1. Net Force Problems There are 2 basic types of net force problems
On a horizontal surface, the formula is:
Fnet = FA Ff
If it is a “up-down” problem, the formula is:
Fnet = Fup - Fdown
Fup = applied force Fdown = weight

2. “Up – Down” Fup If a ball is thrown upward with a force of 148 N
and has a mass of 10 kg, what will it’s acceleration
be?
Fdown

3. “Up – Down” If a ball is thrown upward with a force of 148 N
and has a mass of 10 kg, what will it’s acceleration
be?
Fnet = Fup Fdown
Fup
Fdown

4. “Up – Down” If a ball is thrown upward with a force of 148 N
and has a mass of 10 kg, what will it’s acceleration
be?
Fnet = Fup Fdown
(m)(a) = 148 N - (10 kg)(9.8m/s2)
Fup
Fdown

5. “Up – Down” If a ball is thrown upward with a force of 148 N
and has a mass of 10 kg, what will it’s acceleration
be?
Fnet = Fup Fdown
(10kg)(a) = 148 N - (10 kg)(9.8m/s2)
Fup
Fdown

6. “Up – Down” If a ball is thrown upward with a force of 148 N
and has a mass of 10 kg, what will it’s acceleration
be?
Fnet = Fup Fdown
(10kg)(a)= 148 N - (10 kg)(9.8m/s2)
(10kg)(a)= 148 N N
Fup
Fdown

7. “Up – Down” If a ball is thrown upward with a force of 148 N
and has a mass of 10 kg, what will it’s acceleration
be?
Fnet = Fup Fdown
(10kg)(a) = 148 N - (10 kg)(9.8 m/s2)
a = 148N-98N / 10 kg
a = 5 m/s2
Fup
Fdown

8. “Up – Down” Another type of “up –down” problem is where the
acceleration is given and you need to find the “up” force.
A 2.5 kg ball is thrown upward with an acceleration of
3.5 m/ss. What force is needed to do this?
Fup
Fdown

9. “Up – Down” A 2.5 kg ball is thrown upward with an acceleration of
3.5 m/ss. What force is needed to do this?
Fnet = Fup - Fdown
Fup
Fdown

10. “Up – Down” A 2.5 kg ball is thrown upward with an acceleration of
3.5 m/ss. What force is needed to do this?
Fnet = Fup - Fdown
(m)(a) = Fup - (m)(g)
Fup
Fdown

11. “Up – Down” A 2.5 kg ball is thrown upward with an acceleration of
3.5 m/ss. What force is needed to do this?
Fnet = Fup - Fdown
(m)(a) = Fup - Fdown
(2.5 kg)(3.5 m/s2) = Fup - (2.5kg)(9.8m/s2)
Fup
Fdown

12. “Up – Down” A 2.5 kg ball is thrown upward with an acceleration of
3.5 m/ss. What force is needed to do this?
Fnet = Fup - Fdown
(m)(a) = Fup - Fdown
(2.5 kg)(3.5 m/s2) = Fup - (2.5kg)(9.8m/s2)
8.75 N = Fup N
Fup
Fdown

13. “Up – Down” A 2.5 kg ball is thrown upward with an acceleration of
3.5 m/ss. What force is needed to do this?
Fnet = Fup - Fdown
(m)(a) = Fup - Fdown
(2.5 kg)(3.5 m/s2) = Fup - (2.5kg)(9.8m/s2)
8.75 N = Fup N
Fup = N
Fup
Fdown

14. Horizontal Problems
In order to do these, we need to understand the force of friction first.
Friction.ppt

15. Chapter 7 Forces in 2 Dimensions
There are two types of problems
One type is when the applied force is at an angle with the surface
The other type is when the object is on an inclined plane (“ramp problems”

