# Forces Problem

1. Sep 16, 2007

### bob1182006

I have 2 problems I just can't seem to get so I'll post both here instead of making 2 threads.
Both of these are from Halliday & Resnick 5th ed chapter 3 Problems.

Problem #1.
1. The problem statement, all variables and given/known data

A light beam from a satellite-carried laser strikes an object ejected from an accidentally launched ballistic missile. The beam exerts a force of $2.0 * 10^{-5}$ N on the target.
If the "dwell time" of the beam on the target is 2.4s by how much is the object displaced if it is
a) a 280-kg warhead
b) a 2.1-kg decoy?

(These displacements can be measured by observing the reflected beam)

2. Relevant equations
$$\triangle x=v_0 t +\frac{1}{2}at^2$$

F=ma

3. The attempt at a solution
The missile experiences some uknown horizontal acceleration/velocity so I will just ignore those..
It also experiences a downward acceleration of g.
So does the laser beam?

I need to find the sum of the forces to find the acceleration.
$$\frac{mg+2.7*10^{-5}}{m}$$ for a I get about 9.8 m/s^2 .
Plugging that into the equation I get a displacement of about 56m but the answer should be some micro-meters..

Problem #9.
1. The problem statement, all variables and given/known data

A chain consisting of five links, each with mass 100g, is lifted vertically with a constant a =2.5m/s^2.
Find
a) the forces acting between adjacent links
b) the force F exerted on the top link by the agent lifting the chain
c) the net force on each link

2. Relevant equations
F=ma

3. The attempt at a solution
2.5m/s^2 up
-9.8m/s^2 up

a total of -7.3m/s^2 up

for a.
add up the masses of link+links beneath it and then multiply by 7.3m/s^2 to get a force of:
for the top most link
.5kg*7.3m/s^2 = 3.65 N being exerted on the top link.

Which is completely wrong :/.

for b.
find the total mass and multiply by 7.3m/s^2
F=.5kg*7.3m/s^2=3.65 N
again completely wrong..

for c.
Somehow I got this right...
lowest link force acting upon it is: .1kg*2.5m/s^2 = .25 N
link above it: .2kg*2.5m/s^2=.50 N
...for all others

then subtracting the link force - link below it to get a net force of .25 N on each link.

Any help on either problem is greatly appreciated.

2. Sep 16, 2007

### bob1182006

I'm mainly trying to get problem #9.

I was working backwards from the answers the book gave.
for a the lowest # and thus the lowest link in the chain I think.

has a total force of 1.23N which requires a force of 12.3 m/s^2 (g+2.5) on a mass of 100g.
Isn't this wrong though? since there's a downward force of .1g N but an upward of .1*2.5 N but somehow the in the book they're adding them...

3. Sep 16, 2007

### learningphysics

For the first problem, I don't think you should use mg... we don't really know anything about the missile... it may not be accelerating downwards at g (it may have its own thrust or something). we also don't know that the laser is directed straight upwards...

4. Sep 16, 2007

### learningphysics

For problem 9, part b)

Write this equation out:

$$\Sigma\vec{F} = ma$$

5. Sep 16, 2007

### PhanthomJay

I don't think you are corrrectly applying newton 2 in your free body diagrams. Isolating the bottom link, there are 2 forces acting on it. You have correctly identified the downward force. The upward force is unknown...call it T. Now use newton 2nd law....F_net = ma. a is given, don't mess with it.

6. Sep 16, 2007

### bob1182006

Yea the first problem is wierd especially the "These displacements can be measured by observing the reflected beam" o.o since it's barely chapter 3 and I have no knowledge of optics so far.

So sticking to #9.
for part a, there are 2 forces acting on each link, 1 up and 1 down, I add them together right?
so for the lowest link:
.1 kg * 2.5 m/s^2=.25 N
.1 kg * 9.8 m/s^2=.98 N
total being 1.23 N

the second link:
there are 2 links being pulled up so I add the previous link's force + a new link:
1.23 N + (.1kg * 2.5 m/s^2 + .2kg * 9.8 m/s^2) = 1.23 N + 1.23 N=2.46

and so forth for the rest correct?

for part b)
I would do almost the same as I did for part a correct? part a I stop adding @ the 4th link since it has force exerted by the 5th which is topmost and the 3rd beneath it.

7. Sep 16, 2007

### learningphysics

Not sure what happened above... : .1kg * 2.5 m/s^2 + .2kg * 9.8 m/s^2 is not 1.23... but 2.46 is the correct answer though.

8. Sep 16, 2007

### bob1182006

sorry did it first as 2ma+2mg=2.46 N
but split it up into (ma+mg)+(ma+mg)=2.46N

just forgot to change that .2 to a .1.

But I'm still curious, why is it that when I add ma and mg they are both positive? a is pointing up but g is pointing down so shouldn't they be subtracting?....

9. Sep 16, 2007

### learningphysics

If you're lifting 1 kg object upwards at an acceleration 1.0m/s^2 in a gravityless environment... you only need to exert 1N of force.

If you're lifting 1 kg object upwards at an acceleration 1.0m/s^2 in a 9.8m/s^2 gravity environment... you'll need to exert a greater force... you're exerting a force to compensate for gravity.

Fnet = ma
Fupwards - mg = ma
Fupwards = ma + mg

The greater the gravity... the greater the force required to compensate for gravity, and move the object at the same acceleration.

10. Sep 16, 2007

### bob1182006

ok I think I get it now.
so when I find ma for .1kg m the net force IS .1*2.5=.25N but that is the total of the force I want - mg so that's why I add the mg!

Thanks alot!!

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