Thanksgiving physics problem

Matthew

Matthew

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Matthew
Assume a 5 gallon Homer bucket full of water (in reality, a turkey being brined).

Assume it is in the trunk of a car and the friction between the carpet and the bottom of the bucket is such that the bucket won't slide.

What force would be required to tip the bucket over?

Asking for a friend.
 
The guy ahead of you at the round-about can make it tip over.
 
When the vector sum of gravity and lateral acceleration of the combined mass of bucket + brine + turkey falls outside the base of the bucket, it will tip over.
 
Matthew said:
Assume a 5 gallon Homer bucket full of water (in reality, a turkey being brined).

Assume it is in the trunk of a car and the friction between the carpet and the bottom of the bucket is such that the bucket won't slide.

What force would be required to tip the bucket over?

Asking for a friend.
Click to expand...

It's not super complicated, but is easier to figure with actual numbers.
 
Dana said:
When the vector sum of gravity and lateral acceleration of the combined mass of bucket + brine + turkey falls outside the base of the bucket, it will tip over.
Click to expand...
That's what "my friend" and I were figuring. But putting the actual formula together was more than my turkey addled brain can handle on a Monday. The CG of the bucket needs to move along an arc that has an upward motion, so this might be a little more complicated than I thought...or maybe not.
 
Matthew said:
or maybe not.
Click to expand...
If it doesn't slip, it will tip at the point where effective location of the force holding the bucket up has to be at the edge of the bucket to keep the sum of the moments equal zero. Simple weight and balance problem.
 
This is from the "static equilibrium" chapter of an intro physics book.

Let's say the bucket is a cylinder with radius "R" and height "H" (so the center of mass is at "H/2"). Let's also say there's a rope (or something to provide the a sideways force) connected to the center of mass (this is what you're envisioning, right?). Let's also say that the rope is pulling "just barely" hard enough to start to tip the bucket onto its edge, with mystery force "F". The free-body diagram has four forces: 1) gravity "mg" down from the CM, 2) the rope force "F" from the CM, 3) the normal force up from the table, which will be exerted at where the edge makes contact, and 4) the force of static friction preventing it from sliding, which pushes sideways at the point where the edge makes contact.

If you choose "the point where the edge makes contact" as the pivot point for computing the torques in this problem, then forces (3) and (4) exert no torque, and it's just a matter of balancing the torques from the two other forces. The torque due to gravity is "mg*R", and the torque due to the rope force is "F*H/2". Balancing 'em: F = 2mgR/H. Voila.

If the rope is connected somewhere else (like if you grab it from the top rim and pull), use its height off the floor (H, instead of H/2) as the lever arm instead.

Now, that's if it's just barely lifting. Once it starts to tilt, the lever arm for gravity will go down and the lever arm for the rope force will go up, so it'll only move *easier* if the horizontal force is held constant. Getting that first lift is the hardest part requiring the most force. Once the CM moves to the other side of the pivot point, over it goes.

Now, if the bucket is in the trunk of a car and there's no rope, but you're worried about the thing tipping over because of outrageous driving, then the relevant question is "what acceleration of the car will tip if over?" Accelerations and forces are related through F=ma; the easiest way would be to frame the problem in the "trunk frame of reference" as before, and say that the accelerating reference frame causes the bucket to experience a "fictitious force" on its CM (backward) equal to the force of static friction at the pivot point holding it accelerating together with the car (forward). That force of static friction is "ma", so:
mg*R = (ma)*(H/2)
a = 2gR/H

Edit: this all assumes that static friction is sufficient to prevent slipping. (Maximum mu_s*N, like in the video). And that the bucket is a rigid mass (which water isn't).
 
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As a general rule of thumb, we usually assume that a tank that has a a height at or greater than its diameter will have a tipping moment. It's not hard and fast, but it beats the hell out of a=2gR/Hing yourself to death.
 
The lateral g force that a Honda Civic with average tires can handle before it skids would have to be > than the bucket tipping force, too, wouldn't it? But body lean would also move the CG of the bucket. That will be an exercise left to the reader.
 
