Download Notes 3.7 Net Force with Angles

Unit 3.7: Net Forces with Angles
Problems with net force and angles work the same way that it does when there is not an angle except you must
draw a triangle with the angled vector and break the angled force into the horizontal and vertical components of
the forces motion so that you can cancel out the forces that need to cancel and calculate the net force with the
parts that do not.
Examples:
1.
A boat is drug across the ground at
a constant velocity by a rope that
forms a 50 angle with the
ground. If there is 77 N of tension
in the rope, how much friction is
acting on the boat?
Since the boat is moving at a constant velocity, it
is in equilibrium and thus all the forces must
cancel. Thismeans that the horizontal (x)
component of the tension must equal the force of
friction in order for them to cancel out with each
other. Thus:
x
 x  77 cos 50
77
x  49.5  F f  49.5 N
cos 50 
2.
A 60 kg boat is drug across the
ground by a rope that forms a 50
angle with the ground with 77 N of
tension in the rope and a frictional
force of 25 N. How much
acceleration does the boat
experience?
In this case the boat is not in equilibrium. We must therefore
calculate the horizontal component of the pulling force and
compare it to the frictional force in order to determine the net
force acting on the boat Thus:
pulling foce  cos 50 
frictional force  25 N
thus Fnet  F p  ( F f )  49.5  35  14.5
Fnet  ma
3.
A 120 N lawnmower is pushed by the
handlebars by a force of 100 N which
causes the handlebars to form a 30ᵒ
with the ground.
a) How much is the Normal Force
b) How much friction is acting on the
mower if it has 2.6 m/s² of
acceleration?
x
 x  77 cos 50  49.5 N
77
 14.5  (60)a
 a
14.5
 .24m / s 2
60
a) First we must calculate the horizontal and vertical components
of the pushing force on the handlebars in order to determin the
amount of Normal Force from the ground pushing up on it.
x
horizontal pushing force  cos 30 
 x  100 cos 30  86 N
100
x
vertical pushing force  sin 30 
 x  100 sin 30  50 N
100
then Fg  y  FN  FN  120  50  170 N
b) In this case the mower is not in equlibrium. Using the mass of
the lawnmower and the accelertation we can determine the net
force acting on the mower and then combine it with the
horizontal component of the pushing force in the handle bars to
detemine the friction force.
mass of the mower
Fg  mg  120  m(9.8)
m  12.2kg
Net Force : F  ma  12.2(2.6)  31.7 N
Fnet  x  F f

31.7  86  F f

F f  54.3N