Moments of force on Jenga tower

In summary, the conversation is discussing a problem with a stacking game where the center of mass for layers 8-12 is at the middle of layer 10. The question is whether the tower will topple and how far the last block in layer 7 can be moved to prevent this from happening. The solution involves calculating the net torque on the upper section of layers and determining the maximum distance the last block can be moved to reduce the torque to zero.
  • #1
maserati1969
20
2
Homework Statement
How would you determine the torque on a jenga tower if it was 12 layers how and layer 7 had 2 of its 3 blocks missing, (middle and side blocks).
Would you use that layer as the reference for the Moment formula? How would you know wether it would topple over?
Thanks in advance
Relevant Equations
M = F x d
Centre of mass = Sum of [Individual layers C.O.M's x mass) / Total Mass
M= F x d at layer 7, layers 8 - 12 has its centre of mass at middle of layer 10, which is where the weight of these layers would be acting. Not sure hoe to determine whether it would topple
 
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  • #2
Welcome to PF. :smile:

Please use the "Attach files" link below the Edit window to upload a diagram of your tower for this problem. Thank you.
 
  • #3
Hello @maserati1969 ,
:welcome: ##\qquad##!​

Had to look up what this is about. apparently it's about this stacking game
1678716140119.png

and your problem description leads me to imagine a side view like this

1678716172507.png


am I correct so far ?##\ ##
 
  • #4
BvU said:
Hello @maserati1969 ,
:welcome: ##\qquad##!​

Had to look up what this is about. apparently it's about this stacking game
View attachment 323553
and your problem description leads me to imagine a side view like this

View attachment 323554

am I correct so far ?##\ ##
Hi BvU, absolutely correct so far, many thanks.
 
  • #5
berkeman said:
Welcome to PF. :smile:

Please use the "Attach files" link below the Edit window to upload a diagram of your tower for this problem. Thank you.
Hi Berkeman, sorry for putting it in the wrong place, and thanks for the reply. BvU below has illusrtated exactly what I was asking.
Best regards
 
  • #6
BvU said:
Hello @maserati1969 ,
:welcome: ##\qquad##!​

Had to look up what this is about. apparently it's about this stacking game
View attachment 323553
and your problem description leads me to imagine a side view like this

View attachment 323554

am I correct so far ?##\ ##
So the question is: Calculate the Net Torque on the 8th - 12th layers, and will the top part of the tower fall over?
If it does fall over, find the maximum distance that the final 7th layer block can be from the centre of the tower so that it won't fall over.
 
  • #7
maserati1969 said:
M= F x d at layer 7, layers 8 - 12 has its centre of mass at middle of layer 10, which is where the weight of these layers would be acting. Not sure hoe to determine whether it would topple
Generally speaking, a rigid object is sure to topple if its center of mass lies outside the [convex] outline of its base.

You can arrive at this principle by thinking about what happens if the object were tilted ever so slightly in a particular direction. The object would then be supported only by the point (or points) on the tilted-to side of its base.

You draw your free body diagram and consider all of the forces acting on the slightly tilted object. You figure out what torques are associated with each one. Then you decide whether the resulting net torque:

1. Tends to un-tilt the object. -or-
2. Tends to tilt the object even further.

Un-tilting meant that you have negative feedback. A stable equilibrium is possible. The tower will not topple in that direction unless disturbed more significantly.

Tilting further means that you have positive feedback. The tower will topple in that direction, speeding up as it does so.
 
  • #8
maserati1969 said:
So the question is: Calculate the Net Torque on the 8th - 12th layers, and will the top part of the tower fall over?
If it does fall over, find the maximum distance that the final 7th layer block can be from the centre of the tower so that it won't fall over.
Good. So we have
1678732178929.png

(all I did was remove the two blocks from their 'default position'.)
The center of mass from the upper five layers is at the red dot. Weight of the upper section of five layers is ##F = 15\, mg## when ##m## is the mass of one block.

If we take the top right of the last block in layer 7 as origin (the blue dot), we see that there is a torque ##\tau = \vec r \times \vec F## pointing into the paper (screen...). Acting on the top section of five layers, which will topple to the right (cheating with glue is of course severely prohibited :smile: ).

You can see how far the last block in layer 7 has to be moved to the right to reduce the torque to zero.

