|
So our teacher gave us this website to complete assignments
 it's not entirely hard or anything, but we havn't covered circular motion and stuff in physics C yet,
this is off an assignment about newton's laws and forces/ kinematics
so i don't really have an idea of how to approach this problem
D:
I figure you convert the 130 g to kg
then use F=ma to get the force, but i don't know how the circular motion applies to all of that
A 130 g coin sits on a horizontally rotating
turntable. The turntable makes one revolu-
tion each 1 s. The coin is located 14 cm from
the axis of rotation of the turntable.
The acceleration of gravity is 9.81 m/s2 .
What is the frictional force acting on the
coin? Answer in units of N.
Part 2
The coin will slide off the turntable if it is
located more than 23 cm from the axis of
rotation.
What is the coefficient of static friction?
any help would be great
;edit;
also Sorry to bump/ ask another hw question but im stuck again
A crate of books is to be put on a truck with
the help of some planks sloping up at 31◦. The
mass of the crate is 70 kg, and the coefficient
of sliding friction between it and the planks is
0.4. You and your friends push horizontally
with a force ~F .
The acceleration of gravity is 9.81 m/s2 .
Once the crate has started to move, how
large must F be in order to keep the crate
moving at constant speed? Answer in units of
kN.
i did .4 x F = sin(31) x 70 x 9.81, solving for F i get 884.192 N/ 1000
= .884192 kN
however it's wrong
so im confused
-solved! answer .9~
edit;
i want help drawing free body diagrams/pictures for forces and knowing when to apply kinematics , also for example, the problem states " this motion starts from rest" , then u assume Vo or Vi is 0.
i would still appeiciate any form of physics whatsoever, tips and help drawing digrams, overall picture, mindset, some practice problems - recommend books/ websites
the main goal for me is to get a 4,5 on the AP physics C exam
im not really signed up in the UT university, i don't even go there, i just enrolled in an online class and as you can tell, having difficulties learning the material.
;edit; help
Likewise,
A tired worker pushes with a horizontal force
of 400 N on a 100 kg crate initially resting on a
thick pile carpet. The coefficients of static and
kinetic friction are 0.6 and 0.2, respectively.
The acceleration of gravity is 9.81 m/s2 .
Find the frictional force exerted by the car-
pet on the crate. Answer in units of N.
Ok, i know Ff = u*Fn
I know F=ma
Fg = mg
and i can picture the forces 400->>>> horizontally, and 100kg, 9.81 downwards
, now the coefficients confuse me =/
 
|
Could you use a = V^2/r ?
although not sure what velocity would be
2pi/sec ?
|
United States24612 Posts
Part 1: Overall we know Fnet=ma. The forces acting on it in the x direction are the force of friction inward, and nothing else. Is it accelerating? Yes, since it's going in a circle. So, Fnet is Ff. Hopefully you will be able to figure out what to do.
Part 2: Similar to part 1, except you plug numbers in differently... if you get part 1 you should get part 2 pretty soon after.
Hints:
Yes convert the mass to kg. Also convert cm to m.
Ac = v^2/r yes. Since F=ma you can say Fc = mv^2 / r
Speed is equal to distance over time, or circumference of circle over period :D
|
oh circumference lool im an idiot
so convert to (.14m) x 2 (pi)/1second = v = .879646
v^2/.14 = a = 5.52698
a x ( .13kg ) = F
F= .718507
then do it again , except with .23 m and then divide those totals = co efficient of friction ? D:
which i got 1.1804
so co effecient = .60895 ?
|
yayz i got the first one right :D
but the second one i hath failz
tried .60895/ 1.64285
|
United States24612 Posts
Part 1 looks good.
