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Net Force Problems There are 2 basic types of net force problems On a horizontal surface, the formula is: F net = F A - F f If it is a “up-down” problem,

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Presentation on theme: "Net Force Problems There are 2 basic types of net force problems On a horizontal surface, the formula is: F net = F A - F f If it is a “up-down” problem,"— Presentation transcript:

1 Net Force Problems There are 2 basic types of net force problems On a horizontal surface, the formula is: F net = F A - F f If it is a “up-down” problem, the formula is: F net = F up - F down F up = applied force F down = weight

2 “Up – Down” F up F 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?

3 “Up – Down” F up F 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? F net = F up - F down

4 “Up – Down” F up F 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? F net = F up - F down (m)(a) = 148 N - (10 kg)(9.8m/s 2 )

5 “Up – Down” F up F 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? F net = F up - F down (10kg)(a) = 148 N - (10 kg)(9.8m/s 2 )

6 “Up – Down” F up F 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? F net = F up - F down (10kg)(a)= 148 N - (10 kg)(9.8m/s 2 ) (10kg)(a)= 148 N - 98 N

7 “Up – Down” F up F 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? F net = F up - F down (10kg)(a) = 148 N - (10 kg)(9.8 m/s 2 ) a = 148N-98N / 10 kg a = 5 m/s 2

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

9 “Up – Down” F up F down A 2.5 kg ball is thrown upward with an acceleration of 3.5 m/s s. What force is needed to do this? F net = F up - F down

10 “Up – Down” F up F down A 2.5 kg ball is thrown upward with an acceleration of 3.5 m/s s. What force is needed to do this? F net = F up - F down (m)(a) = F up - (m)(g)

11 “Up – Down” F up F down A 2.5 kg ball is thrown upward with an acceleration of 3.5 m/s s. What force is needed to do this? F net = F up - F down (m)(a) = F up - F down (2.5 kg)(3.5 m/s 2 ) = F up - (2.5kg)(9.8m/s 2 )

12 “Up – Down” F up F down A 2.5 kg ball is thrown upward with an acceleration of 3.5 m/s s. What force is needed to do this? F net = F up - F down (m)(a) = F up - F down (2.5 kg)(3.5 m/s 2 ) = F up - (2.5kg)(9.8m/s 2 ) 8.75 N = F up N

13 “Up – Down” F up F down A 2.5 kg ball is thrown upward with an acceleration of 3.5 m/s s. What force is needed to do this? F net = F up - F down (m)(a) = F up - F down (2.5 kg)(3.5 m/s 2 ) = F up - (2.5kg)(9.8m/s 2 ) 8.75 N = F up N F up = N

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 1. FfFf FAFA 2. θ FAFA FfFf This is what we did last chapter Now the applied force is at an angle

17 Forces in 2 dimensions θ Now, only the horizontal component of the force is used FAFA FfFf FhFh

18 Forces in 2 dimensions θ FAFA FhFh FfFf You find the horizontal component by using the following formula: Cos θ = F h F A

19 Forces in 2 dimensions θ FAFA FhFh FfFf 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

20 Forces in 2 dimensions 2.5 kg 35° Only the horizontal component of the force is used 45 N FfFf 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

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

22 Forces in 2 dimensions Next you need to calculate the force of friction F f = µ(m)(g) F f = (0.3)(2.5 kg)(9.8m/s 2 ) F f = 7.35 N

23 Forces in 2 dimensions Finally, you put it all together into a net force equation: F net = F h - F f (m)(a) = 37 N N (2.5 kg)(a) = N a = 11.7 m/s 2

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. FfFf F II FWFW F┴F┴ FNFN

26 Forces in 2 dimensions Inclined Plane or “The Ramp” - when box is being pushed up the ramp. FAFA F 11 FWFW F┴F┴ FNFN FfFf

27 Forces in 2 dimensions The forces on an inclined plane are: F F Force of Friction- always opposes F A F 11 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 F N Normal Force – it is not equal to the weight - it is equal to the perpendicular force F w Weight force – always points straight down F A 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: F N = F ┴ (always) F 11 = F f If F 11 > F f object will accelerate F 11 = Sin θ F W F ┴ = Cos θ F W F f = µ F ┴ (or F f = F N )

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? F 11 FfFf FwFw F┴F┴

30 “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? F 11 FfFf F w = 562N F┴F┴ 30°

31 First, you have to redraw the parallel force vector. F 11 FfFf F w = 562N F┴F┴ 30° F 11 30°

32 Second, find F 11 and F ┴ F 11 = (Sin 30°)(562N) F ┴ = (Cos30)(562N) F 11 = 281 N F ┴ = 487N F 11 FfFf F w = 562N F┴F┴ 30° F 11

33 Third, find the Force of friction F f = µF N F f = (0.2)(487N) F F = 97.4 N F 11 FfFf F w = 562N F┴F┴ 30° F 11

34 Next, you need to know the mass. F w = mg 562 N = m (9.8 m/s 2 ) m = 57.3 kg F 11 FfFf F w = 562N F┴F┴ 30° F 11

35 Finally, use the net Force equation: F net = F 11 – F f (m)(a) = 281 N N 57.3 kg (a) = N a = 3.2 m/s 2 F 11 FfFf F w = 562N F┴F┴ 30° F 11


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