₦airaland Forum

Morbeta11:
Who dies first if "E" pushes the stone?




Ccccccccccc

D dies first

1. E pushes the stone → it rolls down.


2. D dies first → the stone crushes D at the bottom of the slope.


3. The stone keeps rolling → it hits the smaller stone sitting on the seesaw.


4. The smaller stone is launched → the spike side of the seesaw slams down on C, killing C.


5. The launched smaller stone flies toward B → it smashes into B, killing B.


6. Finally, without B blocking, the chain reaction continues → A is left standing but also at risk depending on the force. If the small stone keeps momentum, A could be hit last.



👉 Order of deaths: D → C → B → (possibly A last).

D

D dies first.

MMempire:
C!
This is where nairaland bot is useless as an answer as simple as an alphabet or digit can not be welcomed anymore without the bot asking for 40 more characters.

the Ali baba and the 40 thieves nairaland spam bot tire me o.

A

Ccccccc for sure

D

B dies first

Fiscus105:
You totally ignored question Op asked but ranting and lamenting, why won't you guys have toxic relationship?


Go back to OP, it depends on the weight and extent of force applied by E, if it stuck in first hole, only D would suffocate to death, but if force applied strong, it may bye pass D and move forward to land on "C", definitely C would die. Meanwhile having landed on plank above C, it would catapult the stone and shift it to any of the them, including the person that pushes the stone "E".

IN CONCLUSION, IT DEPENDS ON THE FORCE APPLIED BY E

ASIDE:
This your signature, I'm assuming it's symbolic representation of an image.
What are we supposed to be looking at...what are we to see...?
What are you communicating?

Morbeta11:
Who dies first if "E" pushes the stone?




Ccccccccccc

No one dies
Help me add the remaining letters

E dies first, classic case of, "arigi se gi, ori ara e lo ma fi gbe"

Sirchef:
1. E pushes the stone → it rolls down.


2. D dies first → the stone crushes D at the bottom of the slope.


3. The stone keeps rolling → it hits the smaller stone sitting on the seesaw.


4. The smaller stone is launched → the spike side of the seesaw slams down on C, killing C.


5. The launched smaller stone flies toward B → it smashes into B, killing B.


6. Finally, without B blocking, the chain reaction continues → A is left standing but also at risk depending on the force. If the small stone keeps momentum, A could be hit last.



👉 Order of deaths: D → C → B → (possibly A last).

1 like 👍 for the effort.
Not really coz you're right.
You seem to be the first to have seen an oscillating motion...that's a wild possibility, but a possibility nonetheless.

Actually looking at it again A or B is likely to die ..

There is curve on that wood that make the stone flip over to where A stands ..

No one dies if the stone is push down by e the first contact would bend down and protect himself the second one will be balance as a form of scale by the two stone are this point the stone will be at rest no one get hot

CoronaVirusPro:
No one dies.

I don’t have 500 characters to type and will update when I have.

You're very correct. No one dies. The weight of the stone is not enough to lift the lever at C. So C stays alive. Also, the stone cannot cross over to B because of the obstacles and D will duck inside.

ROTTWEILER007:
I believe C will die first. If E pushes the stone down the slope, there’s a very high chance (about 99.9%) that the gap in the stone will align with D’s head as it rolls, causing it to land on the crowbar (acting as the effort). The sharp edges under the crowbar would then fatally injure C. However, if the gap doesn’t align, the stone would roll over D’s head, pushing D back into the hole.

How you take measure your 99.9%
If your assessment is right, then the stone stops on his head because it's a flat surface there and the gap has sharp and equal edges that will balance out. It stops there.