For Part 2, you get the force of friction is 1.18 so 1.18=u*Fn and you get .93 for u since Fn = mg = .130*9.8
|
oh whoops lol
it just goes back to normal physics stuff after the circle stuff
thanks micro ! ^.^
*note to self Ff= u*Fn *
|
As usual, physics help = micronesia.
|
ahhs. ill pm ; updated OP
|
ma = Fpush - Fg - Ff
Ff = .4*(cos31)*686.7 = 235.4
Fg = 686.7*sin31 = 353.7
a = 0
0 = Fpush - 353.7 - 235.4
Fpush = 589.1 N
edit: Ah crap it's 9.81 ~_~
edit2: ah crap you're pushing horizontally
Fpush = Fhor - Ff
589.1 = Fhor*cos31 - .4*Fhor*sin31
589.1 = 0.651*Fhor
Fhor = 905 N
or .905 kN
|
o.o it worked
now i feel bad for just copying an answer
|
although you used basic.. F = ma then got the Ff ..
i think im drawing the picture wrong 
hmm alright now i get why you used sin for Fg and cos for Ff, made the triangle
whoa where'd Fpush and Fhor come from/ mean / represent
ohhh hahaha Fhor means horizontal, I'm used to just writing Fn
I'm still not sure what Fpush represents
|
The coefficient of friction (µ) just gives us the maximum force of friction per normal force acting on an object.
This means that F(Slide)=µ*F(Normal) if F(Slide) is the lowest amount of force required to make the object slide. This of course assumes that everything is hitting in clear angles, in more complicated problems you have to use your head instead.
To solve general powers you usually first look at the object of interest isolated from everything else, and see what forces are acting on it. If the object is lying still the sum of forces and momentums must always be zero. In this case of the rotating wheel you can instead look it from the point of view of the coin, there it lies completely still. Now since the reference frame is accelerating you just put the negative of that as an extra acceleration on the coin, just like as if it were math and you substitute in integrals you have to make up for the change of reference frame, after that this is just the definition of friction coefficient.
|
thanks! now to um apply that with more examples
ahh more hw problems D::: i ran out
like this
Is it accelerating? Yes, since it's going in a circle. So, Fnet is Ff.
why?
If the object is lying still the sum of forces and momentums must always be zero.
does that mean the force is always 0 ? , or is it negative... momentum negative, how does this piece of logic apply to solving equations and problems?
To solve general powers you usually first look at the object of interest isolated from everything else, and see what forces are acting on it.
i really need help with this, i've been trying to draw these "free body diagrams" like my teacher is showing the class, but to no avail, 'Hey, look a drawing... what do i do with it now."
|
Let's say, just for example, to anaylze the force there is
A box weighing 630 N is pushed along a hor-
izontal floor at constant velocity with a force
of 250 N parallel to the floor.
alright, you got a force going down, due to weight, and the force due to gravity
you got a force, going across the floor, let's call this Fpush... of 250N
did i miss anything?
alright let's say it asks
What is the coefficient of kinetic friction
between the box and the floor?
Ff = Fn x u ... so solve for coefficient ?
please point out where my logic stopped, i don't want a numeric answer, i just want to understand physics
+ Show Spoiler +
|
United States24612 Posts
On October 27 2008 14:21 HeavOnEarth wrote:thanks! now to um apply that with more examples ahh more hw problems D::: i ran out like this why?
In order for an object to be accelerating, its velocity must be changing. You can change velocity by accelerating forwards (increasing speed), or accelerating backwards (deceleration; decreasing speed). However, if you go in a circle, your speed is constant, but the direction of the velocity is still changing... so you are accelerating.
|
On October 27 2008 14:36 micronesia wrote:Show nested quote +On October 27 2008 14:21 HeavOnEarth wrote:thanks! now to um apply that with more examples ahh more hw problems D::: i ran out like this
Is it accelerating? Yes, since it's going in a circle. So, Fnet is Ff.
why? In order for an object to be accelerating, its velocity must be changing. You can change velocity by accelerating forwards (increasing speed), or accelerating backwards (deceleration; decreasing speed). However, if you go in a circle, your speed is constant, but the direction of the velocity is still changing... so you are accelerating.
what if you go in a circle, while twisting on an axis
do you go supernova?
|
Likewise,
A tired worker pushes with a horizontal force
of 400 N on a 100 kg crate initially resting on a
thick pile carpet. The coefficients of static and
kinetic friction are 0.6 and 0.2, respectively.
