Download Unit 3 – Net Force

Unit 3.8: Net Forces with Inclines
For most motion problems, the object in question is moving horizontally or
vertically but not in both directions at the same time. Objects that are sliding up
or down hills present difficulty due to the number of forces that are acting on the
object as it slides because of the number of forces that are at angles to the axis of
origin (as shown to the right-where the dotted line is the hills surface). As shown
in example 3.7 we would have to find each of the components of all the forces
and then add and subtract the forces that are acting in the same dimension
(horizontally & vertically). This means doing a lot of trigonometry to find all these
components. However, there is an easier fix to this problem. Since friction acts
along the surface of the hill and the normal force acts perpendicular to it, we can
tilt the x-y axis to match the tilt of the hill as shown to the left. When tilting the
axis, rotate the x & y axes until they match the angle of the hill’s tilt. This
reduces the number of components that must be calculated. In this set up, the
force of friction and the x component of gravity are working against each other.
However, since the object is not sinking through the hill nor is it shooting off of
it, the normal force and the y component of gravity cancel each other out.
(Remember that Fg -weight- is found by multiplying mass-kg-by acceleration
2
due to gravity-9.8 m / s )
Examples:
1.
A 300 N box slides down a 20
incline at constant velocity.
a) How much are the horizontal and
vertical components of the box’s
weight?
b) How much is the net force acting
on the box?
c) How much is the normal force
acting on the box?
d) How much is the friction acting on
the box to make its velocity
constant?
Since this box is moving at constant velocity, the net force acting
on the box is zero and thus all the forces in the same directions
must cancel out. This means that x and F f must cancel each
other out and thus must be equal to each other. Furthermore, y
and FN must also cancel each other out and thus must be equal.
a) x component:
y component:
x
 x  300 sin 20  102.6
300
y
cos 20 
 y  300 cos 20  281.9
300
sin 20 
b) Net Force is zero due to the constant velocity
c) Since Net Force is zero, y and FN must be equal so that they
can cancel with each other. Thus FN =281.9 N
d) Again, since the Net Force is zero, x and F f must be equal so
that they can cancel with each other. Thus F f =102.6 N
2.
A force of 200 N is used to push a
27 kg suitcase up a 38 inclined
ramp at constant velocity.
a) How much does the suitcase
weigh?
b) How much is the normal force?
c) How much is the friction?
a) Since weight is mass times by acceleration due to gravity,
Fg  ma  27(9.8)  264.6 N
b) Since the suitcase is moving at constant velocity, the net force is
zero which means that all the forces cancel out. Thus, y
FN must be equal so that they can cancel. Thus
and
y
 y  264.6 cos 38  208.5
264.6
y  FN  208.5 N
cos 38 
c) Again since the net force is zero, x, F p , and F f
cancel with
each other. Because there are 3 forces this time, the two
directed down the incline (x & F f ) must add up to and
equal
the one directed up the incline ( F p ). Thus
x
 x  264.6 sin 38  162.9
264.6
x  F f  FP  162.9  F f  200  F f  37.1N
sin 38 
3.
A 100 kg rock remains stationary on First calculate the weight of the rock
a 50 inclined hill. How much
Fg  mg  100(9.8)  980 N
friction must be acting on the rock to
keep is stationary?
Again Now find the horizontal ( x) component of the weight
sin 50 
x
 x  980 sin 50  750.7 N
980
Since the rock is at rest it is therefore in equilibrium and thus has a
net force of zero. This means that x and F f must be equal in
order to cancel with each other.
F f  x  750.7 N
4.
A man pushes a 200 N box down a
15 inclined hill by pushing with a
force of 35 N. If friction produces 40
N on the box, calculate the amount
of acceleration the box experiences
as it slides down the hill.
First find the x-component of the box’s weight:
sin 15 
x
 x  200 sin 15  51.76 N
200
Now find the net horizontal force acting on the box:
Fnet  x  Fp  F f
Fnet  51.76  35  40  46.76 N
Now find the mass of the box:
Fg  mg  200  m(9.8)
200
 20.41kg
9.8
m
Now using the net force and mass (calculated above) find the
acceleration that the box experiences:
Fnet  ma  46.76  20.41(a)
a
5.
An 18 kg wagon at rest on a 30
inclined hill rolls to the bottom of
the hill in 4.5 s and is going 12 m/s
by the time it reaches the bottom.
a) What is the weight of the wagon?
b) Find the horizontal (x) component
of the wagon’s weight.
c) What is the acceleration that the
wagon experiences as it rolls to
the bottom of the hill?
d) How much is the net horizontal
force acting on the wagon?
e) How much friction is acting on the
wagon?
46.76
 2.29m / s 2
20.41
a) Fg  ma  18(9.8)  176.4 N
b) sin 30 
x
 x  176.4 sin 30  88.2 N
176.4
c) Use v  v0  at thus 12  0  a(4.5)  a 
12
4.5
 2.67m / s 2
d) Since Fnet  ma  18(2.67)  48.06 N
e) Since Fnet  48.06 N this means that one of the two horizontal
forces is this much larger than the other and since, by
definition, net force is the sum of all forces then
Fnet  x  F f
 48.06  88.2  F f
F f  40.14 N