No one dies. If the effort of that E ball lands on the see saw the fulcrum of the see saw will not be equal or balance the see-saw from the image we were not given the weight of the objects. The load and resistance of that other ball looks very high. So no one dies

B dies first and probably the only one to die.
First, diving the ball into 4 quadrants, the sharp edge will be up at the base of the slope. It though will press D's head down but will not crush him.
Once it rolls on the cantilever, it will not have enough weight because close to a quarter of it is off hence, weights less than the the on the other side of the balance. Another reason why it won't through the second balnce off the balance is because, any object in motions losses its weight so it's lighter. It hence rolls over the wedge which gives it a lift and likely drops on the second ball. The second ball now rolls it into the pit where Mr. B is and that marks his end. May God save us from such.

Fiscus105:
You totally ignored question Op asked but ranting and lamenting, why won't you guys have toxic relationship?


Go back to OP, it depends on the weight and extent of force applied by E, if it stuck in first hole, only D would suffocate to death, but if force applied strong, it may bye pass D and move forward to land on "C", definitely C would die. Meanwhile having landed on plank above C, it would catapult the stone and shift it to any of the them, including the person that pushes the stone "E".

IN CONCLUSION, IT DEPENDS ON THE FORCE APPLIED BY E

C can't die. The other ball on the other end of the plank is heavier than the ball that is about to be rolled. So, even if the ball is rolled onto the plank with a heavy force, the plank will remain in it's position because of the heavier ball on the left side of the plank

The image presents a logic puzzle with a diagram showing a large stone balanced on a ledge. Below the stone, there are five individuals labeled: A, B, C, D, and E. The stone is positioned such that if pushed, it would fall directly onto person E. However, the puzzle asks: "Who Dies If 'E' Pushes the Stone?"

The key to solving this puzzle is to recognize that if person E pushes the stone, they would be applying force away from themselves. Since the stone is above E, pushing it would likely cause it to fall in the opposite direction. In the diagram, the stone is on a ledge with E below it. If E pushes the stone forward (away from themselves), it might not fall straight down but could be dislodged to fall sideways or backward.

Looking closely at the diagram, the stone is positioned on a pivot point. If E pushes the stone, it would cause the stone to tilt or roll away from E. Specifically, the stone might fall to the left (from E's perspective), which would be toward the side where persons A, B, C, and D are standing. Given the arrangement, the stone would likely fall onto person D first, as D is directly adjacent to the stone on the left side.

Therefore, if E pushes the stone, it will fall away from E and onto D, causing D to die.

Final Answer:
D dies if E pushes the stone.

It will kill no one, the calibration and the grove on the rolling Stone clearly indicates that D can't be crushed and due the to Grove on the rolling Stone it's weight is less than the Stone on the pivot hence C is also safe. Lastly, the rolling Stone cannot roll over the Stone on the pivot as its speed is impeded by the pivot, hence every one is safe.

Common Sense Should Tell You That ONLY "C" Would Die. "D" Might Sustain Injury However.

Well, Commons Sense Isn't Common Anyway

Nobody dies

E will just squart
D will not be hurt because the weight on the other side of the toothed see-saw is bigger than the one rolling down. Both will descend to the left hand side.

A and B stand no risk of dying at all.

ShoeMarket:
C is dead already.

Others might manage to escape but C has no route or way of escape.

If he sees the danger coming, he can adjust himself in such a way that he won't be affected.

NO ONE WILL DIE! CHECK THE CONSTRUCTION, THEY'RE ALL IN SAFE POSITIONS

All of dem

All of dem

All of dem

40 complete

1) Geometry & what matters

The stone is a circular disk with a sector removed (a crescent). That matters because:

Its center of mass (COM) is offset from the geometric center by a distance .

Its mass distribution is asymmetric, so its moment of inertia about the contact-rotation axis differs from a full disk.

As it rotates/rolls on an incline, gravity produces a torque that depends on the COM offset and the stone’s orientation.


We’ll use:

= mass of the stone

= nominal radius of the disk (distance from geometric center to rim)

= distance from geometric center to COM ( )

= incline angle (slope angle)

= static and kinetic friction coefficients at contact

= moment of inertia about the rotation axis through contact (depends on shape)




2) Push from E — initial conditions

When E pushes, two possibilities occur immediately depending on the push and friction:

A. Small push / strong tendency to lock — stone rocks, rotates a bit and then settles with the missing sector facing downhill so the COM is at a local minimum. It will stop.

