MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.01
W02D2_1 Table Problem: Free Body Force Diagrams Solution
A block of mass m is sliding down an inclined plane with angle  . There is kinetic
friction between the block and the inclined plane with coefficient  k . The gravitational
force on the block is directed in the downward vertical direction.
a) Draw a free-body (force) diagram for the block. Clearly identify your forces with
whatever symbols you find appropriate.
b) For the coordinate system with unit vectors î pointing horizontally to the right
and ĵ pointing vertically upward, what are the vector components of the sum of
the forces in the î -direction and the ĵ -direction?
c) For the coordinate system with unit vectors b̂ pointing along the inclined plane
and ĉ pointing normal (perpendicular) to the plane as shown in the figure, what
are the vector components of the sum of the forces in the b̂ -direction and the ĉ direction?
Solution: There are three forces acting on the block, kinetic friction fk opposing the
r
r
sliding motion, gravity m g pointing downward, and the normal force N perpendicular
to the contact surface. The free body force diagram on the block is shown in the figure
below.
a)
Let’s first choose unit vectors î pointing horizontally to the right and ĵ pointing
vertically upward as shown in the figure below.
With these unit vectors, the vector decomposition of the forces are
r
N  N sin  φ
i  N cos φ
j
r
m g  mg φ
j
fk   fk cos î  fk sin  ĵ .
So the vector sum of the forces on the block is
r
r
r r
FT  N  mg  fk  (N sin  fk cos ) φ
i  (N cos  mg  fk sin )φ
j
(1)
b)
Now let’s choose unit vectors b̂ pointing along the inclined plane and ĉ pointing
normal (perpendicular) to the plane as shown in the figure below.
With these unit vectors, the vector decomposition of the forces are
r
N  N cφ
r
m g  mg sin  bφ  mg cos cφ
fk   fk b̂ .
So the vector sum of the forces on the block is
r
r
r r
FT  N  mg  fk  (N  mg cos ) cφ (mg sin  fk )bφ
(2)
Suppose we want to find the magnitude of the acceleration of the block. Newton’s
Second Law in the b̂ -direction is then
mg sin   f k  mab
(3)
Because the block is constrained to move along the inclined plane ac  0 , then Newton’s
Second Law in the ĉ -direction is then
N  mg cos  0
(4)
The friction force law is f k   k N   k mg cos (where we solved Eq. (4) for N ) so we
can solve Eq. (3) for the b̂ -component of the acceleration
ab  g(sin    k cos )
which is also the magnitude of the acceleration.
Suppose instead we were to us the î and ĵ unit vectors and Eq. (1) for the vector
decomposition of the sum of the forces.
(5)
Because the block is constrained to slide down the inclined plane, the component of the
accelerations are constrained. In order to determine the constraint conditions we
introduce a coordinate system as shown in the figure above.
From the geometry we see that
tan 
Thus
y(t)
.
L  x(t)
y(t)  (L  x(t)) tan 
(6)
We can differentiate Eq. (6) twice and find that
a y  ax tan
(7)
In this coordinate system, using f k   k N , Newton’s Second Law in the î -direction is
then
N (sin    k cos )  max
(8)
Using f k   k N and Eq. (7), Newton’s Second Law in the ĵ -direction is then
N (cos   k sin  )  mg  max tan 
(9)
Let’s solve our equations for N , and a x . We first rewrite Eq. (9) as
N (cos   k sin  )  mg  max tan 
(10)
Let’s start with N . We can multiply Eq. (8) by sin yielding
N(sin2   k sin cos )  max sin
(11)
We can multiply Eq. (10) by cos yielding
N(cos2   k sin cos )  mg cos  max tan cos  mg cos  max sin
(12)
we can now add Eqs. (11) and (12) using the trig identity cos 2   sin 2   1 and solve for
N:
N  mg cos
(13)
which is what we found already (Eq. (4)). We can now substitute Eq. (13) into Eq. (8)
and find that
ax  g sin cos  g k cos2 
(14)
We now use the trigonometric fact that a  ax / cos in Eq. (14) to find the magnitude of
the accelration
a
a  x  g(sin    k cos )
(15)
cos
in agreement with our earlier result (Eq. (5)). We can now use Eq. (7) to find that
a y  (g sin  cos  g  k cos 2  ) tan 
which simplifies to
a y  g sin 2   g  k sin  cos
This exercise should convince you to think carefully about how you choose your
coordinate system.
(16)