TL;DR, just don't drive so damned fast when you have a bucket of saltwater in the trunk. QED.
 
kath said:
This is from the "static equilibrium" chapter of an intro physics book.

Let's say the bucket is a cylinder with radius "R" and height "H" (so the center of mass is at "H/2"). Let's also say there's a rope (or something to provide the a sideways force) connected to the center of mass (this is what you're envisioning, right?). Let's also say that the rope is pulling "just barely" hard enough to start to tip the bucket onto its edge, with mystery force "F". The free-body diagram has four forces: 1) gravity "mg" down from the CM, 2) the rope force "F" from the CM, 3) the normal force up from the table, which will be exerted at where the edge makes contact, and 4) the force of static friction preventing it from sliding, which pushes sideways at the point where the edge makes contact.

If you choose "the point where the edge makes contact" as the pivot point for computing the torques in this problem, then forces (3) and (4) exert no torque, and it's just a matter of balancing the torques from the two other forces. The torque due to gravity is "mg*R", and the torque due to the rope force is "F*H/2". Balancing 'em: F = 2mgR/H. Voila.

If the rope is connected somewhere else (like if you grab it from the top rim and pull), use its height off the floor (H, instead of H/2) as the lever arm instead.

Now, that's if it's just barely lifting. Once it starts to tilt, the lever arm for gravity will go down and the lever arm for the rope force will go up, so it'll only move *easier* if the horizontal force is held constant. Getting that first lift is the hardest part requiring the most force. Once the CM moves to the other side of the pivot point, over it goes.

Now, if the bucket is in the trunk of a car and there's no rope, but you're worried about the thing tipping over because of outrageous driving, then the relevant question is "what acceleration of the car will tip if over?" Accelerations and forces are related through F=ma; the easiest way would be to frame the problem in the "trunk frame of reference" as before, and say that the accelerating reference frame causes the bucket to experience a "fictitious force" on its CM (backward) equal to the force of static friction at the pivot point holding it accelerating together with the car (forward). That force of static friction is "ma", so:
mg*R = (ma)*(H/2)
a = 2gR/H

Edit: this all assumes that static friction is sufficient to prevent slipping. (Maximum mu_s*N, like in the video)
Click to expand...

My "friend" will look this over, but this does seem to be the way to solve this puzzle.
 
Also... which direction are you traveling?
 
EdFred said:
It's not super complicated, but is easier to figure with actual numbers.
Click to expand...



Well let's see...

landscape-1476389673-gettyimages-168351286.jpg


What was the question again..??
 
Matthew said:
Assume a 5 gallon Homer bucket full of water (in reality, a turkey being brined).

Assume it is in the trunk of a car and the friction between the carpet and the bottom of the bucket is such that the bucket won't slide.

What force would be required to tip the bucket over?

Asking for a friend.
Click to expand...

Instead of theoretical calculation, I can report an actual experiment.

I had a full 5 gal bucket of water (with lid) in my car trunk, when I drove slowly just 3 miles, from home to hangar. (I intended to wash bugs off the plane, after my next flight.) Well, it tipped over. From just going around a corner at normal city driving speeds.

And what’s worse, the car battery and some related electronics are located under the trunk floor. (German engineering gone amuck.) Lots of electrical gremlins after that. After much effort to dry it out, the car eventually worked okay, once again.

That experiment was done with clean water from the tap. I’m not about to do another run of the same experiment, especially with brine.

If you’re tempted to try anyway, I can specifically recommend anything other than a BMW 5 series.
 
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The amount of force required to start the tipping is easy to calculate and will vary based on the moment arm. A complication is added in that a round bucket will have a tendency to rotate once it begins to tip if the force is applied as a point load. The maximum amount of force will be at the very beginning. The required force will decrease as water pours from the tipping bucket.
 
alfadog said:
The amount of force required to start the tipping is easy to calculate and will vary based on the moment arm. A complication is added in that a round bucket will have a tendency to rotate once it begins to tip if the force is applied as a point load. The maximum amount of force will be at the very beginning. The required force will decrease as water pours from the tipping bucket.
Click to expand...

The slosh factor can muck things up, but it's more an issue with a closed tank than an open one.
 
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