The acceleration of gravity is 9.81 m/s2 .
Find the frictional force exerted by the car-
pet on the crate. Answer in units of N.
Ok, i know Ff = u*Fn
I know F=ma
Fg = mg
and i can picture the forces 400->>>> horizontally, and 100kg, 9.81 downwards
, now the coefficients confuse me =/
|
On October 27 2008 14:36 micronesia wrote:Show nested quote +On October 27 2008 14:21 HeavOnEarth wrote:thanks! now to um apply that with more examples ahh more hw problems D::: i ran out like this
Is it accelerating? Yes, since it's going in a circle. So, Fnet is Ff.
why? In order for an object to be accelerating, its velocity must be changing. You can change velocity by accelerating forwards (increasing speed), or accelerating backwards (deceleration; decreasing speed). However, if you go in a circle, your speed is constant, but the direction of the velocity is still changing... so you are accelerating.
oh right... direction
heh i was just under the impression the speed had to get higher/lower
zomg speed derivative = acceleration
i blame calculus for making me forget all the basic stuff =[
back to the vectors of my ap physic C book for me!
|
United States2789 Posts
On October 27 2008 14:32 HeavOnEarth wrote:Let's say, just for example, to anaylze the force there is A box weighing 630 N is pushed along a hor- izontal floor at constant velocity with a force of 250 N parallel to the floor. alright, you got a force going down, due to weight, and the force due to gravity you got a force, going across the floor, let's call this Fpush... of 250N did i miss anything? alright let's say it asks What is the coefficient of kinetic friction between the box and the floor? Ff = Fn x u ... so solve for coefficient ? please point out where my logic stopped, i don't want a numeric answer, i just want to understand physics + Show Spoiler +
You need to know the force of friction or some other piece of information in order to solve it I think.
edit:
It helps to draw all of the forces. One force is, as you mentioned, the force due to gravity BY the earth ON the box (630N). Another force (though it's not a real force, consider it that when drawing diagrams), is Force Normal BY the floor, ON the box (630N). That's why the box isn't accelerating straight downwards right? Because something is pushing it back up, which is force normal (one of newton's laws...an applied force must have an equal and opposite force or something like that). The obvious force is the 250 N, which is BY pusher, ON the box. The last force is the force of friction, which is BY the floor, ON the box. So now you have all the forces. Do this for every force problem, even if it's easy.
In the case of this problem, they're not really asking for a net force or anything like that, they're just asking for the coefficient. So in that case, knowing that Fn (or Force Normal) is the force of the surface pushing back on the object, you know that Fn is 630N. If you knew the force of friction you would just solve for u.
Sorry if I don't make sense.
edit: bleh but I didn't read that it was moving at a constant velocity. So that means that Ff=the 250N as well, so the net force on everything is 0. u=250/630 N
|
On October 27 2008 13:19 HeavOnEarth wrote:although you used basic.. F = ma then got the Ff .. i think im drawing the picture wrong hmm alright now i get why you used sin for Fg and cos for Ff, made the triangle whoa Where'd Fpush and Fhor come from/ mean / represent ohhh hahaha Fhor means horizontal, I'm used to just writing Fn I'm still not sure what Fpush represents
Fpush is if you're pushing diagonally ( didn't realize you were pushing horizontally x_x)
I used F = ma because I'm still in B 
|
On October 27 2008 14:32 HeavOnEarth wrote:Let's say, just for example, to anaylze the force there is A box weighing 630 N is pushed along a hor- izontal floor at constant velocity with a force of 250 N parallel to the floor. alright, you got a force going down, due to weight, and the force due to gravity you got a force, going across the floor, let's call this Fpush... of 250N did i miss anything? alright let's say it asks What is the coefficient of kinetic friction between the box and the floor? Ff = Fn x u ... so solve for coefficient ? please point out where my logic stopped, i don't want a numeric answer, i just want to understand physics + Show Spoiler +
ma = Fpush - Ff
|
United States24612 Posts
On October 27 2008 14:38 travis wrote:Show nested quote +On October 27 2008 14:36 micronesia wrote:On October 27 2008 14:21 HeavOnEarth wrote:thanks! now to um apply that with more examples ahh more hw problems D::: i ran out like this
Is it accelerating? Yes, since it's going in a circle. So, Fnet is Ff.