B. Large push / high initial speed — stone overcomes any energy barrier and moves down the slope (either sliding, rolling, or a combination), possibly reaching the seesaw.

Which happens depends on whether gravity + push produce enough torque/energy to escape the nearest stable orientation (determined by COM) and on friction.



3) Forces, torques, and the condition to roll vs. tip or stop

Consider rotation about the instantaneous contact point on the incline. Gravity exerts a torque about that contact proportional to the horizontal/vertical lever arm of the COM. The gravitational driving torque (magnitude) that tends to rotate the stone downhill is roughly:


The stone will start/continue to roll if this torque can be converted into rotational acceleration overcoming static friction constraints and any geometric “potential energy hill” due to asymmetry.

For pure rolling (no slipping), the translational acceleration down the slope is (generalized):


will be higher or lower depending on how mass is missing; and

COM offset changes the way gravity drives rotation (not a symmetric disk).


Friction constraint: static friction must be large enough to produce the torque required for rolling without slipping. Max static friction force is . If the required friction exceeds this, the stone will slip rather than roll.



4) Energy view & stable orientations

Because the COM is offset, the stone has orientation-dependent potential energy. As it rocks/rolls, the COM height varies. There will be local minima where the COM is at the lowest height relative to the contact point — these are stable equilibria where the stone will tend to stop.

The energy condition to roll over a “bump” (escape a local minimum) is:


Because the crescent has a chunk missing, the (energy barrier) to flip it to a different orientation can be non-trivial. Practically, the stone will often lurch a bit then lock with the hollow facing downhill (COM lowest).



5) Two realistic outcomes — and their downstream effects

Outcome A (most likely for a modest push)

The asymmetry produces a stable orientation downhill; friction and the geometry dissipate energy.

The stone lurches and then stops on the slope (or at the step partway). Result: The stone does not reach the seesaw → no chain reaction → nobody dies.


Outcome B (stone reaches the seesaw)

This requires E to give a fairly large impulse (large initial kinetic energy) or a very steep slope and low friction so that the stone overcomes the orientation energy barrier and rolls rapidly down.

If the stone impacts the seesaw with enough linear momentum/kinetic energy, it will apply an impulse that can rotate the seesaw and launch the person C.

Whether C is killed depends on the velocity imparted, the lever geometry, and the distance to the spikes. If C is launched with enough vertical velocity to land on the spikes, then C would die first (as in the initial simplified chain). If the impact is weaker, C might be stunned or thrown harmlessly.




6) Collision / impact mechanics at the seesaw (if the stone reaches it)

Treat the seesaw as a lever pivoted at the fulcrum.

Bottom line: stone speed at impact is the key parameter. That speed is determined by whether the crescent can actually accelerate down the slope (see above).




7) Which is most likely in a real-world scenario?

Given:

The visible crescent shape (significant sector removed) => notable COM offset .

Typical surface friction and small pushes from a human.

The slope in the picture does not look extremely steep.


Conclusion: the crescent will most likely lurch and settle with its hollow side downhill and stop before the seesaw. So the chain reaction will not occur and nobody dies.

LET IT BE KNOWN IMMEDIATELY I SAW THAT CRESCENT I KNEW THAT IT WOULD KILL NO ONE, ROLLING THE BALL DOWN WONT BE EASY WITH THAT CURVE INSIDE

NO ONE DIES EXCEPT IN UNKNOWN CONDITIONS NOT SPECIFIED.

'E' pushes the stone, which is puched on a slope.

The stone will roll down the slope, hitting the cage where 'D' is trapped.

* The impact will either crush 'D' directly or push the cage and 'D' off the cliff, causing them to fall to their death.

The people at positions 'A', 'B', and 'C' are not in the direct path of the falling stone and are not in immediate danger from this specific action.

No one dies