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Tension of the pulley is calculated using........
newtons laws of motion
applying newtons law to the whole system force = mass * acceleration 6*g = 11 * a a = 6*9.8 /11 = 5.35 m s-2
now i know how to find the tension in the string considering the horizontal string applying newtons second law;- Tension = 10 * 5.35 = 53. 5 N
but you require tension in the pulley right?
i think its mg= 6*9.8=tension on the string since the is no effect of 5 kg wt. as surface is frictionless
sorry i dont know where i got 10 from!!!!
if you consider the horizontal string its 5 * 5.35 not 10 * 5.35
hhmm.. i'm not sure I would have thought you would have to take the 5 kg into account. Its been a long time since i did these...
but i'mm considering the whole system 5 kg and 6 kg are part of the whole system
6 g is the downward force (dues to 6 kg mass) but 5kg is part of the mass
the tension will be the same throughout the whole string so i guess thats also the force on the pulley
5kg would have created friction on the surface due to wt. which would hav then created a resistance to allow the whole system to move down....this would hav decreased the tension on the string but here the surface is frictionless...
hmm maybe but i dont think you can ignore the 5 kg...
maybe i'm wrong
I'd have to look it up
https://www.youtube.com/watch?v=e2-yHHSzjTs - this is about a similar problem
no effect of the 2 forces on each other due to surface is frictionless...
if you check out the above link you'll see i'm right The 5kg has to be taken into account
there is a tension between the 2 masses - the tension in the string
may be you are right....i got confused...i hav to think again on it....
its a tricky subject I wasn't sure but the link gives a good explanation.
No you didn't waste my time - I had to make sure I was right also!!
Hey guys thanks for all your insights
This kind is one of the most counterintuitive
Therefore so far to our best of knowledge tension 11kg*5.35m/^2
I am crying right now
yes - its s a bit hard to visualive - If the mass of the object on the table was say 15 kg there would still be motion because the table is frictionlass
according to inertia that is.
So it naturally falls upon us to calculate the acceleration and corresponding weight of the whole system
Because tension is responsible for holding two masses which sum up to kg. Tension holds them from falling apart after all. Given that one side of the tension is pulling the system both masses are involved in the motion of the whole system.
You established that "connection" with me here. Universal law of affection of physics applies to all of usXD
we've found that the acceleration is 5.35 m s-2 but now we have to find the tension
Isn't the tension 11kg*5.35m/s^2?
But tension on the pulley... I don't know
that's the hardest part of this question is "tension on the pulley" instead of tension of the line.
well the units of the tension will be N Newtons
* will be in Newtons
see what i hav checked in my book, (m1+m2)a = m2*g T= m2*g - m2*a
yea that looks good
Your equation assumes that overall force required to move the entire system is equivalent to the force exerted by a mass due to gravity.... Ok. This holds true on frictionless surface I suppose.
*overall force required to accelerate
so its 5 * 5.35 N?
11*5.35m/s^2 assumes normal force is ignored?
yes normal force acts perpendicular to motion so can be ignored
No I think you should consider the "tension" which is induced by the 6kg mass as well. To assume there is only 5kg being responsible for the tension is absurd because it's not the only factor pulling the line.
People say he's a physicist so
can we apply newtons second law to the 5 kg mass in isolation? in which case we can say T = 5 * 5.35 = 26.75 N ?
- why can't we do that?
- that gives the tension in the horizontal string
The prospect of assuming the tension to be "stretched" is somehow absurd?
I think that reason why 6kg object is considered is because it's the source of force.
I'll have to look at another video lol! I'm not saying you guys are wrong but i'm not convinced.
I;ll give you a medal for saying that lol!
So it's not the initial 11kg*5.35m/s^2
but rather the second one.
I'd like to giv medal to Robert also but it wont let me...
Thanks so much physics aficionados I think I understand it now.
yea the more i think about it.... the horizontal tension pulls the 5 kg mass so T must = mass * a = 5 * 5.35
the more I think about it the better it settles in my brainXD
but thats the tension in string what about the pulley? T acts directly on the pulley so that makes T on pulley = 26.75 N also?