why? In order for an object to be accelerating, its velocity must be changing. You can change velocity by accelerating forwards (increasing speed), or accelerating backwards (deceleration; decreasing speed). However, if you go in a circle, your speed is constant, but the direction of the velocity is still changing... so you are accelerating. what if you go in a circle, while twisting on an axis do you go supernova?
Lol what? Serious question nestled in there?
Earth?
|
It helps to draw all of the forces. One force is, as you mentioned, the force due to gravity BY the earth ON the box (630N). Another force (though it's not a real force, consider it that when drawing diagrams), is Force Normal BY the floor, ON the box (630N). That's why the box isn't accelerating straight downwards right? Because something is pushing it back up, which is force normal (one of newton's laws...an applied force must have an equal and opposite force or something like that). The obvious force is the 630 N, which is BY pusher, ON the box. The last force is the force of friction, which is BY the floor, ON the box. So now you have all the forces. Do this for every force problem, even if it's easy.
oh right, 3rd law, heh
Yeah i'll try and kill lots of trees next homework session
In the case of this problem, they're not really asking for a net force or anything like that, they're just asking for the coefficient. So in that case, knowing that Fn (or Force Normal) is the force of the surface pushing back on the object, you know that Fn is 630N. If you knew the force of friction you would just solve for u.
Sorry if I don't make sense.
oh it made perfect sense, Fn is the perpendicular force, so it's Fn, i had more problems identifying the Ff in that problem however.
|
United States2789 Posts
The mgcos(theta) is calculated because it is equal to the Fn, which is the force by the surface on the block. The mgsin(theta) is the force that it is sliding down the slope thanks to gravity.
edit: So the net force of the block sliding down is mgsin(theta)-Ff where Ff=Fn * mgcos(theta). So the force needed to keep it from moving is a force in the opposite direction with the same magnitude.
edit2: crap didn't read question, lemme read it again.
edit3: oh ok, so it's started to move, and you don't want it to accelerate. In that case, you just want the net force to be 0 down the slope so that there's no acceleration. So yea, you have the hypotenuse, which is the same as the net force of the block going down, you have the angle, so all you have to do is use trig to solve for the horizontal component.
|
hehe ur adjacent side looks bigger than the hypotenuse
|
On October 27 2008 14:32 HeavOnEarth wrote:
Let's say, just for example, to anaylze the force there is
A box weighing 630 N is pushed along a hor-
izontal floor at constant velocity with a force
of 250 N parallel to the floor.
Since it is not accelerating the sum of the force vectors must be zero. Have you done linear algebra? F(Gravity) is F(G)(0,0,-1), F(Normal) is F(N)*(Unit vector that is orthogonal against the surface), F(Push) is just a constant force in any direction(Usually given in the example), F friction is always parallel to the plane and always in the opposite direction of the sum of other forces.
Now assume that the plane is orthogonal to the force of gravity and that the pushing force is is parallel to the plane. We know F(G) is in the (0,0,-1) direction and that no other forces are acting in the z-axis except F(N), which since it is also orthogonal to the plane is in the direction (0,0,1). As such we can deduce that F(N)=F(G) and it is easy to see that the sum of them are 0.
To make it easy just assume that the pushing force is acting with a Force f along the positive x-axis, then we get f(1,0,0), since we only got a single force left friction must be f(-1,0,0), Now since the object is moving the friction force is always |friction|=µ|F(N)| which gives µ=|friction|/|F(N)|.
