at what speed do you turn your cart? :)

[reggio]

Hello, I ask for help for a resumption,

Knowing:
weight strength on baricentro fp=200kg
side push force fs=50kgwhat formula allows me to obtain at what speed (direction fs) the cart will bounce bumping the rock "s"?

 

[biz]

just find the balance to the rotation to get fs.
then, find the acceleration (in your deceleration case).
does not overturn with speed but with acceleration/deceleration.

 

[reggio]

just find the balance to the rotation to get fs.
then, find the acceleration (in your deceleration case).
does not overturn with speed but with acceleration/deceleration.

ciao News Thank you.

"rotation balance"? You mean, zero the two pairs fp*730 = fs*1800 or how?

I understand that it will be the difference of speed (deceleration) to turn, but it is how I find it and with what the comparison that escapes me as a concept and as formulas :

Would you be able to complete and simplify the explanation with more details and the "exercise"?
If you can, explain how if you were a 6-year-old boy;-p

 

[Marco F inox]

the design is beautiful but incomplete.
As it is, it cannot be resolved, because the thrust point is missing.
keep in mind that everything revolves around the rock, if we cross the wheel.
around the rock therefore converge both the moment of thrust and resistance.

♪

 

[reggio]

The design is beautiful

ciao @marco ferraroni who is a liar;

As it is, it cannot be resolved, because the thrust point is missing.
keep in mind that everything revolves around the rock, if we cross the wheel.
around the rock therefore converge both the moment of thrust and resistance.

...uhmmm maybe I didn't understand ... I added 1225 but I don't think it's useful
tell me if I'm wrong

I wait for the skewer, in addition to the formula, I really need resumption

 

[biz]

you have to make the balance to the point s, that is:
fp730-fs1800=0
fp is a given, the only unknown is fs.
fs is the inertia force of your body. the force of inertia by definition is:♪
with m mass and acceleration.
the body when decelerated tends to assume the previous configuration (of the previous moment), so on the body acts a force that pushes it in the opposite direction to that of aceleration (negative). the force of inertia has always opposed to aceleration, both negative and positive.then:
fp730-ma1800=0;
a=(fp730)/(m1800)

generally these problems are solved with the principle of d'alembert (study of the dynamics of a material point)
This principle states:dc+fr+fi=0
dc= forza motorfr=resistant strengthfi=force of inertia

for rotating material points:mm+mr+mi=0
con mi=jε
j=moment of mass inertiae= angular acceleration
But, I don't want to dive any further.

 

[cacciatorino]

There is also to consider that if the rock is small and/or the wheel is large it will easily overrun.

 

[meccanicamg]

you have to make the balance to the point s, that is:
fp730-fs1800=0
fp is a given, the only unknown is fs.
fs is the inertia force of your body. the force of inertia by definition is:♪
with m mass and acceleration.
the body when decelerated tends to assume the previous configuration (of the previous moment), so on the body acts a force that pushes it in the opposite direction to that of aceleration (negative). the force of inertia has always opposed to aceleration, both negative and positive.then:
fp730-ma1800=0;
a=(fp730)/(m1800)

generally these problems are solved with the principle of d'alembert (study of the dynamics of a material point)
This principle states:dc+fr+fi=0
dc= forza motorfr=resistant strengthfi=force of inertia

for rotating material points:mm+mr+mi=0
con mi=jε
j=moment of mass inertiae= angular acceleration
But, I don't want to dive any further.

is correct the rotational balance system around the rock.
certainly as hunters say or, it could happen that the wheel jumps off the rock. but this if the deformation of the wheel is sufficiently small and therefore the ground reaction multiplied its deformation center arm is smaller than the torque supplied to the wheel a little from the engine (if present) and partly from the inertias of the system.

 

[reggio]

there is also to consider that if the rock is small and/or the wheel is large it will easily overrun.

hi, let's pass: we don't know if pebble or bone - but if it goes well, I just want to avoid tipping by limiting speed (decelection) critical.

