The contact force therefore can be expressed as a a 27 F 2xÊ / ºr (d-1²) dr = -1°2xr 2xr Ozz dr n 2R giving the force to achieve a given penetration depth, d, as TÊR d² n

Calculus: Early Transcendentals
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Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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can someone explain this intergration please 

?
See
below
learn-eu-central-1-prod-fleet01-xythos.content.blackboardcdn.com
C&F Lectures : Panopto
Content
doing
SNTS
The contact force therefore can be expressed as
a
exÊ
27
=
2xr Ozz dr
п
giving the force to achieve a given penetration depth, d, as
ÊR
ㅈ
2
F
d²
=
n
√ª (dr - 12³) dr
:)
=[dr²_
2R
2
BR]
da²
2
8R
d & Rd
R
Rd²
2
=
+88
Bb https://learn-eu-central-1-prod-fleet01-xythos.content...
Contact & Friction: Lecture 01
a
✓ªr(d-re ke) dr
2R
but an √2Rd¹
2
4R²d²
28R
which
can be substitued
back into F to obtain
the final equation.
Transcribed Image Text:? See below learn-eu-central-1-prod-fleet01-xythos.content.blackboardcdn.com C&F Lectures : Panopto Content doing SNTS The contact force therefore can be expressed as a exÊ 27 = 2xr Ozz dr п giving the force to achieve a given penetration depth, d, as ÊR ㅈ 2 F d² = n √ª (dr - 12³) dr :) =[dr²_ 2R 2 BR] da² 2 8R d & Rd R Rd² 2 = +88 Bb https://learn-eu-central-1-prod-fleet01-xythos.content... Contact & Friction: Lecture 01 a ✓ªr(d-re ke) dr 2R but an √2Rd¹ 2 4R²d² 28R which can be substitued back into F to obtain the final equation.
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