]

These would normally be OK, but you seem to have overlooked that couple of magnitude 200 N-m which wants to rotate the plate around the z-axis.

The presence of this couple complicates things, and changes the tensions in cables A and B.

ok , the @00Nm is the moment about z axis , a i right ? So , my -1.6FX +200 = 0 , FX = 125N , the other reading is not affected , right ?

 

[SteamKing]

ok , the @00Nm is the moment about z axis , a i right ? So , my -1.6FX +200 = 0 , FX = 125N , the other reading is not affected , right ?

It's not that simple, I'm afraid.

After calculating FX above, did you sum all of the forces acting in the x-direction? Was that sum equal to zero? If it was not, then the plate ABDE will rotate about the z-axis, and the cables A and B will no longer remain vertical; there will be some x and y components to the tensions in those cables which necessarily develop to keep the plate in static equilibrium.

 

[goldfish9776]

It's not that simple, I'm afraid.

After calculating FX above, did you sum all of the forces acting in the x-direction? Was that sum equal to zero? If it was not, then the plate ABDE will rotate about the z-axis, and the cables A and B will no longer remain vertical; there will be some x and y components to the tensions in those cables which necessarily develop to keep the plate in static equilibrium.

can you find the other force in X direction to counter the FX , i couldn't find any

 

[SteamKing]

can you find the other force in X direction to counter the FX , i couldn't find any

That's why I wrote what I wrote in Post #8.

The plate can rotate only about the ball and socket joint at C, but it cannot translate in any of the three coordinate directions x, y, or z. If the plate rotates about the joint at C, then cables A and B are no longer vertical, and since each cable has a tension in it, there will be x and y components of the tension created in each cable where it attaches to the plate. The sum of the x and y force components of the reaction at C and cables A and B must both equal zero for the plate to remain in static equilibrium. That's where the other forces come from to counter FX and FY.

 

[goldfish9776]

That's why I wrote what I wrote in Post #8.

The plate can rotate only about the ball and socket joint at C, but it cannot translate in any of the three coordinate directions x, y, or z. If the plate rotates about the joint at C, then cables A and B are no longer vertical, and since each cable has a tension in it, there will be x and y components of the tension created in each cable where it attaches to the plate. The sum of the x and y force components of the reaction at C and cables A and B must both equal zero for the plate to remain in static equilibrium. That's where the other forces come from to counter FX and FY.

that seems very complicated

 

[SteamKing]

that seems very complicated

It doesn't have to be, not if you write the correct force and moment equations for the plate taking these facts into consideration.