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[H] Physics

Blogs > il0seonpurpose
Post a Reply
il0seonpurpose
Profile Blog Joined January 2007
Korea (South)5638 Posts
November 29 2008 18:54 GMT
#1
I'm learning free body diagrams right now and there's a question I'm stuck on:

Jack is sledding with his friends when he becomes disgruntled by one of his friends comments. He exerts a rightward force of 9.13 N on his 4.68 kg sled to accelerate it across the snow. If the acceleration of the sled is .815 m/s/s (or .815 m/s^2), then what is the coefficient of friction between the sled and snow?

So the equation I got was like this

9.13-(u*98)=4.68(.815)
I ended up getting some crazy number and it doesn't seem right and fit back in properly.

Here's another one, the answer I got seemed incorrect:

Edwardoapplies a 4.25 N rightward force to a .765 kg book to acclerate it acoss a table top. The coefficient of friction between the book and the tabletop is .41. Determine the acceleration

Thanks

*
Zherak
Profile Blog Joined November 2007
Norway256 Posts
November 29 2008 18:59 GMT
#2
Hm, these ones looked rather straightforward.

In both cases, your equation should read like:

mass * acceleration = the sum of all forces = pull force - force of friction
or
m * a = F(pull) - F(friction) = F(pull) - u*F(normal) = F(pull) - u * m * g

Where F(normal) is the force which negates the gravitational pull, in both of these cases m*g.

In both cases, use algebra to solve for the unknown, insert the correct numbers, and voila.

The bowsprit got mixed with the rudder sometimes...
OneOther
Profile Blog Joined August 2004
United States10774 Posts
November 29 2008 19:15 GMT
#3
yeah, these are very straightforward...
I'll do the second one for ya.

friction force = coefficient * normal force
first, find out how much excessive force there is.

4.25N - (0.765 kg)(9.8 m/s^2) * (0.41) = force
then divide the answer by the mass, using Newton's F=ma law, to get the acceleration.
Archaic
Profile Blog Joined March 2008
United States4024 Posts
Last Edited: 2008-11-29 19:31:06
November 29 2008 19:23 GMT
#4
(I'm just going to use u as mu)
f = uFn
Fn is usually weight (w = mg)

1.
You are trying to find u.
The force on the sled is only horizontal, so it won't affect the weight.
Fn = 4.68 * 9.8 = 45.86N

Then
F = ma
9.13 = 4.68 * a
On a frictionless surface, the acceleration should be
9.13/4.68 = a = 1.95 m/s/s

He is going 0.815 with friction, so the friction would be lowering 1.95m/s/s to .815m/s/s

Friction is simply the force going backwards, in a way.
The pull is ->, friction would be <-

Then you plug it back into the formula:
F = 4.68*0.815
F = 3.81N

9.13 - f = 3.81
f = 9.13-3.81 = 5.32

Back to the friction formula:
5.32 = uFn
5.32 = u45.86
u = 5.32/45.86
u = 0.116

2.
F = ma
4.25 = 0.765a
f = uFn
f = 0.41*(.765*9.8)
f = 0.41*7.50
f = 3.07

4.25 - 3.07 = 0.765a
1.18 = 0.765a
a = 1.18/0.765
a = 1.542 m/s/s

Notice: I haven't done this in a while so it might not be exactly correct. Also, I know I didn't do this in the quickest way, but it helps to understand every aspect, rather than just to get the answer.