##\ ##
 
  • #9
maserati1969 said:
find the maximum distance that the final 7th layer block can be from the centre of the tower so that it won't fall over.

BvU said:
You can see how far the last block in layer 7 has to be moved to the right to reduce the torque to zero.
Note that those two questions aren’t exactly the same. @maserati1969, are you sure you have stated it correctly? @BvU's phrasing seems more likely.
 
  • #10
jbriggs444 said:
Generally speaking, a rigid object is sure to topple if its center of mass lies outside the [convex] outline of its base.

You can arrive at this principle by thinking about what happens if the object were tilted ever so slightly in a particular direction. The object would then be supported only by the point (or points) on the tilted-to side of its base.

You draw your free body diagram and consider all of the forces acting on the slightly tilted object. You figure out what torques are associated with each one. Then you decide whether the resulting net torque:

1. Tends to un-tilt the object. -or-
2. Tends to tilt the object even further.

Un-tilting meant that you have negative feedback. A stable equilibrium is possible. The tower will not topple in that direction unless disturbed more significantly.

Tilting further means that you have positive feedback. The tower will topple in that direction, speeding up as it does so.
Many thanks for the jbriggs444, very helpful.
Best Regards
 
  • #11
BvU said:
Good. So we have
View attachment 323564
(all I did was remove the two blocks from their 'default position'.)
The center of mass from the upper five layers is at the red dot. Weight of the upper section of five layers is ##F = 15\, mg## when ##m## is the mass of one block.

If we take the top right of the last block in layer 7 as origin (the blue dot), we see that there is a torque ##\tau = \vec r \times \vec F## pointing into the paper (screen...). Acting on the top section of five layers, which will topple to the right (cheating with glue is of course severely prohibited :smile: ).

You can see how far the last block in layer 7 has to be moved to the right to reduce the torque to zero.

##\ ##
Great, that makes perfect sense. Thanks BvU
 
  • #12
haruspex said:
Note that those two questions aren’t exactly the same. @maserati1969, are you sure you have stated it correctly? @BvU's phrasing seems more likely.
Yes, I see, they are more or less the same question. Thanks for the note though.
 
  • #13
BvU said:
Good. So we have
View attachment 323564
(all I did was remove the two blocks from their 'default position'.)
The center of mass from the upper five layers is at the red dot. Weight of the upper section of five layers is ##F = 15\, mg## when ##m## is the mass of one block.

If we take the top right of the last block in layer 7 as origin (the blue dot), we see that there is a torque ##\tau = \vec r \times \vec F## pointing into the paper (screen...). Acting on the top section of five layers, which will topple to the right (cheating with glue is of course severely prohibited :smile: ).

You can see how far the last block in layer 7 has to be moved to the right to reduce the torque to zero.

##\ ##
Hi again BvU, thanks again for this information. Can you help with Question 4 below:

1679252239822.png
 
  • #14
maserati1969 said:
Hi again BvU, thanks again for this information. Can you help with Question 4 below:

View attachment 323834
New question, new thread please.
And in that, please post your answers to 3a and 3b, and your thoughts on q4.
 
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Likes BvU
  • #15
Apologies, I'll post this as new
 
  • Like
Likes BvU

1. What is a moment of force on a Jenga tower?

A moment of force on a Jenga tower refers to the rotational force applied to the tower when a block is removed or when external forces act on the tower. It is a measure of the tendency of the tower to rotate around a specific point.

2. How does the weight of the Jenga blocks affect the moments of force on the tower?

The weight of the Jenga blocks affects the moments of force on the tower by increasing the force of gravity acting on the tower. This can lead to a greater rotational force and increase the likelihood of the tower collapsing.

3. What factors can influence the moments of force on a Jenga tower?

Several factors can influence the moments of force on a Jenga tower, including the weight and distribution of the blocks, the surface on which the tower is built, and external forces such as wind or vibrations.

4. How can moments of force on a Jenga tower be calculated?

Moments of force on a Jenga tower can be calculated by multiplying the force applied to the tower by the perpendicular distance from the point of rotation to the line of action of the force. This is known as the moment arm.

5. What precautions should be taken to minimize moments of force on a Jenga tower?

To minimize moments of force on a Jenga tower, it is important to evenly distribute the weight of the blocks and build the tower on a stable and level surface. Additionally, avoiding sudden movements or external forces can help reduce the risk of the tower collapsing due to moments of force.

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