samoth plz help

[Sub-Zero]

A massive steel cable drags a 20 kg block across a horizontal, frictionless surface. A 100 N force applied to the cable causes the block to reach a speed of 4.0 m/s in a distance of 2.0 m.


What is the mass of the cable?

 

[youngguns]

I failed sex ed too man

 

[Raptorex]

33

 

[Sub-Zero]

33

can i get ur work plz

 

[samoth]

Mass of the cable? Hmm... I don't think I've ever ran into a problem taking into account the cable's mass.

I'd say start with Newton's second law and some force diagrams. Is this a calc problem? Mass of the entire cable, or mass per unit length of the cable?

 

[paradox]

http://www.ajdesigner.com/phpforce/force_equation_m.php

 

[ksharp01]

a who?


lol

 

[Raptorex]

diagram it, that is your answer

 

[youngguns]

a who?


lol

we know you can't

 

[Raptorex]

mass (m) = 25 kilogram

 

[ksharp01]

we know you can't

get off my jock old yeller!!

 

[Sub-Zero]

Mass of the cable? Hmm... I don't think I've ever ran into a problem taking into account the cable's mass.

I'd say start with Newton's second law and some force diagrams. Is this a calc problem? Mass of the entire cable, or mass per unit length of the cable?

i tried solving for F in negative direction going 9.81(Mc+20kg)=F1

then i did for for F2-F1=(20+Mc)(v^2/r)

so i did 100N-9.81(Mc+20)=(20+Mc)(8)

i keep getting negative answers,

does that even make sense?

fuck

 

[youngguns]

^^ Uh ya?

 

[Sub-Zero]

mass (m) = 25 kilogram

u sure?

 

[Raptorex]

u sure?

bro, you gotta do your work, I've been thru it 3 degrees, if you take the easy way out now, you will be fucked later on...I'm close.

 

[youngguns]

^^ Good post man, so good.
#1. Extremly helpful
#2. Easy way out for you
#3. Easy way out for everyone else

 

[Sub-Zero]

bro, you gotta do your work, I've been thru it 3 degrees, if you take the easy way out now, you will be fucked later on...I'm close.

so then would u mind explaining it to me instead of just giving me the answer...

either i fail the homework and dont understand it, or you can help me and i learn it and do well on the homework.

i know its not your job to teach me, but dont give me a dumbass lecture about your degrees and then give me wrong answers to try and fuck me over.

 

[paradox]

so then would u mind explaining it to me instead of just giving me the answer...

either i fail the homework and dont understand it, or you can help me and i learn it and do well on the homework.

i know its not your job to teach me, but dont give me a dumbass lecture about your degrees and then give me wrong answers to try and fuck me over.

Did you take a look at the link I posted for you?

 

[Raptorex]

so then would u mind explaining it to me instead of just giving me the answer...

either i fail the homework and dont understand it, or you can help me and i learn it and do well on the homework.

i know its not your job to teach me, but dont give me a dumbass lecture about your degrees and then give me wrong answers to try and fuck me over.

I wasn't trying to fuck you over, it was an "educated" guess...see paradox post.

 

[samoth]

Well, the distance is actually 1.96m, lol.

I've never done this kind of problem, so I'll have to think about it for a minute.

 

[Sub-Zero]

Did you take a look at the link I posted for you?

i did, i appreciate it but i know the equations and how to manipulate them as far as that link goes

 

[Sub-Zero]

Well, the distance is actually 1.96m, lol.

I've never done this kind of problem, so I'll have to think about it for a minute.

i think it wants me to take into acoun tan equal and opposite force at the connection point of the cable and the weight

 

[Raptorex]

mass (m) = 25 kilogram

Samoth, please check this answer and get back to Elroy for me.

 

[ksharp01]

why ask if you don't believe...google it....hehe

 

[Raptorex]

Example

3. A 20.0 kg mass is pulled by along a surface by a horizontal force of 100 N. Friction is 20.0 N. What is the acceleration of the mass?


Answer to Example 3:



Free body diagram (whenever there are two or more forces):

Sum of forces (vertical forces cancel as evidenced by lack of acceleration in the vertical dimension)
Fnet = T + Ff
Fnet = (100 N) + (-20.0 N) = 80 N


Similar problem here, but solve for mass...

