meccanicamg
Guest
finally after 38 post we have the complete explanation of formulas and comments useful for the solution of the problem.
Marco F inox
Guest
gives
we have never considered friction, for obvious reasons of calculation; to more reason acceleration
can not be of 1,72 m/s^2, as for the formula (f = m*a) if we multiply it for the weight mass of 2000 n, we get a thrust force of 3440 n, while we have a thrust of 500 n.
that is the only force that creates acceleration.
I believe that the use of gravity acceleration 9.8 m/s^2, has led to error.
here we are on a horizontal plane and there is nothing to consider fallen from the sky.
Marco F inox
Guest
because there is impatience to get to the bottom, but also for a complete verification, from my calculations it turns out that the bear that pushes with 500 n, reaches the tipping speed of 0.8 m/sec.
corresponding to 2,88 km/h, in 3,2 sec, running a space of 1,28 meters.
to not turn over, it will be necessary to put the stone a little earlier.
DoctorBomber90
Guest
In reality it is not a mistake to use the acceleration of gravity g=9,81 m/s^2.
because the beginning of the post, @reggio has given the starting data but in a wrong way.
had put as values the Forces espresse in kg (sbagliato).
Unfortunately, at work (and not only), we tend to confuse the weight (which is a force, expressed in n) con la mass (espressa con kg).
often it is said "the piece weighs 50 kg" but, it should be said "the piece weighs 490,5 n (50*9,81).
Either way, if you say about work (or out of work) "the piece weighs 50 kg", it's fine the same. because it has become a common thing to say.
However, there is also to say that, nowadays, it is rather difficult to measure the strength of someone (or a man), then the use of the value expressed in kilograms is used.
I, for example, to evaluate my strength without using any tool, I use the handlebar 10 kg.
If I lift it---> ok, I take the 12 kg handlebar and I keep lifting it (I see that I can lift even the 12 kg handlebar? then I lift that from 14 kg and so on).
if I don't lift it---> I stop (not to hurt me).
so I can estimate the strength I can exercise (for example).
Either way, these are just my ideas.
then, returning to us.
- We can see, for example, the value of the thrust force fs if it is actually 50 kg or if it is 50 n, although I doubt it is expressed in n;
- the weight strength of the cart will be 200*9,81 =1962 n or it is said that the mass of the cart is 200 kg;
- In the rotation equation (around the rock) I multiplied the force of man's push for the acceleration of gravity precisely because the fs was expressed in kg (this could be a mistake). not by chance, at the first point, I said to investigate whether the force of the fs thrust is actually expressed in kg or n.
- always in the equation, the weight force, as it was described by @reggio, must obviously be multiplied by gravity acceleration. also because you are likely to sum/subtract numerical values with flat measuring units. from here comes the importance of dimensional analysis in equations.
DoctorBomber90
Guest
One more thing @marco ferraroni (not take it as a preacher but a correction)
attention to how you use that formula (f=m*a)
you took 2000 n and multiplied by the acceleration I estimated (1,72 m/s^2). from your account comes out 3440 n * m/s^2 ----> is completely wrong.
if I had the mass m expressed in kg and multiplied by acceleration m/s^2, then it was ok.
Anyway, I'm open to criticism and doubt.
Marco F inox
Guest
Agreed on many points, I rounded 9.81 with 10, I used the force in newton,
the weight of the cart in newton, the shift from to to b for the necessary work, of 820 instead of 821mm, no losses of friction, but the strength , the mass and acceleration, are constant and such remain, but they increase the speed.
the formula that binds them in a uniformly accelerated horizontal motion, can only be
f = m*a, where f is the manual push.
I could have brought the strength of 81 kg (810n) of balance, to the level of the center, but this force would have increased with a decrease of the race, so the moment would have been the same.
I have spoken ironically of a little man, because actually finding a person capable of pushing 50 kg horizontally with constancy, it will be very difficult, but these were the data of
problem and they are inconsistible.
evaluating my results, including space and time spent, which I have transmitted lately, I think they can be quite realistic. Hello, everyone.
ceschi1959
Guest
reviewing the results that I have previously exposed this solution.
starting from the balance of the moments I find a resulting moment of 560 nm (approximating g=10 m/s^2).
I can assume this moment as that generated by a force applied to the point of application of fp pairs to 767 n (560nm/.73m) that is as if there were a mass of 76.7 kg applied at that point, the rod to which the force is applied is 1.76 m long for which the work to be carried out to vertically the point of application is pairs to 767n*(1.76-1.6)m i. 122 j'
starting from the balance of the moments I find a resulting moment of 560 nm (approximating g=10 m/s^2).
