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Fun physics question

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TheSiege said:
All I see is a bunch of people who dont know how to answer the question and instead of admitting that, they just want to throw ill conceived jabs at the OP
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The question has already been answered:

GagHalfrunt said:
Impossible to determine with information provided. Fun not found. I demand a refund.
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GagHalfrunt said:
Coefficient of friction of object vs ramp. Or rolling resistance of tires if said object is on wheels.
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benzylic said:
http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Dynamics/InclinePlanePhys.html

Edit: The length of the ramp doesnt matter, and I hope it should be obvious that a steeper ramp will require more force.
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It's impossible to determine with the information provided. Your best bet is to go out with a fish scale and a measuring tape to actually measure the forces involved.

And I'm assuming you're using machines for this task, so why do you care how much work they have to do?
 
Buy a bigger motor. Bigger motors are always the answer. Overspec the motor against the day we have a monorail to space. The motor you buy for your server ramp should be able to pull a train into orbit. Anything less would be unmanly.
 
the smaller ramp would be used to move the populated ramp onto a specialized pallet and then lifted to the height of the location it needs to be in. I am suggesting getting rid of the pallet and just buy a longer ramp straight to the rack location. And either way, people, not machines would be moving the rack. So I want to know what, difference in force it would take to use the longer, taller ramp instead of the shorter ramp. The data floor it would be used in would be a mobile data center, HP and Dell make them and they are data floors in a GIANT connex for lack of better words. Its not homework.
 
You don't need the friction coeffecient per say, I would think napkin math is good enough and just assume the same coefficient.

edit: can't you just compare the difference in angles since it's a straight multiplier? So about 17.5% more force.
 
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If OP is not trying to get physics homework done, I see OP crushed under a toppled rack in the near future.

We warned you.
 
TheSiege said:
the smaller ramp would be used to move the populated ramp onto a specialized pallet and then lifted to the height of the location it needs to be in. I am suggesting getting rid of the pallet and just buy a longer ramp straight to the rack location. And either way, people, not machines would be moving the rack. So I want to know what, difference in force it would take to use the longer, taller ramp instead of the shorter ramp. The data floor it would be used in would be a mobile data center, HP and Dell make them and they are data floors in a GIANT connex for lack of better words. Its not homework.
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Call a warehousing supply company. Someone has to have already come up with a better solution than tipping carts loaded with expensive electronics onto ramps.
 
Also, a 44U rack is not going to weigh 3500 lbs, even fully populated.

The most dense (weight wise) equipment I can think of off the top of my head would be pushing 2300-2400 lbs, and the typical weight is going to be more like 1700-1800 lbs.
 
1zp7dlf.jpg

(not drawn to scale)

x1 = 12
y1 = 156

x2 = 8
y2 = 122

W = 3500

theta1 = 4.399 degrees
theta2 = 3.752 degrees

To push the object up the ramp you need to overcome the weight parallel to the ramp surface (p1 and p2). Since the second ramp has a shallower angle the pusher will experience a lighter equilibrium force, but will encounter a greater resistance thanks to the greater force normal to the surface (n2). A user of the first ramp will have to exert a greater equilibrium force and will have to apply that force over a greater distance, but will experience less resistance as the trade off.

sin(90 - theta1) = n1/W => n1 = 3489.689
cos(90 - theta1) = p1/W => p1 = 268.56

sin(90-theta2) = n2/W => n2 = 3492.498
cos(90-theta2) = p2/W => p2 = 229.033

friction1 = 3489.689*mu
friction2 = 3492.498*mu

Minimum Push force 1 (F1) = 268.56 + friction1
Minimum Push force 2 (F2) = 229.033 + friction2

F1 - F2 = 39.527 + mu(-2.809) = 0 when mu = 14.072

So if mu > 14.072 then ramp 1 will require a lower minimum force.
If mu < 14.072 then ramp 2 will require a lower minimum force.

Since that's a retardedly high value for mu (< 1 for non-gripping substances), it's pretty safe to say ramp 2 will require less muscle to use at any given instant in time; which isn't a surprise since it's not as steep, not as high, and not as long.