Plus I said DIAGRAM THE FUCKER ALREADY

 

[iHulk]

ah, shit, according to your calculations the world is gonna end.

oh, nevermind, u forgot to carry the one

all is fine

 

[youngguns]

ah, shit, according to your calcualtions the world is gonna end.

oh, nevermind, u forgot to carry the one

all is fine

oh you are just adorable

 

[iHulk]

oh you are just adorable

ihulk spelled calculation wrong and you let it go YG.
u fucken completely missed the point
lol

 

[redguru]

You look at the force applied versus the acceleration and you should be able to find the total mass. Once you find the total mass moved then you can figure out the cables mass.

 

[samoth]

Samoth, please check this answer and get back to Elroy for me.

Looks like you're taking F=ma, but dividing the force by the velocity.

 

[redguru]

Oh, and MightyMouse is right, vector diagram every friggin problem.

 

[iHulk]

Looks like you're taking F=ma, but dividing the force by the velocity.

damm, I'm in AP physics all over again... I actually loved that class.
shit, that sounded nerdy

 

[Raptorex]

Oh, and MightyMouse is right, vector diagram every friggin problem.

Thanks, doing just the math without actually seeing the problem (via the diagram), will get you every time.

 

[samoth]

I'm stuck thinking there's an angle somewhere. What is this, statics? lol

I did get ~25 by reversing your F1 and F2 in the last equation, but that seems too odd, as taking F=mv would do the same.

 

[redguru]

First find acceleration of the block

dv = 4m/sec
dt = 2 sec



a = 2 m/sec^2

F = 100N

F = ma

 

[Raptorex]

I'm stuck thinking there's an angle somewhere. What is this, statics? lol

I did get ~25 by reversing your F1 and F2 in the last equation, but that seems too odd, as taking F=mv would do the same.

as would M= F/A

 

[youngguns]

^^ Correct

 

[redguru]

BTW, this is a very simple problem in comparison to the last one you asked. Same course?

 

[Raptorex]

Do I get my official "science forum" badge yet?

 

[samoth]

First find acceleration of the block

dv = 4m/sec
dt = 2 sec



a = 2 m/sec^2

F = 100N

F = ma

I thought he had to use (v^2)/r for accel? That would be 8 m/s^2.

 

[p0ink]

physics 1, baby.

 

[samoth]

as would M= F/A

Huh? F=mv=ma?

 

[redguru]

I thought he had to use (v^2)/r for accel? That would be 8 m/s^2.

lol

I'm drunk, gimme a break.

 

[youngguns]

Huh? F=mv=ma?

moderater

 

[Raptorex]

Mass=Force / Acceleration that is all I used...simple.

 

[samoth]

lol

I'm drunk, gimme a break.

lol

 

[redguru]

I thought time was given but it was dx

 

[samoth]

Mass=Force / Acceleration that is all I used...simple.

Yeah, N2L. But total mass is unknown. Acceleration (I'm assuming) needs to be taken via (v^2)/r = 8 m/s^2. How did you calculate the forces?

 

[samoth]

I thought time was given but it was dx

Ooooh, that's why you had dt=2. I was wondering about that, lol.

 

[Raptorex]

I used 100 N as stated

 

[redguru]

I assume elasticity is negligible?

 

[Raptorex]

frictionless surface

 

[samoth]

Sub, following your equations, you had F1=..., F2-F1=deltaF=..., then put the two together with F2=100N. What would happen if you took F1=100N instead? Is the choice arbitrary?

 

[samoth]

I used 100 N as stated

Won't follow.

RG: Mu sub k = 0.

 

[Subzeero]

i tried solving for F in negative direction going 9.81(Mc+20kg)=F1

then i did for for F2-F1=(20+Mc)(v^2/r)

so i did 100N-9.81(Mc+20)=(20+Mc)(8)

i keep getting negative answers,

does that even make sense?

fuck

are you sure they are asking for the mass of the rope? Or maybe the tension force in the rope. Do us all a favour and elaborate on what you have done here, and we will come up with a solution.

Define the following in your solution.

r, is this radius? Are you trying to figure the radius of the rope?
also all the units are missing. 9.81 what? N/kg??? Common man!
I also don't know what Mc is. So define that as well please.

And last please state the context of your question. Are you doing simple forces and free body diagrams in two dimensions or something a lot more complex?