Let's go ahead, let's try to do it, because I'm getting the doubt that I have misunderstood my bran... or that I haven't understood anything yet.

weight force on baricentro fp=200kg => +/-2000n
side push force fs=50kg => +/-500n
fi (force of inertia)=-m(mass)a(acceleration)quindium:
(2000n*0,730mt)-(2000n*a*1,8mt)=0;
a=(2000n*0,730mt)/(2000n*1,8mt) = 0.73mt/1,8mt = 0,405mt ...

sbaglio? sbaglio? e ora?

 

[meccanicamg]

hi, let's pass: we don't know if pebble or bone - but if it goes well, I just want to avoid tipping by limiting speed (decelection) critical.

Let's go ahead, let's try to do it, because I'm getting the doubt that I have misunderstood my bran... or that I haven't understood anything yet.

weight force on baricentro fp=200kg => +/-2000n
side push force fs=50kg => +/-500n
fi (force of inertia)=-m(mass)a(acceleration)quindium:
(2000n*0,730mt)-(2000n*a*1,8mt)=0;
a=(2000n*0,730mt)/(2000n*1,8mt) = 0.73mt/1,8mt = 0,405mt ...

sbaglio? sbaglio? e ora?

lengths in m and not in mt.
accelerations in m/s not in mt.
from the equation of balance determines what is missing.
but if you already have fs.... we don't get anything. Why? because there is no force of inertia to add to the scheme.
But is fs a not rigid thrust? give us more details?

 

[Marco F inox]

If you put a rock or a mutt, it must be the same thing, because the problem says that the cart must be overturned. in both cases the true rotation, would be around the pivot of the wheel, but for simplicity we also consider a point, that is the rock.
I remember then that in physics, a moment is the product of a force for an arm, but both must make a 90-degree angle.
the decomposition of forces, it serves precisely to deal with these problems.
the 1225 is just the quota that was missing and that I asked.
compliments for precision.

 

[meccanicamg]

if you mean positive moments in the anticlockwise direction....the formula has the wrong signs.

 

[reggio]

again

m=meters all right?
force/weight in newton okay?
acceleration in meters per second, all right?

you have to make the balance to the point s, that is:
fp730-fs1800=0
fp is a data (2000newton), the only unknown is fs.fs is the force of inertia of your body.
the force of inertia by definition is:be me with m mass and acceleration.

in case fi=-m*a becomes fs=-m*a
Is all this above correct?

quindium:
fp730-ma1800=0;
a=(fp730)/(m1800)

So does this on it?

if everything is okay we continue
(fp*0.73metri) - (fs*1.8metri) = 0
(2000newton*0.73metri) - (fs*1.8metri) = must be balanced = 0
♪
= 0
the mass m worth how much the weight force fp?
(2000newton*0.73metres) - (2000newton*a*1.8meters) = 0

a (accelerated)= (2000newton*0.73metri) / (2000newton*1.8metri)
But now, as before, I from newton and meters can not get meters per second (as they ask for acceleration or deceleration) ..

and then, after I have my deceleration, how do I get to the maximum speed of the cart, so that it does not turn it over?

I am increasingly lost, clarified, please...
@marco ferraroni I added 1225mm, but I can't follow you, I don't think you need to, can you give me the solution and explain it to me?
Now, in the next post I try to reformulate and simplify to see if I can help you help me

 

[reggio]

What's there to do?

a man pushes a heavy cart 2000n putting it in motion,
pushing at maximum speed v (m/s to be obtained),
and bumping a obstacle the cart should not turn over,
the man stops pushing,
What maximum speed should I impose on man?

Of course, if the resulting speed is too low, I will go to widen the wheel strap to have greater stability.

the problem could also be seen on the other side, that is, pushing to the maximum speed of 5km/h (man's pass) how much it has to become the 730mm (or 630mm) to reach the tipping sena balance?