Right the tension is essentially moving the 5kg mass on frictionless surface so the tension must only account for the 5kg and greater acceleration implies proportionally valid force being applied. Ha! I think that's our ultimatum
well guys we've learnt something with this session Good
The pulley is more like the center of this system at 90 degree angle. What if it was 180 degrere angle we all know to be the same?
Since frictionless the direction must not matter ??
Mechanical advantage says it takes one half of force if pulley is used in two fold like manner. But I guess this doesn't concern this partiuclar problem...
Its coming back to me now - after all these years - I wont say how many (lol!!)
Physics is the protein bar for the intelligence
Where intuition fails all the time. problem of this kind makes me reflect on my life from more logical poin of view.
thanks all bye!
the pulley must also be frictionless for the tension throughout the string to be uniform
uniform acceleration that is
Where veocity time graph would have a positive slope.
Do you remember that stuff??
some of it yes acceleration = rate of change of velocity = slope
and distance traveled = area under the graph
uniform velocity is indicated by a straight line with a positive slope
* sorry that's uniform acceleration
yeah whereas friction would prevent that
One quick recap on this thread; when we did the calculation on acceleration of the whole system we used the sum of both masses.... is that accurate?
As in 6.0kg*9.8m/s^2=11a as opposed to 5a just like what we did in the last counterpart.
The object being acted upon is the 5.0kg mass accelerating... hmm my brain is such a pelletty equipment.
yes we dealt with the whole system as we have no friction it was just like everything was horizontal with the force of gravity 6g acting on whole system and a weightless rigid body connecting the 2 masses
the double arrows indicate acceleration
the force acting is 6g mass * acceleration due to gravity
Ok. I get it. But how's that related to having only the 6kg object as being responsible for the entire system? I feel like 5.0kg is being "pulled" rather than moving "together" with the 6.0kg object. HmmmI guess I am really dumb
the force has to accelerate the total mass which is 11 kg
yes the 5 kg is being pulled by the tension in the string which is caused by the downward force of the 6 kg weight. I f there was no weight on the table the acceleration would be much greater . The 5 kg slows the whole system down. bit it is part of the system .
If its weight was say 50 kg the system would still be in motion but acceleration would be a lot less.
It is intuitive to a degree because the heavier the mass the acceleration would be a lot slower - because the force is constant at 6 g N but the mass is greater
you can also find the tension in the string by considering the mass of 6 kg without referring to the 5 kg |dw:1439398934381:dw|
Tension can be calculated with 6kg using the acceleratoin?
the tension is working against the force 6g so the net force = 6g - T applying the second law 6*9.81 - T = 6 * 5.35 T = 26.76 N
yes its valid to consider both masses separately
in fact we could have found T and a by using 2 system of equations 6.g - T = 6 a - for the g kg mass T = 5a - for the 5 kg mass
so we could have found T and a together
adding we get 6g = 11a a = 6g/11 = 5.35 then T = 5 8 5.35 = 26.75
* 5 * 5.35 = 26.75
its all coming back now
We all strive to find that momentXD Nostalgia old schoolXD
Yea we had a very good math teacher. Our school was a very good one one - in Swansea UK. II guess its most famous pupil was Dylan Thomas the poet
Awesome! I miss going to american schools:)
Not that he was any good at Math. He was exceptional at English and cared nothing about anything else.
yea At the time you didn't realize how good those days were.
Have you ever heard recordings of Dylan Thomas reciting poetry? Some people don't like the way he reads them others find it awesome.
Dylan Thomas don't gentle into that good night. Old age should rage at close of day. Rage rage rage into that good night.
yea that's his most famous poem.
It's so true though.
Well id better feed the cat. She's in charge!! Yes. His father was the English teacher in the school and its to him the poem was written.
Yeah every time I read his poem it makes me regret and realize so much I missed in life
I guess I am supposed to be raging and raving but every day seems o be the reality of inevitable time passing.
I wish I could have more emotions at work and assign a meaning to every day each day until I die. I guess travelling seems to be the best option for all of us.
I think we waste a lot of time thinking about what we are going to do but never do it. Yea If you can afford it travel is great.
nice talking to you must go
Likewise! Thank so much for your help:) See you around.