Center of gravity again!

1. Mar 20, 2004

lollypop

hey everybody:
here is my problem
A uniform, aluminum beam 9.00 m long, weighting 300 N,
rests symmetrically on two supports 5.00 m apart. A boy
weighing 600 N starts at point A and walks toward the right.
How far beyond point B can the boy walk before the beam tips?

it will be like this right?
Code (Text):

boy
|-----*----------*-----|
A          B

my questions is what am i going to use as a point of
reference,do i use B? how cna i calculate the distances?
I always get confused when plugging in distance for
the formula

D_cm = (m1x1+m2x2+..) / (m1+m2+..)

2. Mar 20, 2004

Chen

The beam will tip when the center of gravity of the system is to the right of point B, so the boy must stop walking when the cente of gravity is at point B. Since the question asks "how far beyond point B..." it would be the easiest to pick point B as the reference point, because then you know the boy must stop when the distance of the center of gravity from point B is zero.

Where is the beam's center of gravity? In its middle, 2.5 meters to the left of point B. Where is the boy's center of gravity? Wherever he is standing, let's say X.

$$X_{cg} = 0 = \frac{2.5m_{\mbox{beam}} + Xm_{\mbox{boy}}}{m_{\mbox{beam}} + m_{\mbox{boy}}}$$

You can cancel the denominator, since it cannot be zero:

$$2.5m_{\mbox{beam}} + Xm_{\mbox{boy}} = 0$$
$$X = -2.5m \frac{m_{\mbox{beam}}}{m_{\mbox{boy}}} = -2.5m \frac{m_{\mbox{beam}}g}{m_{\mbox{boy}}g}$$

$$X = -2.5m \frac{W_{\mbox{beam}}}{W_{\mbox{boy}}}$$
The answer is negative, because by saying that the center of gravity of the beam is 2.5 to the left of point B we decided that the positive direction of the X axis is to the left. So when the boy is moving right, his X position is negative relative to point B.

3. Mar 20, 2004

lollypop

thank you chen
now, if i want to move support B to the right
so that the boy could walk without tipping the beam
how far from the right end would i put it?
I'll have to use 4.5 as the cg of the beam, do i have
to include the boy's distance and weight here?

4. Mar 20, 2004

Chen

You mean, at least how much you need to move point B to the right so the boy could walk through the beam without tipping it?

Take the right end of the beam as the reference point, and let X be the distance of point B from there. The center of gravity still needs to be at point B, i.e X. The distance of the boy's mass from the reference point is zero, and the distance of the beam's center of gravity from the reference point is indeed 4.5 meters.

$$X = \frac{4.5m_{\mbox{beam}}}{m_{\mbox{beam}} + m_{\mbox{boy}}}$$

You can expand the fraction by $$g$$ so that you have the objects' weight and not their mass.

5. Mar 20, 2004

lollypop

great!
so every point of refernece that i use, i have to make it zero?

here is another problem that I really have no clue as to how to set it up. pls help again
Your dog Nikita has a length of 0.93 m (nose to hindlegs). Her forelegs are located a horizontal distance 0.15m behind her nose, her center of gravity is located a distance 0.29 m horizontally in front of her hindlegs, and she has weight 195 N.
.How much force does a level floor exert on each of her hind feet?
.How much force does a level floor exert on each of Nikita's front feet?
.If Nikita picks up a bone of weight 23 N. and holds it in her mouth (directly under her nose), what is the force exerted by the floor on each of her front feet?

6. Mar 20, 2004

Chen

Think of Nikita as two vertical beams (front legs and back legs) connected with a massless horizontal beam. You know where the center of gravity is, you know its location in reference to the two beams and you know the weight of the two beams together, so you can easily find the weight of each beam on its own:

Let $$m_1$$ be the mass of the front legs; let $$m_2$$ be the mass of the back legs; let $$m_T$$ be the total mass of the dog. Use the center of gravity as the point of reference.

$$0 = \frac{(0.93 - 0.15 - 0.29)m_1g + (0 - 0.29)m_2g}{m_1g + m_2g}$$

$$m_Tg = 195N = m_1g + m_2g$$
$$0.49m_1g = 0.29m_2g$$

Two equations, two variables... solvable. Once you have the weight of each pair of legs you can know the force they exert on the floor, which is equal to the force the floor exerts on the legs.

If the dog picks up a 23N bone, her center of gravity changes (it moves forward). You need to find the new center of gravity, using two objects - the dog, whose weight and center of gravity you already know, and the bone, whose weight and position you now know. Once you have the new center of gravity of the dog+bone, repeat what you did above but replace 0.29m with the new position of the center of gravity, to find the new $$m_1$$ and $$m_2$$ (and don't forget $$m_T$$ has already changed!).

7. Mar 20, 2004

lollypop

chen:
on your second equation .49 m1g=.29m2g, i get them both to be negative when i work out the formula, i did the calcualtions both ways and the result is wrong.

8. Mar 21, 2004

Chen

Why is that... how did you solve it?

$$m_1g = 195N - m_2g = \frac{0.29}{0.49}m_2g$$

The answer should be $$m_1g = 72.5N$$ and $$m_2g = 122.5N$$.

9. Mar 21, 2004

lollypop

yes i get 122.6 and 72.4, but it says that i need
to check over my signs for 122.6?? and the other one 72.4
is just wrong. i then change it to negative
and still tells me the same thing

10. Mar 21, 2004

Chen

Who tells you what? [?]

The book is asking to find the force that the floor exerts on the legs. This is the normal force, and it's equal to the weight of legs but opposite in direction (i.e it goes up). Maybe that's why the answer is negative?

11. Mar 21, 2004

lollypop

Hey , i got it!! jeje, is on each leg, so i had
to divide by 2
i used 195=n1+n2
0=0.49n1-0.29n2

solving for it i get for n1=71.87 nad n2=123.13,
so for each leg 36 and 62 respectively.