16. Forces in 2 dimensions Ff FA 1. This is what we did last chapter FA 2.θ
Now the applied force is at an angle

17. Forces in 2 dimensions FA Ff Fh θ
Now, only the horizontal component of the force is used

18. Forces in 2 dimensions
You find the horizontal component by using the following
formula:
Cos θ = Fh FA FA Ff θ Fh

19. Forces in 2 dimensions A sled is pulled with a force of 45 N and makes
an angle of 35º with the surface. If the sled has a mass
2.5 kg, what is it’s acceleration if the µ is 0.30 FA Ff θ Fh

20. Forces in 2 dimensions A sled is pulled with a force of 45 N and makes
an angle of 35º with the surface. If the sled has a mass
2.5 kg, what is it’s acceleration if the µ is 0.30
45 N
2.5 kg Ff
35°
Only the horizontal component of the force is used

21. Forces in 2 dimensions
First you have to determine the horizontal component.
Cos 35° = Fh
45 N
Fh = 37 N

22. Forces in 2 dimensions
Next you need to calculate the force of friction
Ff = µ(m)(g)
Ff = (0.3)(2.5 kg)(9.8m/s2)
Ff = N

23. Forces in 2 dimensions
Finally, you put it all together into a net force equation:
Fnet = Fh - Ff
(m)(a) = 37 N N
(2.5 kg)(a) = N
a = m/s2

24. Forces in 2 Dimensions
The second type of problem is the inclined plane
or “Ramp” problems.
It easy to confuse these because they tend to
look alike.

25. Forces in 2 dimensions
Inclined Plane or “The Ramp” - when box is sliding down ramp. Ff FN
F II F┴ FW

26. Forces in 2 dimensions
Inclined Plane or “The Ramp” - when box is being pushed
up the ramp. FA FN
F 11 Ff F┴ FW

27. Forces in 2 dimensions The forces on an inclined plane are:
FF Force of Friction- always opposes FA
F11 Parallel Force- Force that causes box to slide
down the ramp.
It always going down the ramp
F ┴ Perpendicular Force – It is the force that the ramp
pushes up with. It is equal but opposite to Normal
Force
FN Normal Force – it is not equal to the weight
- it is equal to the perpendicular force
Fw Weight force – always points straight down
FA Applied force – when box is being slid up the ramp

28. Forces in 2 dimensions
If the ramp is stationary or moving at a constant speed,
the following is true:
FN = F ┴ (always)
F11 = Ff
If F11 > Ff object will accelerate
F11 = Sin θ • FW
F ┴ = Cos θ • FW
Ff = µ F┴ (or Ff = FN )

29. “The Skier”
A skier goes down a mountain that makes a 30° angle
with the horizontal. If he weighs 562 N and the snow has
a coefficient of friction of 0.20, how fast will he accelerate? Ff
F11 F┴ Fw

30. A skier goes down a mountain that makes a 30° angle
“The Skier”
A skier goes down a mountain that makes a 30° angle
with the horizontal. If he weighs 562 N and the snow has
a coefficient of friction of 0.20, how fast will he accelerate? Ff
F11 F┴
30°
Fw = 562N

31. First, you have to redraw the parallel force vector.Ff
F11
30° F┴
30°
Fw = 562N
F11

32. F11 = (Sin 30°)(562N) F ┴ = (Cos30)(562N) F11 = 281 N F ┴ = 487N
Second, find F11 and F ┴
F11 = (Sin 30°)(562N) F ┴ = (Cos30)(562N)
F11 = 281 N F ┴ = 487N Ff
F11
30° F┴
30°
Fw = 562N
F11

33. Third, find the Force of friction Ff = µFN Ff = (0.2)(487N)
30° F┴
30°
Fw = 562N
F11

34. Next, you need to know the mass. Fw = mg 562 N = m (9.8 m/s2)
m = 57.3 kg Ff
F11
30° F┴
30°
Fw = 562N
F11

35. Finally, use the net Force equation: Fnet = F11 – Ff
(m)(a) = 281 N N
57.3 kg (a) = N
a = 3.2 m/s2 Ff
F11
30° F┴
30°
Fw = 562N
F11