# Net Force Problems There are 2 basic types of net force problems

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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

“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

“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

“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

“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

“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

“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

“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

“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

“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

“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

“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

“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

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

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”

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

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

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

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

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

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

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

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

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.

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

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

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

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 )

“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

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

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

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

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

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

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

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