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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
  
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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.
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United States10774 Posts
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.
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(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.
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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)
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YOUR MATH IS ALL WRONG!
YOU FORGOT TO CARRY THE ONE LOL!
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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
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is awesome32277 Posts
hahaha sick diagram slayer
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United States24750 Posts
Okay I'm not god. I haven't even been watching tl since thanksgiving haha
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I do not really understand the fapping part though:
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17025 Posts
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.
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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
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EPT![[image loading]](http://www.glenbrook.k12.il.us/gbssci/Phys/Class/newtlaws/u2l2c1.gif)
