Finding normal force from center of gravity

In summary: Finally, you can plug in the value of Fh from the first equation into the second equation to get the value of Ff.
  • #1
thatgirlyouknow
58
0

Homework Statement



The figure shows a person whose weight is W = 725 N doing push-ups. Find the normal force exerted by the floor on (a)each hand and (b)each foot, assuming that the person holds this position.

Homework Equations


center of gravity = cg
Xcg = (W1x1 + w2x2 +...)/(w1 + w2...)
net torque = w1x1 + w2x2 +...

The Attempt at a Solution



Placing the axis of rotation at the far left gives:
W1(0) + W2(.84) + W3(1.25) / 725 = .84
Placing it at the far right gives:
W1(1.25) + W2(.41) + W3(0) / 725 = .41

Solving these yields W1 = [297.25 - W2(.41)]/1.25
or
W2 = 297.25 - W1(1.25)
and
W2(.84) + W3(1.25) = 609

However, there are still too many unknowns. Where do I take it from here?
 

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  • #2
thatgirlyouknow said:
Placing the axis of rotation at the far left gives:
W1(0) + W2(.84) + W3(1.25) / 725 = .84
Not quite sure what you're doing here. Looks like you're trying to calculate the center of gravity? But you are given the center of gravity.

Instead, set the net torque equal to zero for equilibrium.

Don't forget that the net force on the person must also be zero.

That will give you two equations and two unknowns.
 
  • #3
So w1x1 + w2x2 + w3x3 = 0?

w1*.840 + w2*0 + w3*.41 = 0

w1+w2+w3 = 725

I get that the first equation has two unknowns, but what about the second?
 
  • #4
thatgirlyouknow said:
So w1x1 + w2x2 + w3x3 = 0?

w1*.840 + w2*0 + w3*.41 = 0
Realize that the torques will have different signs.
w1+w2+w3 = 725

I get that the first equation has two unknowns, but what about the second?
Are all three Ws unknown?
 
  • #5
They appear to be, unless I should assume that the middle one is 725 (the total weight.) Is this assumption correct?
 
  • #6
Ok, I got it:
Set up as Fn1(0) + Fn2(.84) + Fn3(1.25) = 0
Fn2 = 725
so Fn3 = 487.2
Divided by 2 gives the hands, and the remainder is split between the feet.
 
  • #7
Good.

Here's how I look at this problem. There are three forces acting on the person:
(1) The upward force at his feet: Ff
(2) The upward force at his hands: Fh
(3) The downward force of his weight: W (which is known)

So your torque equation becomes:
Ff(0) + W(.84) -Fh(1.25) = 0
Which you can solve for Fh:
Fh = W(.84)/(1.25)
(Note that clockwise and counterclockwise torques have different signs.)

And your net force equation is:
Ff -W + Fh = 0
Which you can solve for Ff in terms of W and Fh (which are known):
Ff = W - Fh
 

1. What is normal force?

Normal force is the force exerted by a surface on an object that is in contact with it. It is always perpendicular to the surface and acts to prevent the object from passing through the surface.

2. How is normal force related to center of gravity?

Normal force is directly related to the center of gravity of an object. The center of gravity is the point at which all the weight of an object can be considered to act. The normal force acts through this point, providing support and balancing out the weight of the object.

3. Can normal force be greater than the weight of an object?

Yes, normal force can be greater than the weight of an object. This can happen if the object is accelerating or if it is on an inclined surface. In these cases, the normal force must be greater to counteract the additional forces acting on the object.

4. How do you calculate normal force from center of gravity?

To calculate normal force from center of gravity, you need to know the weight of the object and the angle of any incline that it is on. The normal force can then be calculated using the formula N = mg cos(theta), where N is the normal force, m is the mass of the object, g is the acceleration due to gravity, and theta is the angle of the incline.

5. Why is it important to find normal force from center of gravity?

Finding normal force from center of gravity is important because it helps us understand the stability and equilibrium of an object. It also allows us to calculate the forces acting on an object and ensure that it is properly supported and balanced. This information is crucial in fields such as engineering, physics, and architecture.

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