Find the normal force exerted by the floor

In summary: This is frustrating me. i know that whatever answer i get for the hands must be divided by two..i get that.But right now i have an equation...but no idea what i am solving for.
  • #1
spidey12
37
0

Homework Statement



The drawing shows a person (weight W = 581 N, L1 = 0.829 m, L2 = 0.399 m) doing push-ups. Find the normal force exerted by the floor on each hand and each foot, assuming that the person holds this position.

http://img401.imageshack.us/img401/3044/p911zz6.gif [Broken]




Homework Equations



I know this has something to do with net torque. Wd=wd

The Attempt at a Solution



Ok i tried figuring this out in sveral ways. first, i tried multiplying L1 with 581, the persons weight, and then i divided by 2. This did not work. so tried multplying times l1+l2 and that did not work either. What am I doing wrong? i was sure i had it right the first time around.
 
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  • #2
spidey12 said:

Homework Statement



The drawing shows a person (weight W = 581 N, L1 = 0.829 m, L2 = 0.399 m) doing push-ups. Find the normal force exerted by the floor on each hand and each foot, assuming that the person holds this position.

Homework Equations



I know this has something to do with net torque. Wd=wd

The Attempt at a Solution



Ok i tried figuring this out in sveral ways. first, i tried multiplying L1 with 581, the persons weight, and then i divided by 2. This did not work. so tried multplying times l1+l2 and that did not work either. What am I doing wrong? i was sure i had it right the first time around.

First off you should note that he is "holding his position". This suggests that he is not in motion. If he is not in motion in any direction then he is at balance about his center of mass isn't he? That is if you jammed a fulcrum into his stomach he would have no weight on hands or feet and still not be in motion (though maybe in a great deal of pain).

You could treat it as a torque problem then with there being no torque about what convenient point? You look like you have enough knowns to give you an answer.
 
  • #3
LowlyPion said:
First off you should note that he is "holding his position". This suggests that he is not in motion. If he is not in motion in any direction then he is at balance about his center of mass isn't he? That is if you jammed a fulcrum into his stomach he would have no weight on hands or feet and still not be in motion (though maybe in a great deal of pain).

You could treat it as a torque problem then with there being no torque about what convenient point? You look like you have enough knowns to give you an answer.

I tried using the center of gravity as the fulcrum but it did not work. I used L1xW/2 for the feet and this did not work.

also i tried adding the two lengths together and just getting the midpoint and multiplying that times the Weight but I do not think this is correct.
 
  • #4
spidey12 said:
I tried using the center of gravity as the fulcrum but it did not work. I used L1xW/2 for the feet and this did not work.

also i tried adding the two lengths together and just getting the midpoint and multiplying that times the Weight but I do not think this is correct.

Why are you dividing the weight by 2?

Let [tex]W_T_O_T = W_H_A_N_D_S + W_F_E_E_T[/tex]

If it is balanced then: [tex]W_H_A_N_D_S * L_2 = W_F_E_E_T * L_1[/tex]

All you have to do is solve. (Remember they want the answer for EACH hand and foot. Just divide those answers by 2.)
 
  • #5
LowlyPion said:
Why are you dividing the weight by 2?

Let [tex]W_T_O_T = W_H_A_N_D_S + W_F_E_E_T[/tex]

If it is balanced then: [tex]W_H_A_N_D_S * L_2 = W_F_E_E_T * L_1[/tex]

All you have to do is solve. (Remember they want the answer for EACH hand and foot. Just divide those answers by 2.)


Ok i already had that equation but i dumped it because there is not unknown...we know l1 and l2, we know the weight of the hands and feet (both are 581)..so what am i solving for?
 
  • #6
spidey12 said:
Ok i already had that equation but i dumped it because there is not unknown...we know l1 and l2, we know the weight of the hands and feet (both are 581)..so what am i solving for?

The question is what is the Force on EACH hand and EACH foot.?

They all add to the TOTAL weight, yes that is given, but what is the weight on each? How is that weight DISTRIBUTED. That is what you were asked.
 
  • #7
LowlyPion said:
The question is what is the Force on EACH hand and EACH foot.?

They all add to the TOTAL weight, yes that is given, but what is the weight on each? How is that weight DISTRIBUTED. That is what you were asked.

This is frustrating me. i know that whatever answer i get for the hands must be divided by two..i get that.

But right now i have an equation in which i have four variables and i already know ALL FOUR variables. What am i solving for?
 
  • #8
spidey12 said:
This is frustrating me. i know that whatever answer i get for the hands must be divided by two..i get that.

But right now i have an equation in which i have four variables and i already know ALL FOUR variables. What am i solving for?

[tex]W_H_A_N_D_S = ?[/tex]
[tex]W_F_E_E_T = ?[/tex]

Because trust me [tex]W_H_A_N_D_S[/tex] is NOT equal to [tex]W_F_E_E_T[/tex]

All you know is that they add to 581N.
How that weight is distributed by solving for them using the 2 equations is what the problem wants.

And yes [tex]W_H_A_N_D = 1/2 *W_H_A_N_D_S[/tex]
and [tex]W_F_O_O_T = 1/2 * W_F_E_E_T[/tex] but they are not all equal to each other.
 

What is the normal force exerted by the floor?

The normal force exerted by the floor is the force that a surface exerts on an object that is in contact with it. It is a reaction force that is equal in magnitude and opposite in direction to the force that the object exerts on the surface.

Why is it important to find the normal force exerted by the floor?

Finding the normal force exerted by the floor is important in many situations, such as calculating the weight of an object, determining if an object is in equilibrium, or designing structures that can withstand certain forces. It is also a fundamental concept in physics and engineering.

How do you calculate the normal force exerted by the floor?

The normal force exerted by the floor can be calculated by using the equation FN = mg, where FN is the normal force, m is the mass of the object, and g is the acceleration due to gravity. This equation assumes that the object is not accelerating and is in equilibrium.

Does the normal force exerted by the floor always equal the weight of the object?

In most cases, the normal force exerted by the floor will be equal to the weight of the object. However, if the object is on an inclined plane or experiencing other external forces, the normal force may not be equal to the weight. In these cases, the normal force can be calculated by resolving the forces acting on the object.

What factors can affect the normal force exerted by the floor?

The normal force exerted by the floor can be affected by various factors, such as the weight and mass of the object, the angle of the surface, and the presence of other external forces. The normal force will also change if the object is accelerating or in motion.

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