Help With Physics Problem
 

My kid is struggling with this one...I put it to the brain trust to help him out a bit HOW to solve? And, correct answers I can use to check his work..

- - - begin - - -
A chain consisting of five links, each of mass 0.094 kg, is lifted vertically with a constant acceleration of 2.42 m/s2, as shown

http://forums.pelicanparts.com/uploa...1189636164.gif

(a) Find the force acting between the bottom link and the next one up.
_______________

Find the force acting between the second link from the bottom and the one above it.
_______________

Find the force acting between the third link from the bottom and the one above it.
______________

Find the force acting between the fourth link from the bottom and the one above it.
_______________

(b) Find the force F exerted on the top link by the person lifting the chain.
_________________

(c) Find the net force accelerating each link.
________________

- - - - end - - -

I'm pretty sure it's spelled "fysics".
No thanks necessary ;)

F = mA

the force at each link is the sum of the masses in the links below it *A

I believe gravity on the Earth and ignoring air resistance is 9.8m/s2 regardless of other forces.

.094kg * (9.8+2.42)m/s^2
link 1 1.15N up (minus 9.8*.094=.92) --> .23N net You get the same thing from 2.42*.094

so, do you need the forces, or just the net force. If just the net force, then it's exactly what Bill said

I'm guessing .23, .46, .69, .92 is the net for each link
and for the gross, 1.15, 2.3, 3.45, and 4.6

as the chain is lifted, all links will stay together....but it might swing a little bit from side to side

acceleration is the change of velocity with respect to time.

D O P E !

;)

What if it's made of Jello? ....warm Jello, w/ added gin.

How's that saying go? "You have an uncanny talent for stating the obvious." :D

Free body diagrams, enough said. Upwards force is the total mass * vertical acceleration, weight is total mass * 9.81 m/s^2

There are 2 parts to the resolution of this problem
the static case(before upwards acceleration) and the dynamic case (during the upwards acceleration)

assumption; this is taking place near the surface of the earth in a total vacuum chamber.

since we only care about the forces during the acceleration we add the acceleration due to earths gravity to the imposed acceleration
net A = 9.8 + 2.42 = 12.22m/s²
from bottom to top force in Newtons at each link
F = .094 * 12.2 = 1.1468N
F = 2 * .094 * 12.2 = 2.2936N
F = 3 * .094 * 12.2 = 3.4404N
F = 4 * .094 * 12.2 = 4.5872N
F = 5 * .094 * 12.2 = 5.734N

ok, so, wouldn't that be the gross F in the upward direction, and the gross force in the downward direction would be .92, 1.84, ...

So then wouldn't the net force be the mass * the acc which in this case is actually 2.42m/s^2? The Force of gravity down minus the Normal Force (the chain links holding the bottom links up) should cancel each other out and the net force would be the upward acceleration which is only 2.42?

Like this

http://forums.pelicanparts.com/uploa...1189691238.jpg

As I said 2 parts
static case, just hangeing there each link exerts a force to to counteract the force of gravity. up = down because it's not moving.

from lowest to highest
F = .094 * 9.8 = .9212N
F = 2 * .094 * 9.8 = 1.8424N
F = 3 * .094 * 9.8 = 2.7636N
F = 4 * .094 * 9.8 = 3.6848N
F = 5 * .094 * 9.8 = 4.606N

now add the imposed force that adds the additional 2.42m/s² acceleration
again from lowest to highest
F= .094 * 2.42 = .22784N
F= 2 * .094 * 2.42 = .45496N
F= 3 * .094 * 2.42 = .68244N
F= 4 * .094 * 2.42 = .90992N
F= 5 * .094 * 2.42 = 1.1374N

add the 2 components( I rounded in the previous response hence the dicrepancy)

.9212N + .22784N = 1.14904N
1.8424N + .45496N = 2.29736N
2.7636N + .68244N = 3.44608N
3.6848N + .90992N = 4.59472N
4.606N + 1.1374N = 5.7434N

It's a lot easier to do the math if you first add the acceleration components then multiply, but it can be done either way.