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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 =/
 
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Could you use a = V^2/r ?
although not sure what velocity would be
2pi/sec ?
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United States24734 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
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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 ?
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yayz i got the first one right :D
but the second one i hath failz
tried .60895/ 1.64285
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United States24734 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
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oh whoops lol
it just goes back to normal physics stuff after the circle stuff
thanks micro ! ^.^
*note to self Ff= u*Fn *
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As usual, physics help = micronesia.
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ahhs. ill pm ; updated OP
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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
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o.o it worked
now i feel bad for just copying an answer
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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
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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.
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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."
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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 +
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United States24734 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.
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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?
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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 =/
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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!
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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
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EPT