 

[samoth]

r is actually delta r, the change in distance.

Mc is M sub c, mass of the cable.

9.8 is gravity.

 

[redguru]

You are given initial velocity, final velocity and distance traveled. There is a derived equation for acceleration

Then you just have to solve for mass of the cable

 

[redguru]

QED, bitches.

 

[redguru]

Can't believe I am doing this kid's homework.

 

[redguru]

Oh, and if an engineer wanted to know the mass of the cable, he'd do it empirically, not go through this rigmarole at midnight with a buzz.

 

[samoth]

Ah, (v^2)/2r.

I was thinking that it would make no sense if the rope was significantly heavy(er than the block). I didn't think to look up the accel from velocity independently.

Nice

 

[beefcake28]

Ah, (v^2)/2r.

I was thinking that it would make no sense if the rope was significantly heavy(er than the block). I didn't think to look up the accel from velocity independently.

Nice

Slacker...

 

[Subzeero]

r is actually delta r, the change in distance.

Mc is M sub c, mass of the cable.

9.8 is gravity.

Thanks. This is what i think Sub should do.

Find the total work done by the rope. This is easy.

W = Mass/2(V2(squared)-v1(squared) where initial velocity is zero.

but Total work done equals change in kinetic energy.

W = 1/2 Mass X Velocity(squared), use v2 for velocity.

Solve for mass

(2 X change in kinetic energy)/(V(squared)) = mass.

And thaddaahh... there you have it. I am confident that this is right.

There is so need to involve gravity, its cancelled by normal force.

 

[samoth]

Okay, I looked up accel/vel in Serway, and (v^2)/r is centripital/tangential velocity, depending on sign.

 

[redguru]

The only good thing about threads like these is it allows me to brush up on pre-calc physics I haven't used in years.

 

[redguru]

Okay, I looked up accel/vel in Serway, and (v^2)/r is centripital/tangential velocity, depending on sign.

Stop circling the problem.

 

[Subzeero]

so what do you think about my solution?

 

[samoth]

Stop circling the problem.

LOL!

Damn engineers!!

 

[redguru]

so what do you think about my solution?

I've already solved it, the proof is about 12 posts below. Yours is elegant, but over the scope of his course work so far.

 

[Subzeero]

I've already solved it, the proof is about 12 posts below. Yours is elegant, but over the scope of his course work so far.

thank you man.

I read your solution, simple and easy to understand.

I love physics!

 

[Lestat]

wow samote was stumped??

 

[Sub-Zero]

mightmouse had the right idea, total mass was 25, then subract the 20 from the block, gives u 5kg for the rope.

i thought it shoulda been something like 100N=m(8)

which gives you 12.5, which was the first thing i tried, but its gotta be multiplied by 2 for some reason???

theres alot more gay ass impossible problems if u guys are enjoying this

 

[Sub-Zero]

Ah, (v^2)/2r.

I was thinking that it would make no sense if the rope was significantly heavy(er than the block). I didn't think to look up the accel from velocity independently.

Nice

i see....

whys it over 2r? i thought it was just r, thats whats been fucking me over. whys it 2r instead of r?

 

[heatherrae]

oh oh oh!!!!


I know!



the answer is:







ELEPHANT!

 

[blut wump]

RedGuru has the answer. I was about to post it but clicking on his jpgs pretty much mirrored my own calculations.

5Kg.


The only formulae I bother to remember for this type of problem are:

F = m * a
v = u + a * t
s = (u + 0.5 * a * t) * t - or, equivalently, s = ut + 1/2 a t^2

s = distance, t = time, a = accel., v = velocity, u = initial velocity

With some algebra, other relevant equations can be derived.

 

[Raptorex]

Thank you, I will offer you some advice, there is a principal that works over and over, diagram a problem and NEVER make it more complicated than it needs to be, especially in the intro courses.

Good luck!

mightmouse had the right idea, total mass was 25, then subract the 20 from the block, gives u 5kg for the rope.

i thought it shoulda been something like 100N=m(8)

which gives you 12.5, which was the first thing i tried, but its gotta be multiplied by 2 for some reason???

theres alot more gay ass impossible problems if u guys are enjoying this

 

[pintoca]

lots of engineers in this thread

(and samoth)

 

[redguru]

For future reference, subz, here are the derived equations for constant acceleration in a straight line for any missing variable.