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How to compute for Pressure without the acceleration on the Force?

  1. Sep 15, 2014 #1
    1. The problem statement, all variables and given/known data

    Compute for the value of pressure beneath the foot of a person after stomping. The person has a weight of 50kgs, supposed that there is a sensor beneath the foot of the person and after stomping the foot it records a 51.5 kg of weight. The area of the sensor is 2.5 square inch.

    Stationary weight = 50 kgs
    measured weight after stomping = 51.5 kg
    Area of the pressure where change of weight is detected = 2. 5 square inch


    2. Relevant equations


    Pressure = Force / Area
    Force = mass x acceleration


    3. The attempt at a solution

    Computing for the pressure upon stomping on the sensor is equivalent for Pressure = Force / Area.

    I have an Area which is 2.5 square inch now,the problem is what is the applied force. Since I have a change of rate of the weight which 51.5 kgs from 50kgs. Can I consider this as the Force as 51.5kg/s/s.

    Or Force equal to mass x acceleration. The mass would be the default weight that is 50kgs. Then the force is computed as (measured weight - actual weight) / actual weight. Leaving the acceleration as the unknown variable. If I proceed on this method... it would be

    acceleration = force / mass
    a = (51.5kg/s/s - 50 kg/s/s )/ 50kg
    a =.03 s/s.

    Then going back to the pressure formula P = F / A, P = (mass x acceleration) / area.

    P = (51.5 kg /s/s x .03 s/s) / 2.5 square inch. Pressure is .618 Kg/ square inch.

    Please help anyone.
     
    Last edited: Sep 15, 2014
  2. jcsd
  3. Sep 15, 2014 #2
    "I have an Area which is 2.5 square inch now,the problem is what is the applied force. Since I have a change of rate of the weight which 51.5 kgs from 50kgs. Can I consider this as the Force as 51.5kg/s/s."

    No, just look at your units for the next part if this was the case. You make force kg/s^2 when force is Kgm/s^2.

    Are you sure your not supposed to just use gravity? Because the sensor detects the force and divides by 9.8 (g) to give you your mass value. If you were to apply force, it would increase the mass by F/g since it doesn't care whether you are pushing or standing. For example if you went on a bathroom scale and you stood on it, it gives your Fg/g. However if you were to push on it, it won't go Fg/g + Fapp/a = mass, it would just treat the Fapp as if it was Fg. Even from a logical perspective think, is your mass changing? The question says after stomping your weight became higher, where did the extra weight come from? The force you applied to it divided by a g because of what I just said.

    To find the pressure after the stomp, you are missing P and a. The first scenario (standing still) doesn't provide a way to link these two equations and help you solve for a variable if the P and a values are different as well (you would really only be able to find a ratio if this was the case) so I believe your approach on the question is wrong. The standing weight is negligible in this question I believe.

    This is my take on the question, might be wrong haha. :tongue:
     
    Last edited: Sep 15, 2014
  4. Sep 15, 2014 #3
    Hi happysmiles36,

    Thank you for taking time to read my post. Yeah I'm quite figuring out that this approach is wrong. Let us try to focus on getting the Force. Here is what I've been through lately.

    Alright, what I think I am missing here is the ratio when standing still. Now what I did....

    Get the force when standing so, Force = mass x acceleration(g); 50 kg x 9.8 m/s^2. Which will give me a value of 490 Newtons

    Now, that I know with the default acceleration which is gravity is equal to 50kg and 490 newtons. I tried to do a ratio if I measured a weight of 51.5 kg. That brings me to this equation;

    50kg /( 9.8 m/s^2 ) = 52 kg / (unknown acceleration)

    Which then if I compute for the unknown acceleration it would give me a value of 10.192 m/ s^2.

    Knowing that I have that right acceleration when the person stomp the feet I can compute for the total force exerted on the ground.

    Force = 51.5 kg x 10.192 m/ s^2

    Force = 524 Newtons

    Then it would give me way to acquire the pressure... Please can you confirm if this approach is correct.. thank you
     
    Last edited: Sep 15, 2014
  5. Sep 15, 2014 #4
    Well I still think the acceleration wouldn't change. I think it would just be 51.5kg*9.81m/s^2 for the force because the sensor can't detect acceleration, only force.

    Like if I gave you the equation a*b = 100 and asked you what is a and b? Well there is a ton of possibilities, it could be 4*25, 50*2, ect... Same concept can be applied here to the sensor. If I gave b a fixed number of 5, then a is 20 and if I gave you a*b=150 you could still figure out what a is.

    Unless you have been instructed that the acceleration of the force of the stomp is really unknown and not gravity, (title says unknown acceleration but question sort of says the opposite) I think your approach is still wrong.

    Note: The sensor is NOT measuring the MASS (Kg). If this was the case, you would only be getting 50Kg. It measures the FORCE and divides it by gravity to give you MASS because it cannot measure the acceleration. And sorry it took so long to reply, I went to sleep :biggrin:
     
    Last edited: Sep 15, 2014
  6. Sep 15, 2014 #5

    PhanthomJay

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    It is unfortunate that the problem uses the non standard SI unit of Kg of weight as the force unit, and to make matters worse, it gives the area in the Imperial units of inches squared. So I am not sure if they are looking for the answer for pressure in kg of force per square inch, or Newtons per square something. One kg of force is 9.8 N.

    On the assumption that they want the pressure in kg of force per square inch, the force on the scale when stomping and the area of the sensor are given quantities. So you can compute the pressure on the sensor. No need to get accelerations involved, as happysmiles36 indicated. I don't know how one would record the pressure on the foot unless its area were given.
     
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