My first Mechanics Problem

Classical Mechanics Level 5

A long pliable carpet is laid on ground as shown in figure. One end of the carpet is bent back and pulled backwards with constant velocity 16 m/s . Find the minimum force needed to pull the moving part

Details

$\frac{M}{L}$ of carpet is 1 kg/m

Also try He is Invisible

The answer is 128.

2 solutions

Satvik Pandey
Dec 26, 2014

The force that is required is to accelerate a small part of the carpet ( $dm$ ) from rest to final velocity 16m/s.

So $dF=\frac{dp}{dt}$

or $dF=dm\frac{dv}{dt}$

or $dF=\frac{M}{L} dx\frac{dv}{dt}$

or $dF=\frac{M}{L} dx\frac{dv}{dx} \frac{dx}{dt}$

or $dF=\frac{M}{L} vdv$

or $F=\frac{v^{2}}{2}$ on putting value we get F=128N

Really good solution, i did using momentum conservation which is much larger cause you have to prove that when hand moves x distance, the length of chain in motion shortens by only x/2, but in your method, no such confusion is there,

Mvs Saketh - 6 years, 5 months ago

Thanks bro! :)

satvik pandey - 6 years, 5 months ago

I did Same as Satvik Posted , But Can Saketh I didn't Getting You , Can You Please Post Your's , So that I can Learn from it !

Deepanshu Gupta - 6 years, 5 months ago

yeah, you see when the hand (tip of rope) shifts by x, one might think that the moving part of rope has extended by x too, but infact, its just x/2

it can be proved if you carefully observe, the motion of the tip of rope, is not only causing more part of the rope to move but also the exchange point to shift forward (it would have been so if we fixed a clamp to make sure that the portion in space where the exchange (rest portion shifts to moving portion) was at rest),

so we have pvdx/2dt=pv^2/2=128

otherwise one may think that dm/.dt=pv so F=pv^2=256, and it might also cause one to think that newtons second law is not applicable for whole rope at once but we need to consider a dm portion like u did,, but no we can also do like this, we just need to identify how the length of rope in motion is actually changeing,,

replace in ur mind all ropes with carpets, sorry for the inconvenience

Mvs Saketh - 6 years, 5 months ago

@Mvs Saketh – I solved it the same way. At first, I answered 256N, but then I realized my mistake:

If you start at the edge of the carpet and walk back 16 meters (in 1 second) you will only have peeled 8 meters of carpet. So only 8 kg of carpet per second is being accelerated to 16 m/s (requiring a force of 8*16=128)

Nathanael Case - 6 years, 5 months ago

@Mvs Saketh – I got it , Thanks !

Deepanshu Gupta - 6 years, 5 months ago

@Aaron Jerry Ninan Isntit wrong to write df=dp/dt and then next step..it doesn't follow the product rule. While df=dm × a is ok. Isnt i

Md Zuhair - 3 years, 3 months ago

Chew-Seong Cheong
Dec 27, 2014

We can do this by equating the work done with the kinetic energy of the moving carpet. Therefore, we have:

$Fv = \frac {1}{2}m(t)v^2 = \frac {1}{2} \lambda v\dot{}v^2\quad \Rightarrow F = \frac {1}{2} \lambda v^2 = \frac {1}{2}\dot{}1\dot{}16^2 = \boxed{128}$

M(t) should be λx right! Can you pls explain?

Ashwin Gopal - 6 years, 5 months ago

My first Mechanics Problem

The answer is 128.

2 solutions

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