Example 5-1 Friction in Joints
The wrist is made up of eight small bones called carpals that glide back and forth as
you wave your hand from side to side. A thin layer of cartilage covers the surfaces
of the carpals, making them smooth and slippery. In addition, the spaces between
the bones contain synovial fluid, which provides lubrication. During a laboratory
experiment, a physiologist applies a compression force to squeeze the bones together
along their nearly planar bone surfaces. She then measures the force that must be
applied parallel to the surface of contact to make them move. (Figure 5-7 shows the
contact region between these two carpal bone surfaces.) When the compression force
is 11.2 N, the minimum force required to move the bones is 0.135 N. What is the
coefficient of static friction in the joint?
Interface between two nearly
planar carpal bone surfaces
Figure 5-7 ​​Static friction in the wrist This X-ray image shows the carpal bones of the wrist.
Set Up
We begin by drawing the free-body diagram
for one of the two carpals in question, which
we call carpal A, when it is just about to slip
relative to carpal B. The physiologist exerts
two of the four forces acting on carpal A:
the compression force Fscompression (of magnitude
11.2 N) and the force Fsslide applied parallel to
the contact surface (of magnitude 0.135 N)
to make bone A slide relative to bone B. The
other two forces acting on carpal A are the
s and the static friction force
normal force n
sf s, max, both exerted by Carpal B on Carpal A.
(This friction force is the maximum force of
static friction because the carpals are just about
to slip.) Our goal is to find the value of ms, so
we’ll also use the relationship between fs, max
and the normal force given by Equation 5-1b.
Solve
Newton’s second law for carpal A:
Fslide
s
a Fext on A
s + sf s, max
= Fscompression + Fsslide + n
sA
= mAa
Fcompression
A
fs,max
Magnitude of the static friction force:
fs, max = msn
n
B
(5-1b)
Newton’s second law for carpal A
in component form:
Fslide
Let the positive x axis point up, and the
positive y axis point to the left. Since we are
considering the situation in which the carpals
have not yet begun to slide, carpal A is at rest
and aA, x = aA, y = 0. We use this information
to write Newton’s second law for carpal A in
component form.
x: a Fext on A, x
= 0 + Fslide + 0 + (2fs, max)
= mA aA, x = 0
To find the value of ms using Equation 5-1b,
we need to know the values of n and fs, max.
We find these quantities by using Newton’s
second law equations and the known values
Fcompression = 11.2 N and Fslide = 0.135 N.
Solve Equation 5-1b for ms:
fs, max
ms =
n
Solve Newton’s second law equations for fs, max and n.
From the x component equation:
–Fcompression
y: a Fext on A, y
= (2Fcompression) + 0 + n + 0
= mA aA, y = 0
n
A
–fs,max
x
y
fs, max = Fslide = 0.135 N
From the y component equation:
n = Fcompression = 11.2 N so
ms =
Reflect
0.135 N
= 1.21 * 10-2 = 0.0121
11.2 N
This coefficient of static friction is very small, which means that very little effort is required to make your wrist move.
The lubricating fluid between the bones is the key to keeping the friction so small. The fluid also helps reduce wear on the
joints. If the bones or even the cartilage surfaces were in direct contact, our joints would wear down relatively quickly.