Download Inclined Planes

» On an inclined plane the normal force is not opposite the
weight
» Therefore it is necessary to break one of them into its x and y
components so you can find the net force
N
θ
mg
» Since the box is speeding up down the incline, the acceleration is in
that direction, and therefore the net force is in that direction as well
N
a
θ
mg
» It is easier to break the weight into components because the acceleration is
down the incline which is perpendicular to the normal force
» Now some trigonometry
y
N
x
a
θ
90- θ
θ
mg
» Replace the force of gravity with its components.
» Perpendicular to the ramp will be Fg cosƟ.
» Parallel to the ramp will be Fg sinƟ.
N
a
θ
θ
mg
» Use Newton’s second law for both the parallel (y) and
perpendicular (x) directions
x
y
N
a
θ
θ
mg
A person pushes a 30-kg shopping cart up a 10 degree incline with a force of 85
N. Calculate the coefficient of friction if the cart is pushed at a constant
speed.
Fa  Ff  mg sin 
Fa
Fn
Ff  k FN
Fa  k FN  mg sin  FN  mg cos 
Fa  k mg cos   mg sin 
Fa  mg sin   k mg cos 
mg cos 

Ff
mg
mg sin 

Fa  mg sin 
k 
mg cos 
85  (30)(9.8)(sin10)
k 

(30)(9.8)(cos10)
0.117
What is the s required to
prevent a sled from
slipping down a hill of
slope 30 degrees?
s = 0.577
» Example - a block of weight
180 N is pulled along a
horizontal surface by a
force of 80 N acting at an
angle 30 with the
horizontal at a constant
speed. What is the normal
force equal to?
The man pushes/pulls with a force of 200 N. The child and sled combo has a mass of
30 kg and the coefficient of kinetic friction is 0.15. For each case:
What is the frictional force opposing his efforts?
What is the acceleration of the child?
f=59 N, a=3.80 m/s2
/
f=29.1 N, a=4.8 m/s2