I can assume this moment as that generated by a force applied to the point of application of fp pairs to 767 n (560nm/.73m) that is as if there were a mass of 76.7 kg applied at that point, the rod to which the force is applied is 1.76 m long for which the work to be carried out to vertically the point of application is pairs to 767n*(1.76-1.6)m i. 122 j'
mamobono
Guest
Good morning, I follow this exercise with interest from the beginning.
I agree with those who face it using the energy budget (without considering friction).
but I have a doubt, the development by equating the final potential energy with the initial kinetic energy, in my opinion, would answer the question:
I have a cart on a tilted rail, at what speed it has to go instantly 0 altitude h0 because it stops instantly 1 at altitude h1.
In order to respond to this: I have a cart that is pushed horizontally by an omino weighing 50 kg, this encounters an obstacle that blocks the cart, at what maximum speed can go the cart and the omino because the same does not rebound to the impact?
I thought so.
always using energy conservation (without considering friction) the cart instantly 0 has velocita0 (direct horizontally) that is the unknown, mass constituted by the sum of that of the cart + that of the omino, and therefore would have the resulting kinetic energy.
at the moment of impact the kinetic energy is transformed into part rotational inertia energy of the mass of the cart + omino and in part is absorbed by the cart (deformation / heat).
rotational speed w will be the ratio between the normal vtg component to the rotation arm of the speed0 and the rotation arm.
this because I believe that a part of kinetic energy is absorbed by the cart / obstacle system and only a part remains for the rotation of the cart.
at this point having the moment of inertia of the cart/omino masses reported to the point of rotation (huygens-steiner theorem) I would have the rotational inertia energy of the cart.
this will generate a rotation of the cart bringing the baricentro from altitude h0 to altitude h1 (which corresponds to the vertical rotation arm which is the maximum non-turning condition.
this requires a potential energy mg(h1-h0).
By matching the energy back and knowing the moments of inertia and the geometry of the impact point compared to the cart we would first find the value of rotational speed w.
then the value of the tangential component vtg = w x arm .
then knowing the value of the alpha inclination of the rotation arm compared to the horizontal the value of the horizontal speed of the cart vo = vtg / sen(alpha).
This explanation seems to me sensible because in this way the higher the obstacle the more the energy for the rotation will be lower and therefore need a greater starting speed.
the speed is also linked to the position of the obstacle regarding the center of the cart this thanks to the moments of inertia referred to the point of impact.
If the impact point was at the same level as the centre of gravity, in theory, all energy should be dissipated from the cart by deforming.
The reasoning I have made is based on certain conditions:
-the omino push the cart from speed 0 to the impact speed and then it becomes also a part of the cart system that tilts at the moment of impact.
-you must know the geometry of the cart and the position of the omino in order to calculate the moments of inertia.
- deformation energy could in turn be evaluated/stimata as a difference between initial kinetic energy and dissipated energy for inertial rotation and therefore know if the cart structure resists shock
I would like to know the vs opinion .
greetings
I agree with those who face it using the energy budget (without considering friction).
but I have a doubt, the development by equating the final potential energy with the initial kinetic energy, in my opinion, would answer the question:
I have a cart on a tilted rail, at what speed it has to go instantly 0 altitude h0 because it stops instantly 1 at altitude h1.
In order to respond to this: I have a cart that is pushed horizontally by an omino weighing 50 kg, this encounters an obstacle that blocks the cart, at what maximum speed can go the cart and the omino because the same does not rebound to the impact?
I thought so.
always using energy conservation (without considering friction) the cart instantly 0 has velocita0 (direct horizontally) that is the unknown, mass constituted by the sum of that of the cart + that of the omino, and therefore would have the resulting kinetic energy.
at the moment of impact the kinetic energy is transformed into part rotational inertia energy of the mass of the cart + omino and in part is absorbed by the cart (deformation / heat).
rotational speed w will be the ratio between the normal vtg component to the rotation arm of the speed0 and the rotation arm.
this because I believe that a part of kinetic energy is absorbed by the cart / obstacle system and only a part remains for the rotation of the cart.
at this point having the moment of inertia of the cart/omino masses reported to the point of rotation (huygens-steiner theorem) I would have the rotational inertia energy of the cart.
this will generate a rotation of the cart bringing the baricentro from altitude h0 to altitude h1 (which corresponds to the vertical rotation arm which is the maximum non-turning condition.
this requires a potential energy mg(h1-h0).
