Moments of force on Jenga tower

[maserati1969]

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

 

[berkeman]

Welcome to PF.

Please use the "Attach files" link below the Edit window to upload a diagram of your tower for this problem. Thank you.

 

[BvU]

Hello @maserati1969 ,

$\qquad$!​

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

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

am I correct so far ?$\$

 

[maserati1969]

Hello @maserati1969 ,

$\qquad$!​

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.

 

[maserati1969]

Welcome to PF.

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

 

[maserati1969]

Hello @maserati1969 ,

$\qquad$!​

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.

 

[jbriggs444]

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.

 

[BvU]

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

(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 ).

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

$\$

 

[haruspex]

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.

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.

 

[maserati1969]

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

 

[maserati1969]

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 ).

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

 

[maserati1969]

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.

 

[maserati1969]

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 ).

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:

 

[haruspex]

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.

 

[maserati1969]

Apologies, I'll post this as new