What is the most efficient way to solve a wheel and axle system problem?

[chikis]

Homework Statement

Hello folks, here is a problem: The effciency of a wheel and axle system is 80% and the ratio of radius of wheel of axle is 4:1. In other to lift a mass of 20 kg, the effort required is
A.60N
B.62.5N
C.32.5N
D.250N
E.50N

Homework Equations

I made use of the equation:
V.R= distance moved by effort/distance moved by load = 2pieR/2pier =R/r.
Bearing in mind that in the absence of friction, VR=MA
Thus,
MA=VR=R/r = radius of wheel/radius of axle

The Attempt at a Solution

I started by bringing out the data in the question: Effiency, E.f=80%. Radius of bigger wheel, WR=4. Radius of axle of smaller circumfrence wr=1 since their are in the ratio 4:1. Mass lifted=20kg.
Using Velocity ratio, V.R= distance moved by effort/distance moved by load =circumfrence of bigger wheel of R/circumfrenc of axle of smaller radius r =2piR/2pir=R/r.
But in the absence of friction, V.R, R/r =mechanical advantage, M.A.
putting my data in equation:
R/r=load, L/effort, E=M.A. They only give the mass of the object lifted as 20kg. There is no instruction that acceleration due gravity, g which is 10ms-2 should be used. But knowing that load and effort has it units in Newton, N and that force, F=mass, m times g I decided to use it.
Thus: R/r= L/E
4/1=20*10/E
4E=200N
E=200/4=50N.
You can see clearly that the effort required is 50N which is option E. But the book gives the answer as B. i.e 62.5N.
.Can this be true? If it is true then how?

 

[tiny-tim]

hi chikis!

Thus: R/r= L/E
4/1=20*10/E
4E=200N
E=200/4=50N.

erm …

where's the 80% ? ​

 

[PeterO]

Homework Statement

Hello folks, here is a problem: The effciency of a wheel and axle system is 80% and the ratio of radius of wheel of axle is 4:1. In other to lift a mass of 20 kg, the effort required is
A.60N
B.62.5N
C.32.5N
D.250N
E.50N

Homework Equations

I made use of the equation:
V.R= distance moved by effort/distance moved by load = 2pieR/2pier =R/r.
Bearing in mind that in the absence of friction, VR=MA
Thus,
MA=VR=R/r = radius of wheel/radius of axle

The Attempt at a Solution

I started by bringing out the data in the question: Effiency, E.f=80%. Radius of bigger wheel, WR=4. Radius of axle of smaller circumfrence wr=1 since their are in the ratio 4:1. Mass lifted=20kg.
Using Velocity ratio, V.R= distance moved by effort/distance moved by load =circumfrence of bigger wheel of R/circumfrenc of axle of smaller radius r =2piR/2pir=R/r.
But in the absence of friction, V.R, R/r =mechanical advantage, M.A.
putting my data in equation:
R/r=load, L/effort, E=M.A. They only give the mass of the object lifted as 20kg. There is no instruction that acceleration due gravity, g which is 10ms-2 should be used. But knowing that load and effort has it units in Newton, N and that force, F=mass, m times g I decided to use it.
Thus: R/r= L/E
4/1=20*10/E
4E=200N
E=200/4=50N.
You can see clearly that the effort required is 50N which is option E. But the book gives the answer as B. i.e 62.5N.
.Can this be true? If it is true then how?

I think you have ignored the 80% efficiency.

That 50N is what you need in a perfect system. With 80% efficiency, you need an extra quarter of 80% to get the 100% equivalent, so an extra 1/4 of 50.

 

[chikis]

hi chikis!


erm …

where's the 80% ? ​

I think you have to neglect the efficiency = 80% since is rare to see practical machine having such efficiency due to presence of friction. How about that?

 

[chikis]

I think you have ignored the 80% efficiency.

That 50N is what you need in a perfect system. With 80% efficiency, you need an extra quarter of 80% to get the 100% equivalent, so an extra 1/4 of 50.

So are you saying am correct in computing the effort required to lift such load as N50?

 

[tiny-tim]

I think you have to neglect the efficiency = 80%

then why did they tell you?​

try it again, using the 80°

 

[chikis]

then why did they tell you?​

try it again, using the 80°

Then you have to invent the formula for such calculation.

 

[tiny-tim]

The effciency of a wheel and axle system is 80%

this means that for every 100 units of force you put in, you only get 80 units of force out

 

[chikis]

What happened to the other 20 units? They are probably lost through friction.

 

[tiny-tim]

yup! friction, noise, vibration …

 

[chikis]

then why did they tell you?​

try it again, using the 80°

Is it every data in physics that one should use whenever a calculation is involved? Certianly not all.

 

[PeterO]

Is it every data in physics that one should use whenever a calculation is involved? Certianly not all.

You do have to use all the applicable data - and the 80% is vital in this situation. Sometimes additional, unnecessary information is given, and you can "ignore" the unnecessary data, but you cannot choose to ignore something which is crucial to the section of the problem you are solvng.

Usually when "too much" information is given, it flags a multi-part problem where eventually all supplied data is used, but various bits are not used during every single section of the problem.

 

[chikis]

You do have to use all the applicable data - and the 80% is vital in this situation. Sometimes additional, unnecessary information is given, and you can "ignore" the unnecessary data, but you cannot choose to ignore something which is crucial to the section of the problem you are solvng.

Usually when "too much" information is given, it flags a multi-part problem where eventually all supplied data is used, but various bits are not used during every single section of the problem.

But the 80% is unecessary in the calculation. Reason being that if other datas like the ratio of wheel to axle, mass involved and the acceleration due to gravity which equals 10ms^-2, while excluding the efficiency; the same answer will be gotten whenever the problem is solved.

 

[PeterO]

But the 80% is unecessary in the calculation. Reason being that if other datas like the ratio of wheel to axle, mass involved and the acceleration due to gravity which equals 10ms^-2, while excluding the efficiency; the same answer will be gotten whenever the problem is solved.

I am pretty sure that if someone gave you the figures for the amount of energy in the fuel fed to the motor on a crane, it could be important that the Diesel motor is only 31% efficient when working out what sized load you could lift from the ground to the top of a building if only 1 litre of diesel was burned by the crane motor.

 

[tiny-tim]

But the 80% is unecessary in the calculation.

then why is your answer 80% of the correct answer? …

You can see clearly that the effort required is 50N which is option E. But the book gives the answer as B. i.e 62.5N.

 

[chikis]

then why is your answer 80% of the correct answer? …

It will be wise that you solve it in your own way and let me see what you get. You will get nothing far from what I got, unless you did your working wrongly (not following the governing principle).