Calculating Forces in Light Beam Interactions

In summary: If you have 2 forces pulling up (the force of the link below, and the force of the link below that)... how much force do you need to exert to compensate for gravity?If you have 2 forces pulling down (the force of the link above, and the force of the link above that)... how much force do you need to exert to compensate for gravity?2.46 is the correct answer though.sorry did it first as 2ma+2mg=2.46 Nbut split it up into (ma+mg)+(ma+mg)=2.46Njust forgot to change that .2 to a .1.But I'm still curious, why is it that when I add ma
  • #1
bob1182006
492
1
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. Homework Statement

A light beam from a satellite-carried laser strikes an object ejected from an accidentally launched ballistic missile. The beam exerts a force of [itex]2.0 * 10^{-5}[/itex] 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)


Homework Equations


[tex]\triangle x=v_0 t +\frac{1}{2}at^2[/tex]

F=ma

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.
[tex]\frac{mg+2.7*10^{-5}}{m}[/tex] 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. Homework Statement

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

Homework Equations


F=ma

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.
 
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  • #2
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
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
For problem 9, part b)

Write this equation out:

[tex]\Sigma\vec{F} = ma[/tex]
 
  • #5
bob1182006 said:
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...
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
Yea the first problem is weird 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
bob1182006 said:
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

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
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
bob1182006 said:
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?...

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
learningphysics said:
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.

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!
 

1. What is a light beam force problem?

A light beam force problem is a physics problem that involves calculating the force exerted by a beam of light on an object. This type of problem typically involves using the principles of optics and Newton's laws of motion to determine the magnitude and direction of the force.

2. How is light beam force related to other types of forces?

Light beam force is an example of a non-contact force, meaning that it does not require physical contact between objects to exert a force. It is also related to electromagnetic force, as light is a form of electromagnetic radiation.

3. What factors affect the magnitude of light beam force?

The magnitude of light beam force depends on several factors, including the intensity of the light beam, the distance between the light source and the object, and the properties of the object (such as its size, shape, and reflectivity).

4. How is light beam force measured?

Light beam force is typically measured in units of newtons (N), which is the standard unit of force in the International System of Units (SI). This unit is based on the amount of force required to accelerate a mass of one kilogram at a rate of one meter per second squared.

5. What real-world applications involve light beam force problems?

Light beam force problems have many practical applications, such as in laser cutting and welding, optical tweezers used in biology and chemistry research, and the design of precision instruments such as telescopes and microscopes. They are also important in understanding phenomena such as radiation pressure in space and the behavior of light in fiber optic communication.

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