Disclaimer: this is my first use of free body diagrams in seven years.
 
Jaepheth said:
1zp7dlf.jpg

(not drawn to scale)

x1 = 12
y1 = 156

x2 = 8
y2 = 122

W = 3500

theta1 = 4.399 degrees
theta2 = 3.752 degrees

To push the object up the ramp you need to overcome the weight parallel to the ramp surface (p1 and p2). Since the second ramp has a shallower angle the pusher will experience a lighter equilibrium force, but will encounter a greater resistance thanks to the greater force normal to the surface (n2). A user of the first ramp will have to exert a greater equilibrium force and will have to apply that force over a greater distance, but will experience less resistance as the trade off.

sin(90 - theta1) = n1/W => n1 = 3489.689
cos(90 - theta1) = p1/W => p1 = 268.56

sin(90-theta2) = n2/W => n2 = 3492.498
cos(90-theta2) = p2/W => p2 = 229.033

friction1 = 3489.689*mu
friction2 = 3492.498*mu

Minimum Push force 1 (F1) = 268.56 + friction1
Minimum Push force 2 (F2) = 229.033 + friction2

F1 - F2 = 39.527 + mu(-2.809) = 0 when mu = 14.072

So if mu > 14.072 then ramp 1 will require a lower minimum force.
If mu < 14.072 then ramp 2 will require a lower minimum force.

Since that's a retardedly high value for mu (< 1 for non-gripping substances), it's pretty safe to say ramp 2 will require less muscle to use at any given instant in time; which isn't a surprise since it's not as steep, not as high, and not as long.


Disclaimer: this is my first use of free body diagrams in seven years.
Click to expand...

So is the force rated in newtons?
 
I used the values you provided.

Since the same mass is on both ramps the weight, and what units it's measured in, becomes irrelevant. The result is more a function of the ratios of the angles involved.

If you want an exact answer as to just how much harder it is to use ramp 1, then you need to go measure the coefficient of friction using a pull scale, a fully loaded rack, and a level surface.


EDIT:
Although with a mu <= 1 you're looking at less than an additional 40 pounds of force to use the larger ramp.

If mu = .3 (approx. worst case from here) then you're looking at minimum push weights of 1316 lbs and 1277 lbs respectively
 
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Jaepheth said:
I used the values you provided.

Since the same mass is on both ramps the weight, and what units it's measured in, becomes irrelevant. The result is more a function of the ratios of the angles involved.

If you want an exact answer as to just how much harder it is to use ramp 1, then you need to go measure the coefficient of friction using a pull scale, a fully loaded rack, and a level surface.
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Thank you for your help
 
TheSiege said:
Thank you for your help
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Note what he did for you. He presented useful information and methodology and nicely done. What he could not do is provide a precise answer. That however doesn't really matter.

If you look at the problem you see a quite small difference in angles and the difference in his calculations reflects this. In other words don't toss Newtons. Just give them a base figure and percentage differences for various angle options.
 
Jaepheth said:
I used the values you provided.

Since the same mass is on both ramps the weight, and what units it's measured in, becomes irrelevant. The result is more a function of the ratios of the angles involved.

If you want an exact answer as to just how much harder it is to use ramp 1, then you need to go measure the coefficient of friction using a pull scale, a fully loaded rack, and a level surface.


EDIT:
Although with a mu <= 1 you're looking at less than an additional 40 pounds of force to use the larger ramp.

If mu = .3 (approx. worst case from here) then you're looking at minimum push weights of 1316 lbs and 1277 lbs respectively
Click to expand...
OK that's what I thought. He's vastly over complicating things. Weight is the same, friction is the same. All that matters is angle for force required to move the object.

Total force output, which wasn't asked but is relevant if the task is repeated frequently, also changes with angle but now distance is important.
 
Its a first approximation type problem.

So you need to find the force required to move the block up the ramp with only gravity acting as your resistance, no friction. Then find the distance needed to apply that force. Solve for the hypotenuse distance I guess.

The difference in potential energy is going to be mgh no matter what. I guess they want to see if you understand the effect of ramps on the normal force.
 
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