Anyhow, the hard part comes when you have forces which are not orthogonal. For example if we have a tilted plane F(N) could go in the (0,-sina,cosa) direction were a is the tilt degree around the x-axis, and thus (If every other force is orthogonal to the G force, which is usually not the case) cosa*F(N)=F(G) as long as the object is not accelerating.
However if we have an acceleration along the plane we can add the negative of that force on the force equations.
Hmm, maybe I am going overboard...
Anyway doing problems like this is all about linear algebra. And F=ma just means that the force vector is equal to the acceleration vector divided by the mass. If the acceleration is orthogonal to the velocity no change of speed is made. For example if I go 100m/s north, and accelerates 10m/ss to the east in this instant my speed do not change. Speed is equal to the length of the velocity vector, which is sqrt(A^2+B^2+C^2) were A,B,C are the components of the velocity. Now if we take the a-derivative of that we get 2*A*(lots of shit), assuming that the other terms are not zero and that A is zero we get that a small change of A do not change the speed if the A component of the velocity vector is zero. That goes for all directions since you can always transform the components of each vector to a new coordinate system were that direction would be one of the unit vectors.
However the vector still changes, and thus the velocity, but the length do not.
|
whoaa wall of confusing* text o,o gimme a sec
|
By the way, if you want it in simpler terms just do one equation for every dimension of freedom like this:
Sum of forces in x-axis=(Acceleration in the x-axis)*mass
Sum of forces in y-axis=(Acceleration in the y-axis)*mass
Sum of forces in z-axis=(Acceleration in the z-axis)*mass
Are you doing momentum yet? Then you also need to check those equations to see how much its rotation is accelerating/deaccelerating.
|
Since it is not accelerating the sum of the force vectors must be zero. Have you done linear algebra? F(Gravity) is F(G)(0,0,-1), F(Normal) is F(N)*(Unit vector that is orthogonal against the surface), F(Push) is just a constant force in any direction(Usually given in the example), F friction is always parallel to the plane and always in the opposite direction of the sum of other forces.
Fg ,down on the z axis
Fn, perpendicular = orthogonal ?
Fpush, kk noted
Ff, noted
Now assume that the plane is orthogonal to the force of gravity and that the pushing force is is parallel to the plane. We know F(G) is in the (0,0,-1) direction and that no other forces are acting in the z-axis except F(N), which since it is also orthogonal to the plane is in the direction (0,0,1). As such we can deduce that F(N)=F(G) and it is easy to see that the sum of them are 0.
alright, Fn is just pushing back on Fg so they're equal (in magnitude )?
To make it easy just assume that the pushing force is acting with a Force f along the positive x-axis, then we get f(1,0,0), since we only got a single force left friction must be f(-1,0,0), Now since the object is moving the friction force is always |friction|=µ|F(N)| which gives µ=|friction|/|F(N)|.
alright friction moves against Fpush , however, it's not nessecary equal , is it? unless it comes o a complete stop, even then friction might > force, unless then co-oridinate plane you're writing just represents the direction and not magnitude, then nevermind heh.
|
On October 27 2008 15:17 Klockan3 wrote:
By the way, if you want it in simpler terms just do one equation for every dimension of freedom like this:
Sum of forces in x-axis=(Acceleration in the x-axis)*mass
Sum of forces in y-axis=(Acceleration in the y-axis)*mass
Sum of forces in z-axis=(Acceleration in the z-axis)*mass
Are you doing momentum yet? Then you also need to check those equations to see how much its rotation is accelerating/deaccelerating.
z - axis = Fg+Fn ,
x axis = Ff+ Fpush and
y axis = ?
|
United States2789 Posts
Personally, I think it's more confusing it learning to think about it in that way right away.
|
On October 27 2008 15:23 HeavOnEarth wrote:Show nested quote +On October 27 2008 15:17 Klockan3 wrote:
By the way, if you want it in simpler terms just do one equation for every dimension of freedom like this:
Sum of forces in x-axis=(Acceleration in the x-axis)*mass
Sum of forces in y-axis=(Acceleration in the y-axis)*mass
Sum of forces in z-axis=(Acceleration in the z-axis)*mass
Are you doing momentum yet? Then you also need to check those equations to see how much its rotation is accelerating/deaccelerating. z - axis = Fg+Fn , x axis = Ff+ Fpush and y axis = ?
y axis was zero in my example.
