[Guest]

If you weigh 80 kg, which maximum number will you see on the scale, if you jump on to it from 1 meters height.

[Guest]

you might want to reduce the effect of variables for this

[Guest]

you might want to reduce the effect of variables for this

How many variables are there?

A person weighing 80 kg.

A simple digital scale.

One meters of height, for jumping.

I don't see any variables to reduce???

Really, this is not a riddle. But I know that such "simple logic" physics problems are treated here before.

[Guest]

Really, this is not a riddle. But I know that such "simple logic" physics problems are treated here before.

Puzzles certainly don't have to be riddles here so don't worry about that, but LIS is right that there is no definite solution that I can see.

The scales need to slow him down until he's stationary. This means an acceleration greater than -g, which I assume is your reasoning for the question.

The difficulty is that we know nothing about the scales. If they are very stiff, they may stop him very quickly, in which case the acceleration and associated force are high. If they are 'soft' and springy then the force required is much less because the acceleration has a considerably longer period to affect him.

Alternatively, if you prefer, you could think about it in terms of energy/distance or impulse/momentum, but you still have the problem regarding the distance/time over which you apply the 800J (=mg, the GPE he starts with) of energy they need to expend to stop him.

[Guest]

Puzzles certainly don't have to be riddles here so don't worry about that, but LIS is right that there is no definite solution that I can see.

The scales need to slow him down until he's stationary. This means an acceleration greater than -g, which I assume is your reasoning for the question.

The difficulty is that we know nothing about the scales. If they are very stiff, they may stop him very quickly, in which case the acceleration and associated force are high. If they are 'soft' and springy then the force required is much less because the acceleration has a considerably longer period to affect him.

Alternatively, if you prefer, you could think about it in terms of energy/distance or impulse/momentum, but you still have the problem regarding the distance/time over which you apply the 800J (=mg, the GPE he starts with) of energy they need to expend to stop him.

Thank you, foolonthehill and LIS. The variable that LIS asked was the resistance (stiffness of the springs) of scale, I didn't think of it.

[Guest]

since f=ma and you are using the metric system your force is 784 Newtons but the scale would show a max value of 213kg

[Guest]

This what you have in mind?

One can't "weigh" 80 kg. kg is a unit of mass, weight is a force and therefore in SI units would be Newtons.

[Guest]

It's not just the spring constant but also the damping (viscous friction) coefficient. Assuming the spring constant and damping are linear, the dynamic equation is ma= w+Kx+Dv=f where m is the mass of the body, a is the acceleration of the body once it is in contact with the scale, w is the weight of the body in contact with the spring, K is the spring constant of the scale, x is the displacement of the spring from rest position, D is the viscous friction coefficient, v is the instanteous velocity of the body in contact with the spring, and f is the instantaneous force indicated on the scale. For this equation, x, v and a are defined as positive down. Since v is the first time derivative of x and a is the second time derivative of x, this is actually a linear differential equation with constant coeffiecients, a type of equation that is easily solved when the coefficients and initial conditions are known.

The resulting maximum force depends on both the spring force and the viscous friction coefficient. If D is relatively large the maximum indication will be w, the actual weight of the person. If the D is small enough to be negligle, the maximum indication will be dependent on the velocity of the body at contact and the spring constant only, but this would be a very poor scale since it would continue to bounce up and down for a long time.

[Guest]

It's not just the spring constant but also the damping (viscous friction) coefficient. Assuming the spring constant and damping are linear, the dynamic equation is ma= w+Kx+Dv=f where m is the mass of the body, a is the acceleration of the body once it is in contact with the scale, w is the weight of the body in contact with the spring, K is the spring constant of the scale, x is the displacement of the spring from rest position, D is the viscous friction coefficient, v is the instanteous velocity of the body in contact with the spring, and f is the instantaneous force indicated on the scale. For this equation, x, v and a are defined as positive down. Since v is the first time derivative of x and a is the second time derivative of x, this is actually a linear differential equation with constant coeffiecients, a type of equation that is easily solved when the coefficients and initial conditions are known.

The resulting maximum force depends on both the spring force and the viscous friction coefficient. If D is relatively large the maximum indication will be w, the actual weight of the person. If the D is small enough to be negligle, the maximum indication will be dependent on the velocity of the body at contact and the spring constant only, but this would be a very poor scale since it would continue to bounce up and down for a long time.

Correction: The equation should be ma = w - Kx - Dv. The indicated value should be f = -Kx. Sorry, I wrote too fast.

[Guest]

Yes some variables in the scales and how high did he jump - if dropped then the force can be detrmined (kinetics) - also a tiny variable is sea level - mount everest .. too many conditions that are not constant. We are suposed to take the riddle as meant but we now find that the scales are digital and not a pointer. btw mass and weight were not the question - it was the 'reading of the scale' or 'digits' - which would also be difficult to 'see'

Im not picking it apart - just saying its variable for now

based on 9.8m/s of gritational force in the first second basically 1m, a 10th of a second then take the variable of the scales resistance mechanically- which is unknown??

Im still curious as to weather the jumping/descending individual si meant to be the viewer as written in the OP - head first will help in this case, particaularly with poor eyesight.

[Guest]

OK If i doubted that there are A LOT of smart people in here.....

I can only stand in the memory that my first post was me solving Itachi's riddle!

So its good to be right brained too!

[Guest]

Assuming that jumping results to falling on to the scale, and that there is an earth-like gravitational field, i would say that you will get to see something between 120 and 135 as maximum, since most scales show a maximum of that number.

[Guest]

assuming this is not a digital scale... then the maximum number that you would "see" is the maximum number on the scale. It does not ask the maximum number that the scale would reach just the max number you would see.

If the scale ranges from 0-300 then the max number you will see is 300.

[bonanova]

two nice things happen.

1. dv/dt = delta[v] / delta[t]
   
2. F is constant during delta[t]
   

Then, F = m delta[v] / delta[t]

delta[v] is is known - it's the velocity acquired by falling 1 meter.

m is known.

delta[t] is the length of time during which you decelerate to rest.

It depends on some particulars.

As several have pointed out, delta[t] depends on the mechanics of the scale.

It also depends on your footwear.

Running shoes, for example, increase delta[t], softening the impact.

So F can range from a shattering infinity [instantaneous stop]

to a barely noticeable increase over your weight [the 1 meter fall distance filled with sponge rubber]

[Guest]

If it is a proper set of scales It will meassure mass ( in kg) which WILL NOT CHANGE no matter how much you jump on it..

What a shame most scales meassure force..

[Guest]

If you weigh 80 kg, which maximum number will you see on the scale, if you jump on to it from 1 meters height.

if i weigh 80kg and jumps a meter high on the scale, what max number will i see?

considering the type of scale i8s a bathroom scale... then the max number i will see is also the max number of the scale.. because of your heaviness.. the scale might broke. lol