EDIT: Just wait for micronesia =D.
Klockan3
Profile Blog Joined July 2007
Sweden2866 Posts
Last Edited: 2008-11-29 19:29:30
November 29 2008 19:27 GMT
#5
On November 30 2008 03:54 il0seonpurpose wrote:
9.13-(u*98)=4.68(.815)

wtf is that one?
Firstly I hope you realize that g=9.8N/kg and not 98, and also it is not a force it is just a coefficient to find the force. The force is mg, or 9.8*4.68= force downwards which is the same as the normal force.
Thus you get
9.13-u*9.8*4.68=4.68(.815)
u=(9.13-4.68(.815))/(9.8*4.68)
RoieTRS
Profile Blog Joined July 2008
United States2569 Posts
November 29 2008 19:58 GMT
#6
YOUR MATH IS ALL WRONG!
YOU FORGOT TO CARRY THE ONE LOL!
konadora, in Racenilatr's blog: "you need to stop thinking about starcraft or anything computer-related for that matter. It's becoming a bad addiction imo"
Slayer91
Profile Joined February 2006
Ireland23335 Posts
Last Edited: 2008-11-29 20:47:10
November 29 2008 20:26 GMT
#7
Question 1:
2 Possible equations, forces parallel, and forces perpendicular. You should normally draw a diagram but thats retarded on a forum so meh. No resolving required so its an easy question.

1: R - W = 4.68(0) (No upward acceleration) g = 9.8 m/s^2, R = normal reaction, u = co of friction
==> R = mg
==> R = 4.68(9.8)
R = 45.864N

2: 9.13 - uR = 4.68a
==> 9.13 - u(45.864) = 4.68(.815)
==> 9.13 - 45.864u = 3.8142
==> 45.864u = 5.3158
u = 0.116

Question 2:
Again, 2 possible equations and no resolving required.

1: R = W (no upward acceleration)
==> R = .765(9.8)
R = 7.497N

....Total F....=ma
2: 4.25 - uR = ma
==> 4.25 - (.41)(7.497) = .765a
==>4.25 - 3.07377 = .765a
==>1.17623 = .765a
a = 1.573m/s^2

EDIT: Thats a rather long and confusing way of doing it Archaic makes more sense to draw a diagram/visualise and put all the forces in the same equation because thats how it will work for harder problems anyway.
............................^R
........................_____
_________uR<---|____|--->4.25N__________________
.............................v .765g
Thats a force diagram btw. See how hard it is to draw on forums T_T
IntoTheWow
Profile Blog Joined May 2004
is awesome32274 Posts
November 29 2008 23:18 GMT
#8
hahaha sick diagram slayer
Moderator<:3-/-<
HeavOnEarth
Profile Blog Joined March 2008
United States7087 Posts
November 29 2008 23:38 GMT
#9
http://www.glenbrook.k12.il.us/gbssci/Phys/Class/newtlaws/u2l2c.html
"come korea next time... FXO house... 10 korean, 10 korean"
il0seonpurpose
Profile Blog Joined January 2007
Korea (South)5638 Posts
Last Edited: 2008-11-30 05:51:37
November 30 2008 03:14 GMT
#10
micronesia
Profile Blog Joined July 2006
United States24676 Posts
December 01 2008 19:14 GMT
#11
Okay I'm not god. I haven't even been watching tl since thanksgiving haha
ModeratorThere are animal crackers for people and there are people crackers for animals.
Klockan3
Profile Blog Joined July 2007
Sweden2866 Posts
December 01 2008 20:49 GMT
#12
On November 30 2008 08:38 HeavOnEarth wrote:
http://www.glenbrook.k12.il.us/gbssci/Phys/Class/newtlaws/u2l2c.html

I do not really understand the fapping part though:
[image loading]
Empyrean
Profile Blog Joined September 2004
16984 Posts
Last Edited: 2008-12-01 21:18:50
December 01 2008 21:18 GMT
#13
So whenever I read the words "fapping" I think "masturbation," mainly since it's internet-speak for such action.

You really caught me off guard there.

EDIT: And to answer your question, it means the applied force.
Moderator
HeavOnEarth
Profile Blog Joined March 2008
United States7087 Posts
Last Edited: 2008-12-02 00:38:33
December 02 2008 00:38 GMT
#14
tutorials really help me a lot but
its such apain when ur trying to learn and the teaccher goes bwahaha idiot IDIOT
LEARN FASTER
or whenever u say something wrong teach justgoes...

LOLLLOLOL
"come korea next time... FXO house... 10 korean, 10 korean"
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