And the important thing to note here is that if the object is not accelerating then all of these must equal zero, no exceptions at all.
On October 27 2008 15:27 nevake wrote:
Personally, I think it's more confusing it learning to think about it in that way right away.
He should already have learned doing it with only one dimension of freedom since these problems needs vectors.
|
Anyhow, the hard part comes when you have forces which are not orthogonal. For example if we have a tilted plane F(N) could go in the (0,-sina,cosa) direction were a is the tilt degree around the x-axis, and thus (If every other force is orthogonal to the G force, which is usually not the case) cosa*F(N)=F(G) as long as the object is not accelerating.
AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
damn, spamming caps didn't make it easier to understand
However if we have an acceleration along the plane we can add the negative of that force on the force equations.
Hmm, maybe I am going overboard...
Anyway doing problems like this is all about linear algebra. And F=ma just means that the force vector is equal to the acceleration vector divided by the mass. If the acceleration is orthogonal to the velocity no change of speed is made. For example if I go 100m/s north, and accelerates 10m/ss to the east in this instant my speed do not change.
not understanding that, why does it not change? actually , what does orthogonal mean heh
Speed is equal to the length of the velocity vector, which is sqrt(A^2+B^2+C^2) were A,B,C are the components of the velocity.
I learned this one a while back.
Now if we take the a-derivative of that we get 2*A*(lots of shit), assuming that the other terms are not zero and that A is zero we get that a small change of A do not change the speed if the A component of the velocity vector is zero. That goes for all directions since you can always transform the components of each vector to a new coordinate system were that direction would be one of the unit vectors.
However the vector still changes, and thus the velocity, but the length do not.
with... X -> V - > A deriv. / 2nd deriv. / intergals ?
i shall balance redox and half reactions and turn on my memorization hacks for chem solubility/oxidation rules as i waits :O chem is so easy compared to physics ^.^
|
I take it as you have not learned linear algebra, everything gets much easier if you had done that.
http://en.wikipedia.org/wiki/Orthogonality
In other words, orthogonal is just a more general term for perpendicular.
Vector transformations are done by applying an orthogonal matrix to the whole field, but since you have not done linear algebra you wont understand that...
Sorry, but I thought that maybe you would have read such things when you did a physics question like this. I assume that you only work with forces that are perpendicular then, in that case all you would need are those 3 equations I mentioned earlier.
|
I take it as you have not learned linear algebra, everything gets much easier if you had done that.
it's not that i didn't learn it at all, it's just our teacher just gives us the essentials to solve the problem , shortcuts to calculus, then proceeds in teaching us physic II
is this a good site to learn linear algebra from?
http://www.egwald.ca/linearalgebra/vectors.php
|
On October 27 2008 15:44 HeavOnEarth wrote:Show nested quote +
I take it as you have not learned linear algebra, everything gets much easier if you had done that.
it's not that i didn't learn it at all, it's just our teacher just gives us the essentials to solve the problem , shortcuts to calculus, then proceeds in teaching us physic II is this a good site to learn linear algebra from? http://www.egwald.ca/linearalgebra/vectors.php
I need to go now, but I can help you when I get back. Since I live in sweden my timezone is a bit different :p
|
good night or stuffs `
ill be here in about 15hours, right nows.. 2am so ill get home at around 5
|
On October 27 2008 14:38 HeavOnEarth wrote:
Likewise,
A tired worker pushes with a horizontal force
of 400 N on a 100 kg crate initially resting on a
thick pile carpet. The coefficients of static and
kinetic friction are 0.6 and 0.2, respectively.
The acceleration of gravity is 9.81 m/s2 .
