Forces in 2 Dimensions

1
Forces in 2 Dimensions

• Forces are not always acting in one direction
  (same or opposite).
• The forces along the x-axis and y-axis may not be
  in equilibrium.
• We use Pythagorean theorem to resolve the net
  force acting on an object.
• Example
• An object is being pulled by a 3 kiloNewtons
  force towards the north and a 4 kiloNewtons force
  eastward on a frictionless surface. What is the
  net force that will accelerate this object?

2
Forces in 2 Dimensions

• Solution We have to find the net force using
  Pythagorean theorem.
• The magnitude of the net force is 5 kN.
• The direction is O

4 kN, East
5 kN, O
3 kN, North
O
3
Forces involving angles.

• We will use trigonometric functions to determine
  the forces at an angle.

h hypotenuse
b opposite side
O
a adjacent side
4
Trigonometric Functions

• sine O opposite side sin O b
• hypotenuse
  h

h hypotenuse
b opposite side
O
5
Trigonometric Functions

• cosine O adjacent side cos O a
• hypotenuse
  h

h hypotenuse
O
a adjacent side
6
Trigonometric Functions

• tangent O opposite side tan O b
• adjacent side
  a

b opposite side
O
a adjacent side
7
Trigonometric Functions

• Example How long is the base of the 2.5 m ramp
  that is elevated 30o from the ground? How high is
  the ramp?

h 2.5 meters
b opposite side y
O 30o
a adjacent side x
8
Use cosine to solve for the base. Cosine O
adjacent side
hypotenuse cos 30 o __x___ X
(cos30o)( 2.5 m) 2.17 m

(2.5 m)
h 2.5 meters
O 30o
a adjacent side x
9
Use sine to solve for the height. Sine O
opposite side
hypotenuse sin 30 o __y___ y (sin
30o)( 2.5 m) 1.25 m

(2.5 m)
h 2.5 meters
b opposite side y
O 30o
10
Trigonometric Functions

• Use Pythagorean theorem to verify the answer.

h hypotenuse 2.5 m
b opposite side 1.25 m
O 30o
a adjacent side 2.17 m
11
We will use trigonometric functions to determine
the forces at an angle.

• Solve for the Fa applied force along the x axis
  Fa(x)

• What is the applied force acting against a
  frictional force of 10 N, if an object is pulled
  with a force of 200 N at angle of 60o from the
  ground?
• What is the net force?

200 N
60O
Fa(x)
Ff 10 N
Use cosine to solve for Fa(x) Cos 60 o Fa(x) /
200N Rearrange the equation Fa(x) (cos60 o)(
200 N) Fa(x) 100 N
The force applied opposite frictional force is
100 N and not 200 N. We can solve for the net
force Fnet then. Fnet Fa(x) Ff
100 N 10 N 90N
12

• What is the normal force Fn acting on a 180 N
  object on the ramp that made an angle of 60o from
  the ground?

• We will solve this problem using similar
  triangles.
• Take note Fn Fg , but opposite in direction.
• We will solve Fg using cosine.
• Our hypotenuse is the weight Fg 180 N.
• Fg is the adjacent side with respect to the
  angle 60o.
• cos O adjacent side / hyp.
• cos 60o Fg / Fg
• Fg (cos 60o)(180 N)
• Fg 90 N
• Since Fg Fn
• Fn 90 N

Fn
60 o
Fg
Fg
60 o
13

• What is the normal force Fn acting on a 180 N
  object on the ramp that made an angle of 60o from
  the ground?

• We can also solve for the Fa down the ramp.
• We will solve Fa using sine.
• Our hypotenuse is the weight Fg 180 N.
• Fa is the opposite side with respect to the angle
  60o.
• sin O opposite side / hyp.
• sin 60o Fa / Fg
• Fa (sin 60o)(180 N)
• Fa 155.88 N
• If there is no Friction the object will slide
  down with a Fnet Fa.

60 o
Fa
Fg
Fa
60 o
14

• A crate is being pulled by cables along a
  frictionless surface with a force of 500 kN
  eastward and by another force of 400 KN _at_ 120o.
  What is the net force Fnet acting on the crate?

400 kN _at_ 30o WoN
30o
500 kN
90o
15

• We can calculate the Fnet by adding all the
  forces along each axis and use Pythagorean
  theorem to calculate Fnet.

400 kN _at_ 30o WoN
30o
500 kN
90o
16

• Assign the proper values for each vector and be
  aware of the direction along the x and y axis.

Sin 30o - Fx / 400 kN Fx - (sin30o)(
400kN) Fx - 200 kN
Cos 30o Fy / 400kN Fy (cos 30o)( 400kN) Fy
346.41 kN
- Fx
Fy
400 kN _at_ 30o WoN
30o
90o
Fx 500 kN
17

• Add all the vector forces along the x-axis.
• Fx total Fx Fx 500 kN ( - 200 kN )
• Fx total 300 kN
• Add all the vector forces along the y-axis.
• Fytotal Fy 346.41 kN
• Fytotal 346.41 kN
• Use Pythagorean Theorem to solve for Fnet
• Fnet (Fx2total Fy2total )1/2
• (300 kN)2 ( 346.41 kN)21/2
• 458.23 kN - the magnitude of the Fnet

18

• To find the direction, we will use the tangent
  function.
• We have a () Fx total and () Fy total so the
  direction should be on the first quadrant.
• Tangent O opposite side Fytotal
• adjancent side Fx
  total
• Tan O 346.41 kN
• 300 kN
• O (Tan -1) ( 346.41 kN)
• (300 kN)
• O 49.11o
• Fnet 458.23 kN _at_ 49.11o

346.41 kN
O ?
300 kN
19
Light bulb moment!

• If we know how to solve for net force, we will be
  able to solve for the mass and acceleration of an
  object.