Brain Teasers ? or 50 Years On........... ?

Pulleys !

Naim’s new warehouse will have a pulley system installed for loading Statements onto despatch lorries.

It comprises a LIGHT, INEXTENSIBLE rope connecting 50kg and 100kg masses suspended over a light frictionless pulley. (It’s an A-Level Maths question, not reality, remember !) The system is stationary, and has just been released from rest at the instant shown in the diagram.

Are the tensions T1 and T2 the same ?

What is the magnitude and direction of the force on the pulley ?

Trucking Classic Update Facility.

Naim Audio is in the process of re-arranging production of its updated Classic and 500 power amps. The production and warehouse facility is a long straight building, fronting onto a tarmac surfaced area.

Paul S would like to fence-in the largest possible rectangular area of the tarmac as a secure “Goods Outwards” lorry park. He has “acquired” a decent (but fixed) length of Super Lumina fencing to use as the three sides of the  lorry park, the fourth side being formed by the production/warehouse itself. What ratio x/y should he choose for the lengths of the sides x and y of the fenced-in lorry park ?

As a mathematically challenged person who never got beyond O-Level, I'm going to say that T1 and T2 must, according to Isaac Newton, be the same, and the pulley has a vertical force of 150kg. Now perhaps somone who understands the laws of motion, or whatever, will come alomg and correct me...

I agree the tension in T1 and T2 will be equal, but the downward force on the pulley will be 100kg. (Both sides exerting 50Kg.)

The area of the lorry park will be x * y, the length of the fence 2 * x + y. The optimal ratio of x to y should therfore be 2 * x = y.

Remember, although the system is stationary, it has just been released from rest and the two masses will accelerate. The pulley is light and frictionless, so has no inertia and the rope is light and inextensible so cannot store energy.

If the pulley was prevented from rotating and the rope prevented from slipping around the pulley, then the force on the pulley would be 150kg and you would be correct.

You are right to say that T1 = T2, and you are also right to say that perhaps someone who understands the laws of motion will come along and correct you. the load on the pulley isn't 150kg.

Well done for starting the ball rolling and having a crack at it.

Hi Frank, yes, T1 will equal T2, the rope is light and inextensible. The system has been released from rest. The system is no longer at rest nor is it exhibiting uniform motion in a straight line. It is different to how the system would be if the pulley and rope were braked.

But again, a good start, especially recognising that T1 has to be the same as T2

Well done Mulberry. I don't quite follow your logic, but the answer is certainly right.

Could you or someone else expand on the logic ?

yeah, i remember that one, because IIRC, i solved it in the end.

but i have a question for you --what software did you use to produce those very nice diagrams? -- i happen to be looking for something like that. thanks.

excellent thread by the way!

enjoy

ken

You don't need logic, just the back of an envelope!

Nice one Chris. Iterations and approximations - the hall-mark of a true Engineer.

Any pure mathematicians in the house ?

If L is teh length of fence then 

L=2x+y  or rearranging y=L-2x

If the area is A then

A=xy

Substituting the rearranged form in above 

A=x(L-2x) 

A = Lx - 2x**2

In order to find teh maximum we must differentiate so

dA/dx = L-4x

The maximum for A is where this is zero so

0=L-4x

Substituting for L from teh first equation we get

0 = 2x+Y -4x

0=y-2x

y = 2x

I think, but it was rather a long time ago.....

Regards,

Willy.

Much more rusty on applied mechanics but is the load on the pulley 1309N, (133.3kg)?

Doesn't look right but that's what the calculation spat out.

Willy.

Thanks Don, the solution seemed somehow obvious to me. Explaining the obvious has never been one of my strengths and I have a language barrier to jump as well .

First class Willy. Especially if it was 50 years ago or more !! The old grey matter is still working fine.

Well, that calculation spat out the right answer Willy, so it seems its the old eyesight ("Doesn't look right ) is what need attending to.

Care to share your calculation with the others who have given it a good shot ?

Only 38 years ago with some more recent revision putting kids through A level maths.

Regards,

Willy.

No worries Mulberry, you provided us with the right answer. Willy has added the usual A-Level method of using a differential equation.

Its possible to avoid the differential route (and the iteration route) using symmetry and algebra.

Oh ho !! That's cheating

Mind you, on second thoughts, I'm ramping up to help my grandchildren !!

Ok here goes....

Tension in the cable is T

50Kg mass is m

100Kg mass is M

Gravity is G

Force acting on m is

T -mg       therefore applying Force = Mass X Acceleration we get

T-mg=ma  or rearranging  (T-mg)/m=a

Where a is the acceleration that m will experience

Similarly for M the force is

Mg- T  (Since it's being retarded by the Tension and is moving in the opposite direction to m)

and 

Mg-T=Ma    or (Mg-T)/M = a

Since both masses are tethered to each other they can only have the same acceleration so

(T-mg)/m = (Mg-T)/M

Multiply both sides by Mm and we get

(T-mg)M = (Mg-T)m

or

TM -mMg = mMg-Tm

or

TM+Tm = mMg + mMg

or

T(M+m) = 2mMg

or

T = 2mMg/(m+M)

T = 2*50*100*9.82/(50+100) = 654Newtons

Pulley has T acting on it at each side so experiences 2T = 1309.3N

Divide by g to get 133.3Kg

Regards,

Willy.

My exams were just 27 years ago. So this might have been an unfair advantage...

Do yourself a favour and discourage them from doing Physics. A level maths hasn't changed much in 35 years. A level physics has!

Regards,

Willy.

Nicely done Willy.

My workings are very similar and using your notation :-

T - mg = ma    eq(1)

Mg - T = Ma    eq(2)

Adding (1) + (2)

(M - m) g = (M + m) a

Re-arranging

a = g (M - m)/(M + m)

Substituting values

a = g(50/150) = (1/3) g    eq(3)

Substituting eq(3) in either (1) or (2)

T = (200/3) g = 666.7 Newtons (assuming g = 10m/sec/sec)

T = 66.7 kg

2T = 133.3 kg  = upward force on pulley

Basically we are doing the same thing. Hope your kids enjoy maths and physics.