- #1

d.arbitman

- 101

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dT*sin(dθ/2)

source: http://www.jrre.org/att_frict.pdf

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- Thread starter d.arbitman
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- #1

d.arbitman

- 101

- 4

dT*sin(dθ/2)

source: http://www.jrre.org/att_frict.pdf

- #2

Mark44

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dT*sin(dθ/2)

source: http://www.jrre.org/att_frict.pdf

Differentials are assumed to be close to zero, so sin(dθ/2) is close to zero, as is dT. The product of two very small numbers is even smaller, so can be neglected.

- #3

d.arbitman

- 101

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Differentials are assumed to be close to zero, so sin(dθ/2) is close to zero, as is dT. The product of two very small numbers is even smaller, so can be neglected.

That makes perfect sense, but with that reasoning we can neglect a single differential as well since it is close to 0 and then we will have no use for calculus.

- #4

SteveL27

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That makes perfect sense, but with that reasoning we can neglect a single differential as well since it is close to 0 and then we will have no use for calculus.

Another way to look at this intuitively is that if a variable x goes to zero, the variable x

[itex]\displaystyle \lim_{x \to 0} \frac{x^2}{x} = 0[/itex]

Even though x

You can see this numerically. When x is 1/2, x

But your question's a good one. Newton invented calculus in 1680 or so, but the rigorous theory of limits wasn't developed till two hundred years later. So your question shows good insight.

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- #5

d.arbitman

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Another way to look at this intuitively is that if a variable x goes to zero, the variable x^{2}goes to zero much faster. In other words

[itex]\displaystyle \lim_{x \to 0} \frac{x^2}{x} = 0[/itex]

Even though x^{2}and x both go to zero, x^{2}goes to zero so much faster than x, that the limit of their quotient evaluates to zero.

You can see this numerically. When x is 1/2, x^{2}is 1/4. When x = 1/10, x^{2}is 1/100. As the magnitude of x gets smaller, the magnitude of x^{2}gets alotsmaller.

But your question's a good one. Newton invented calculus in 1680 or so, but the rigorous theory of limits wasn't developed till two hundred years later. So your question shows good insight.

That's a great example. Thank you. I guess what I take from that is that the product of two differentials changes so much slower than a single differential that we can almost forget about the product as it will have a miniscule effect. Maybe mathematically it can have an effect on the outcome, but in the context of friction it will have little to no effect.

- #6

HallsofIvy

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- #7

Studiot

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However the question of

There are many areas of physics and engineering where we make this sort of assumption in the interests of a simple formula.

However sometimes we need to take second order (products of two small quantities) or even higher orders into account for more acurate work.

An example would be in beam theory where we obtain simple formulae applicable to long narrow beams (most are like this) by considering differentials of first and second order and disregarding higher orders.

Please note that there is a difference between saying

[itex]{\left( {\frac{{dy}}{{dx}}} \right)^2}[/itex] is small

and

[itex]\frac{{{d^2}y}}{{d{x^2}}}[/itex] is small.

for sharply curving function the latter may not be true at all.

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