Example 4-3 Let Sleeping Cats Lie
A cat that weighs 40.0 N has fallen asleep on a sloping ramp that is tilted by an angle of 15.0° from the horizontal. Three
forces act on the cat: the downward gravitational force, a normal force perpendicular to the ramp, and a friction force
directed uphill parallel to the ramp. The uphill friction force prevents the cat from sliding downhill, so the cat remains at
rest while sleeping. Determine the magnitude of each force that acts on the cat.
Set Up
We are told that the cat’s weight is w = 40.0 N,
which is just the magnitude of the gravitational
s cat). We also know the direcforce on the cat (w
s (perpendicular
tions of the normal force n
to the ramp) and the friction force sf (uphill
parallel to the ramp). Our task is to find the
s and sf . We’ll do this using the
magnitudes of n
s cat, n
s , and sf must be
idea that the sum of w
zero because the cat remains at rest (Newton’s
first law). So we can solve this problem by
­using vector addition.
Solve
Newton’s first law of motion:
n normal force
s
s cat + n
s + sf = 0
a Fext on cat = w
(4-6)
f friction force
15.0°
Like most vector addition problems, this one
is most easily solved using components. If
s
a Fext on cat = 0, then each of the components
of a Fsext on cat must also be equal to zero. It’s
convenient to choose the x axis to be along the
tilted ramp and the y axis to be perpendicular
to the ramp. (The x and y axes must be perpendicular to each other but don’t have to be
horizontal or vertical.) Using this choice, two
s and sf , lie either directly along
of the ­vectors, n
or directly opposite to one of the coordinate
axes.
Newton’s first law in component
form:
Write down the x and y components of each
of the external force vectors in terms of their
magnitudes w, n, and f and the angle u = 15.0°
s cat has a positive x
of the ramp. Note that w
component (down the ramp) and a negative y
s has only a posicomponent (into the ramp), n
tive y component (perpendicular to the ramp),
and sf has only a negative x component (up the
ramp).
Components of gravitational
s cat:
force w
y
a Fext on cat , x = wcat , x + nx + fx
a Fext on cat , y = wcat , y + ny + fy
x
15.0°
ny = n
fx = –f
wcat , x = + wcat sin u
wcat , y = 2wcat cos u
Components of normal
s:
force n
nx = 0
ny = + n
O = 15.0°
Components of sf :
fx = 2f
fy = 0
Substitute the expressions for wcat , x,
wcat , y, nx, ny, fx, and fy into Newton’s first law
and solve for the magnitudes n and f.
wcat gravitational force
Substitute into Newton’s first law
in component form:
a Fext on cat , x = wcat sin u + 0 + (2f ) = 0
a Fext on cat , y = 2 wcat cos u + n + 0 = 0
From the y equation:
n = wcat cos u = (40.0 N) cos 15.0° = 38.6 N
From the x equation:
f = wcat sin u = (40.0 N) sin 15.0° = 10.4 N
wcat
O
wcat,y = –wcat cosO
wcat,x = +wcat sinO
Reflect
n = 38.6 N
The downward gravitational force has magnitude 40.0 N, the normal force
acting perpendicular to the ramp has magnitude 38.6 N, and the friction force
acting uphill has magnitude 10.4 N. Note that no single force “balances” any of the
other forces: All three forces are needed to mutually balance each other and keep the
cat at rest.
f = 10.4 N
wcat = 40.0 N