 

[meccanicamg]

acceleration is measured in m/s2.
speed doesn't hit anything by itself. acceleration is the speed variation from maximum to zero in a time interval. it can also go to 1000km/h but if it stops in 30 minutes you have an acceleration nothing or almost. if you go to a mm/s that is a ridiculous speed....if you stop in 1millisecond you have a nice acceleration.

 

[cacciatorino]

make an energy equivalence between the kinetic energy possessed by the cart and the one necessary to overturn it. from here you can get the v max that tilts the cart.

the problem is not schematizable with a shock between rigid bodies, which by their nature would provoke an infinite (de)celeration.

 

[PIETRO2002]

lengths in m and not in mt.
accelerations in m/s not in mt.
from the equation of balance determines what is missing.
but if you already have fs.... we don't get anything. Why? because there is no force of inertia to add to the scheme.
But is fs a not rigid thrust? give us more details?

acceleration in m/sec^

 

[F_Ingrasciotta]

is not a question of acceleration in m/s^2

Riprovo

m=meters of your bene?force/weight in newton okay?acceleration in meters per second, all right?


in case fi=-m*a becomes fs=-m*a
Is all this above correct?


So does this on it?

if everything is okay we continue
(fp*0.73metri) - (fs*1.8metri) = 0
(2000newton*0.73metri) - (fs*1.8metri) = must be balanced = 0
♪
(2000newton*0.73metres) - (-m*a*1,8metres = 0
the mass m worth how much the weight force fp?
(2000newton*0.73metres) - (-2000newton*a*1,8metri) = 0

the problem is that you are confusing the weight force with the mass (kg).

if you correct this information you will see that also the units of measurement return.. .

 

[Marco F inox]

What's there to do?

a man pushes a heavy cart 2000n putting it in motion,
pushing at maximum speed v (m/s to be obtained),
and bumping a obstacle the cart should not turn over,
the man stops pushing,
What maximum speed should I impose on man?

Of course, if the resulting speed is too low, I will go to widen the wheel strap to have greater stability.

the problem could also be seen on the other side, that is, pushing to the maximum speed of 5km/h (man's pass) how much it has to become the 730mm (or 630mm) to reach the tipping sena balance?
square.View attachment 68433

the share of 1225 served for a completion of the drawing, but it also served to me for
show the breakdown of forces.
using fp2 efs2, from me you find graphically, it proved that the two calculations,
my and proposed by others, lead to the same result of 81 kg, which is the tipping force applied on the same line of the fs force.
pushing the cart with a constant force of 50 kg, you create a bike evenly
accelerated, with a speed that will be increasing, but with a constant acceleration,
which will be given by the formula f = m*a.
I remember that speed is measured in m/sec, acceleration, which is the change of speed
in the unit of time, it is measured in square m/sec ( m/sec/sec). the mass in Newton.
from the formula is obtained: a = f/ m = 500 / 2000 = 0.25 m/sec square.
Now it is only necessary to calculate the speed at which the cart develops an inertia force of 81 kg.

 

[reggio]

Good morning, thank you all, but if you explain to me that
"You have to learn how to fish well" or "you don't have to fish hippopotamus" or "fish, cook, eat" is not enough, I don't come alone to fish and cook it, it doesn't help me solve ...

Now, the question remains and only this:What maximum speed should I impose on man?orhow much must become the 730mm altitude to reach the balance without turning to 5km/h?I need, before reaching a numerical result (I also ask numerically so I will be able to understand the various passages - for example what mass value and what weight-force value) con una (or more) formulas, poi di capire i passsaggithe result I look for is indicative, that is, it can afford to neglect the analysis of deformation or frictions to the smallest details because, once found a result, I will keep a discreet margin of safety.

If you need other data, please ask me, but then please leave a summary of the procedure complete from start to end and without jumping anything... otherwise I lose myself in the street