Answer to Essential Question 5.2: The two components of the contact force are the upward
normal force and the horizontal force of kinetic friction. These two components are at right angles
to one another, so we can use the Pythagorean theorem to find the magnitude of the contact force:
.
5-3 Measuring the Coefficient of Friction
Let’s connect the force ideas to the one-dimensional motion situations from Chapter 2.
EXPLORATION 5.3 – Measuring the coefficient of static friction
Coefficients of static friction for various pairs of materials are given in Table 5.1. Here’s
one method for experimentally determining these coefficients for a particular pair of materials.
Take an aluminum block of mass m and a board made from a particular type of wood (we could
also use a block of the wood and a piece of inflexible aluminum). Place the block on the board
and slowly raise one end of the board. The angle of the board when the block starts to slide gives
the coefficient of static friction. How?
To answer this question, let’s extend the general method for solving a problem
that involves Newton’s laws.
Step 1 – Draw a diagram of the situation. The diagram is shown in Figure 5.5a.
Step 2 – Draw a free-body diagram of the block when it is at rest on the inclined
board. Two forces act on the block, the downward force of gravity and the upward
contact force from the board. We generally split the contact force into components, the
normal force perpendicular to the incline, and the force of static friction acting up the
slope. This free-body diagram is shown in Figure 5.5b
Step 3 – Choose an appropriate coordinate system. In this case, if we choose a
coordinate system aligned with the board (one axis parallel to the board and the other
perpendicular to it), as in Figure 5.5c, we will only have to split the force of gravity into
components.
Step 4 – Split the force of gravity into components. If the angle between the board and
the horizontal is ! , the angle between the force of gravity and the y-axis is also !. The
component of the force of gravity acting parallel to the slope has a magnitude of
. The perpendicular component is .
These components are shown on the lower free-body diagram, in Figure 5.5d.
Step 5 – Apply Newton’s second law twice, once for each direction. Again, we break a
two-dimensional problem into two one-dimensional problems. With no acceleration in
the y-direction we get: . Looking at the
lower diagram in Figure 5.5 for the y-direction forces:
, which tells us that
.
Chapter 5 – Applications of Newton’s Laws Page 5 - 6
Figure 5.5: (a) A diagram showing the block
on the board. (b) The initial free-body diagram
of the block. (c) An appropriate coordinate
system, aligned with the incline. (d) A second
free-body diagram, with the forces aligned
parallel to the coordinate axes.