Find the frictional force exerted by the car-
pet on the crate. Answer in units of N.
Ok, i know Ff = u*Fn
I know F=ma
Fg = mg
and i can picture the forces 400->>>> horizontally, and 100kg, 9.81 downwards
, now the coefficients confuse me =/
|
On October 27 2008 16:05 HeavOnEarth wrote:Show nested quote +On October 27 2008 14:38 HeavOnEarth wrote:
Likewise,
A tired worker pushes with a horizontal force
of 400 N on a 100 kg crate initially resting on a
thick pile carpet. The coefficients of static and
kinetic friction are 0.6 and 0.2, respectively.
The acceleration of gravity is 9.81 m/s2 .
Find the frictional force exerted by the car-
pet on the crate. Answer in units of N.
Ok, i know Ff = u*Fn
I know F=ma
Fg = mg
and i can picture the forces 400->>>> horizontally, and 100kg, 9.81 downwards
, now the coefficients confuse me =/
I would guess that 0.6 is the static friction and 0.2 is the kinetic friction.
Since the crate is initially resting and 400 N is not enough to go above the static friction the answer must be: ???????
Edit: And for what to read to learn the linear algebra needed, read roughly everything in this that is not about matrixes:
http://en.wikipedia.org/wiki/Dot_product
The dot product is usefull all the time in mechanics since you need to constantly project forces on to surfaces or split them up to get the equations straight etc.
Might be a bit abstract but if you plan on doing something that requires college level math then there is no reason not to learn about it now, it is a powerful tool at the same time as the definition of it is really simple.
|
On October 27 2008 14:51 micronesia wrote:Show nested quote +On October 27 2008 14:38 travis wrote:On October 27 2008 14:36 micronesia wrote:On October 27 2008 14:21 HeavOnEarth wrote:thanks! now to um apply that with more examples ahh more hw problems D::: i ran out like this
Is it accelerating? Yes, since it's going in a circle. So, Fnet is Ff.
why? In order for an object to be accelerating, its velocity must be changing. You can change velocity by accelerating forwards (increasing speed), or accelerating backwards (deceleration; decreasing speed). However, if you go in a circle, your speed is constant, but the direction of the velocity is still changing... so you are accelerating. what if you go in a circle, while twisting on an axis do you go supernova? Lol what? Serious question nestled in there? Earth?
you'll know when to take me seriously
|
On October 28 2008 00:28 Klockan3 wrote:Show nested quote +On October 27 2008 16:05 HeavOnEarth wrote:On October 27 2008 14:38 HeavOnEarth wrote:
Likewise,
A tired worker pushes with a horizontal force
of 400 N on a 100 kg crate initially resting on a
thick pile carpet. The coefficients of static and
kinetic friction are 0.6 and 0.2, respectively.
The acceleration of gravity is 9.81 m/s2 .
Find the frictional force exerted by the car-
pet on the crate. Answer in units of N.
Ok, i know Ff = u*Fn
I know F=ma
Fg = mg
and i can picture the forces 400->>>> horizontally, and 100kg, 9.81 downwards
, now the coefficients confuse me =/
I would guess that 0.6 is the static friction and 0.2 is the kinetic friction. Since the crate is initially resting and 400 N is not enough to go above the static friction the answer must be: ??????? Edit: And for what to read to learn the linear algebra needed, read roughly everything in this that is not about matrixes: http://en.wikipedia.org/wiki/Dot_productThe dot product is usefull all the time in mechanics since you need to constantly project forces on to surfaces or split them up to get the equations straight etc. Might be a bit abstract but if you plan on doing something that requires college level math then there is no reason not to learn about it now, it is a powerful tool at the same time as the definition of it is really simple.
thanks :0
the class didn't even attempt this UT based thing except for like 3 of us, so it's ok.
i thought i just failed it horribly, but no one knows whats going on for the most part
our teacher is like "sighh i have to teach"
|
|
|
|
|
|
EPT![[image loading]](http://img530.imageshack.us/img530/5513/physda5.jpg)
