Calculating Tension and Reaction Forces in a Ball and Socket Joint

In summary, the problem involves determining the tension in cables A and B and the x, y, z reactions of the ball and socket joint at C. Using the given equations and attempting a solution, the initial values of TA and TB were found to be incorrect. After making some adjustments, the correct values were found to be TA = 180N, TB = 120N, and FZ = 100N. However, the presence of a couple of magnitude 200 N-m complicates the problem and may result in the rotation of the plate ABDE. Further calculations are needed to determine the x and y components of the tensions in cables A and B to keep the plate in static equilibrium.
  • #1
goldfish9776
310
1

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


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  • #2
goldfish9776 said:

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


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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?
 
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  • #3
SteamKing said:
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 ?
 
  • #4
goldfish9776 said:
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.
 
  • #5
SteamKing said:
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 ?
 
  • #6
goldfish9776 said:
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.
 
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  • #7
SteamKing said:
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 ?
 
  • #8
goldfish9776 said:
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.
 
  • #9
SteamKing said:
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
 
  • #10
goldfish9776 said:
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.
 
  • #11
SteamKing said:
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
 
  • #12
goldfish9776 said:
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.
 

Related to Calculating Tension and Reaction Forces in a Ball and Socket Joint

1. What is tension in negative value?

Tension in negative value refers to the force applied in a direction opposite to the natural tendency of an object. This results in a negative value for the tension force.

2. How is tension in negative value calculated?

Tension in negative value is calculated by considering the forces acting on an object in a given situation. The negative value is assigned when the direction of the tension force is opposite to the direction of the object's motion.

3. What are some examples of tension in negative value?

One example of tension in negative value is a rope pulling a block of wood upwards. The tension force applied by the rope is in the opposite direction of the block's natural tendency to fall due to gravity. Another example is a person holding a heavy object and walking in the opposite direction of the object's natural motion.

4. How does tension in negative value affect an object?

Tension in negative value can affect an object by slowing down its motion or preventing it from moving in a certain direction. In some cases, it can also cause the object to change its direction or come to a complete stop.

5. What is the significance of understanding tension in negative value in scientific research?

Understanding tension in negative value is crucial in scientific research as it helps in predicting the behavior of objects in different situations. It also plays a key role in the design and analysis of various structures, machinery, and systems. Additionally, understanding tension in negative value can aid in identifying potential risks and improving safety measures in various applications.

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