By matching the energy back and knowing the moments of inertia and the geometry of the impact point compared to the cart we would first find the value of rotational speed w.
then the value of the tangential component vtg = w x arm .
then knowing the value of the alpha inclination of the rotation arm compared to the horizontal the value of the horizontal speed of the cart vo = vtg / sen(alpha).
This explanation seems to me sensible because in this way the higher the obstacle the more the energy for the rotation will be lower and therefore need a greater starting speed.
the speed is also linked to the position of the obstacle regarding the center of the cart this thanks to the moments of inertia referred to the point of impact.
If the impact point was at the same level as the centre of gravity, in theory, all energy should be dissipated from the cart by deforming.
The reasoning I have made is based on certain conditions:
-the omino push the cart from speed 0 to the impact speed and then it becomes also a part of the cart system that tilts at the moment of impact.
-you must know the geometry of the cart and the position of the omino in order to calculate the moments of inertia.
- deformation energy could in turn be evaluated/stimata as a difference between initial kinetic energy and dissipated energy for inertial rotation and therefore know if the cart structure resists shock
I would like to know the vs opinion .
greetings
cacciatorino
Guest
In theory, it is enough to study the limit condition that is the one in which the cart is in balance with the center of gravity on the vertical point of support, with the components of linear kinetic energy and rotation zeroed and completely transformed into potential energy (i.e. the lifting of the baricentre). under these assumptions you can avoid the effort to calculate the energy absorbed by the rotation of the cart.
ceschi1959
Guest
I'm sorry. It is misleading to consider that there is an omino that pushes as it is misleading to consider fs as a force of thrust. always according to me, if you schematize the problem without thinking about trolleys, omini and so on, in this way and asks what is the work necessary to turn the system as schematizzo below, you get to the conclusions and results I have exposed.
cacciatorino
Guest
In fact, I think we should consider (unless we are faced with a deliberate sabotage) that the operator stops pushing once he realizes that he has hit a hindrance.
mamobono
Guest
thanks for the answers
I try to explain myself, the omino only needs to increase the speed from 0 to that of the impact point.
to the impact the omino does not push, does not want to sabotage, but its mass will go to load the cart seen the surprise stumbling. (sing)
Then in the "classical" method it is not taken into account that part of the kinetic energy of departure is lost due to deformation of the cart/wheel/ obstacle, therefore the limit speed, to the eye, should be greater than that calculated precisely because part of its energy is dissipated by the shock.
shock that could then be seen as an energy given by the stiffness of the cart x its shift (like spring that compresses but does not release turning its energy into heat for example)
Obviously, mine is a supposition and I gladly accept the constructive comments of the forum.
Hi.
I try to explain myself, the omino only needs to increase the speed from 0 to that of the impact point.
to the impact the omino does not push, does not want to sabotage, but its mass will go to load the cart seen the surprise stumbling. (sing)
Then in the "classical" method it is not taken into account that part of the kinetic energy of departure is lost due to deformation of the cart/wheel/ obstacle, therefore the limit speed, to the eye, should be greater than that calculated precisely because part of its energy is dissipated by the shock.
shock that could then be seen as an energy given by the stiffness of the cart x its shift (like spring that compresses but does not release turning its energy into heat for example)
Obviously, mine is a supposition and I gladly accept the constructive comments of the forum.
Hi.
Marco F inox
Guest
I believe that the data of the problem, clear and complete, cannot be changed to your liking.
not wanting to consider this and that, how can you calculate a speed?
the rock then, is a hindrance that stops the advancement of the cart, it is useless to turn around assuming that the wheel is crushed, the rock flies away, you need a higher wall, you have to calculate only the tipping forces etc. so you do not go anywhere.
on the rotation energy due to the effect of the tipping, I would like to recall that the equality of the two moments of balance, were shared by everyone as calculation of a horizontal or vertical force, multiplied by his arm.
only I wanted to show that even with the moments of rotation, the result does not change. .
Which is why it is useless to pull out the rotation energy, much more than the thrust is horizontal, as horizontal is the force of inertia that develops.
in rotation you have a greater path, but a lesser force, so the result does not change.
not wanting to consider this and that, how can you calculate a speed?
the rock then, is a hindrance that stops the advancement of the cart, it is useless to turn around assuming that the wheel is crushed, the rock flies away, you need a higher wall, you have to calculate only the tipping forces etc. so you do not go anywhere.
on the rotation energy due to the effect of the tipping, I would like to recall that the equality of the two moments of balance, were shared by everyone as calculation of a horizontal or vertical force, multiplied by his arm.
only I wanted to show that even with the moments of rotation, the result does not change. .
