EQUILIBRIUM CASE VI - Translatory Equilibrium up an inclined plane.
EQUILIBRIUM1 CASE 6
Moving a body up an inclined plane @ constant
velocity indicates the body is in a state of
Translatory equilibrium and the 1ST

Condition of Equilibrium holds:
As on all inclined
planes, make the


  0

  0

P
N

fk
X-axis through
the CG of the
body parallel to
plane.


F

w=mg

The effective weight vector F (or F‫ )װ‬always
“pulls down the plane” and its value is given as:
F‫װ‬
The Normal N on the inclined Plane is always:
N
W sin
=
W cos
Since the object moves up the plane, friction will always act down the
plane since it is a degrading force (opposes motion). On inclined planes:


f = Wcos
Moving friction is sliding or kinetic friction, and since the body is
moving:

fk = kW cos 
and since the body
moves upward at constant velocity, it is in equilibrium and:
ALL FORCES UP THE PLANE = ALL FORCES DOWN THE PLANE
P = F‫ װ‬+ fK
PUSH/PULL
EFFECTIVE WT. PLUS KINETIC
FRICTION
example A 100N crate is slid up an inclined plane, tilted at 30, @ constant
speed. The coefficient of kinetic friction k , is 0.250. What force is required?
W = 100N
s = .287
k = .25
EQUALS
P = F‫ װ‬+ k
P = Wsin + kWcos  (100N)sin30+ (.25)(100N)cos30
P = 50N + 21.65N = 71.65N
EQUILIBRIUM CASE VII - Translatory Equilibrium down an inclined
plane.
EQUILIBRIUM1 CASE 7
Moving down a plane @ constant velocity
also constitutes equilibrium.


y
x
  0
P

N
  0
fk
Since the body slides
down the plane,
riction acts opposite
of motion. To produce
Equilibrium, a push
P must be directed up
the plane.


F

w=mg

The effective weight F (F‫ )װ‬is always
directed down the plane:
F‫װ‬
The Normal on the inclined Plane is always:
Again, moving friction is sliding or kinetic
friction, and since the body is moving:
=
W sin
W cos

fk = kW cos 
ALL FORCES UP THE PLANE = ALL FORCES DOWN THE PLANE
P + fK = F‫װ‬
example A 100N crate is to be lowered down an incline @ constant speed.
The coefficient of s is 0.287 and the coefficient of k is 0.250. What force
is needed?
W = 100N
s = .287
k = .25
P = F1 - k
1st let’s find the slip  from: slip  = tan-1 s
 slip  = tan-1 0.287 = 16
This is the angle of an inclined plane needed to give an
impending state of motion. The plane in this problem exceeds
this 16 angle. The crate by its effective weight, will slide down
the plane. To slide down the plane @ constant speed requires a
push P , up the plane:
P + k = F1
= Wsin - kWcos  (100N)sin30 - (.25)(100N)cos30
P = 50N - 21.65N =
28.35N