Which is why it is useless to pull out the rotation energy, much more than the thrust is horizontal, as horizontal is the force of inertia that develops.
in rotation you have a greater path, but a lesser force, so the result does not change.
mamobono
Guest
Thanks for the answer, but I see the theoretical point of view if there is no data is that I can solve the problem regardless.
If the stone flies away then we must say that the cart stops to have the balance and be able to calculate the minimum speed, and it is as if it went against a wall .
I understand your reasoning as a balance of moments and I understand that there must also be balance of energies at stake but I do not understand the results you get
post 43 you find the tipping speed of 0.8 m/sec
post 47 baskets1959 v=(122*2/76.7)^0.5=1.78 m/s
That's what doesn't make me understand, if you please clarify, thank you.
If the stone flies away then we must say that the cart stops to have the balance and be able to calculate the minimum speed, and it is as if it went against a wall .
I understand your reasoning as a balance of moments and I understand that there must also be balance of energies at stake but I do not understand the results you get
post 43 you find the tipping speed of 0.8 m/sec
post 47 baskets1959 v=(122*2/76.7)^0.5=1.78 m/s
That's what doesn't make me understand, if you please clarify, thank you.
cacciatorino
Guest
Then, when the cart is on the road, its center is 1600 mm from the ground. when he stands there to turn forward his center of gravity is located in radq (1600^2 + 730^2) = 1760 mm (approximately) from the ground.
from that the work to succeed is equal to 0.16mt*200kg*9,81m/sec^2 = 315 joule (approximately)
this work comes from the inertia of the cart accumulated in its travel at constant speed (always admitting that the operator is not a saboteur):
ec = 1/2 * m * v^2 --> v = radq (2*ec/m) = 1,8 mt/sec (approximately).
from that the work to succeed is equal to 0.16mt*200kg*9,81m/sec^2 = 315 joule (approximately)
this work comes from the inertia of the cart accumulated in its travel at constant speed (always admitting that the operator is not a saboteur):
ec = 1/2 * m * v^2 --> v = radq (2*ec/m) = 1,8 mt/sec (approximately).
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Marco F inox
Guest
if you mix the results of one, with the results of the other, you can not understand anything, but support
that lack data or even that are wrong, is misleading, because not true.
until the calculation of the speed, which was the request, I clearly exposed my calculations.
the time spent and the space traveled, I calculated them, exposing only their value,
to give and have an element of comparison.
when the applied force is constant, you do not have a uniform straight motion, but a motion evenly accelerated. and that is why it increases speed and there is acceleration.
if the bike was uniform, (but you do not know at which speed), the cart could never turn over.
as the initial thrust would only serve to win friction resistances, then it would decrease
to keep it only in motion.
cacciatorino
Guest
I imagine the truck driver pushing the whole plant, and after a hundred meters he travels faster than an Olympic centimeter!
Mattymecc
Guest
I don't understand reasoning in energy terms. I would go to see what is the maximum horizontal force that the operator can apply without the cart bouncing.
the horizontal force should be assessed by two contributions:
1) acceleration
2) force necessary to overcome the obstacle
foriz=ma + fostacolo
the horizontal force should be assessed by two contributions:
1) acceleration
2) force necessary to overcome the obstacle
foriz=ma + fostacolo
ceschi1959
Guest
the work shifts the strength of dz, i.e. of (1,76-1.6) m = 0.16 m the apparent mass, considering the force of 500 n that continues to act, is of 76,7 kg from which 76,7 kg * 10 m/sec^2 * 0.16 m=122 j from which the speed I calculated.
in case the so-called push force no longer act when the system meets the rock, the work would be 200 kg * 10 m/sec^2 * 0.16 m= 320 j with a tipping speed still of 1,78 m/s. no one has noticed that the mass is influential because it appears on both sides of the equation. . mgh=1/2mv^2 and so far neither I!!!! !РУ£££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££££
your calculations @cacciatorino , while considering how my work equality is wrong because the dl is not 1,76 m but 0,76 m then 1,76*200*9.81=(they would be) 3453 j
writing this answer I realized that That post before is wrong. and then I repeat the result of my first post #25. In fact, we suppose that the thrust force balances exactly the weight force, then it would suffice a speed just over 0 m/s to turn it over.
cacciatorino
Guest
I was wrong to write then I corrected, the difference in height is 0.16 mm.