forces and the laws of motion newton’s laws: chapter 4 buoyant force: chapter 9.1
TRANSCRIPT
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Forces and the Laws of Motion
Newton’s Laws: Chapter 4
Buoyant Force: Chapter 9.1
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Quiz Given: Force of 30N, east
Force of 40N, north
Find: resultant force: magnitude and
direction
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Changes in Motion Force: a push or a pull A push or pull changes the velocity of an
object whether at rest or moving Force increases directly as the mass of an
object increases Force increases directly as an object
accelerates to a higher speed In other words, Force = mass x acceleration
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SI unit for Force Force = mass (kg) x acceleration(m/s2)
= kg m/s2 = Newton (N)
Or Force = mass (g) x acceleration (cm/s2)
= g cm/s2) = dyne
1lb. = 4.448N
1N = 0.225 lb.
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Classes of Forces Contact force: results from physical
contact between two objects Field force: does not require physical
contact: ex: gravity, electrical charges
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Force: a push or a pull
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Force Diagrams
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Try this one, part of homework1. You are walking to class at a constant
velocity. To move forward at 0 degrees you push back at 180 degrees with a force of 15N. If you weigh 600N (140lbs) draw a vector diagram of the forces acting on your body as you walk.
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If Forces are balanced
If Forces are balanced
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Head to Tail Method
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Parallelogram method
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Try this one: also part of homework
Given 3 vectors
Vector A: 15N at 70 degrees
Vector B: 20N at 150 degrees
Vector C: 4kg at 270 degrees
Diagram it and determine the resultant force by finding the horizontal and vertical components
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Everyday Forces: Chapter 4-41. Weight = Fg = mg = kg x 9.8m/s2 on earth
2. Normal force = Fnorm = Fn = opposite in direction to contact surface = mgcosΘ where Θ is the angle between the normal force and the vertical line.
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Practice: Solve the following Mr. Trotts is standing on a ramp that has a
15 degree slope to the ground. If Mr. Trotts weighs 100kg what is the normal force acting on him.
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The Force of Friction Static friction:(Fs) a force resisting objects
at rest from moving by opposing forces applied attempting to set them in motion
Fs = -Fapplied
Kinetic friction:(Fk) retarding frictional force on an object in motion
Net external force: force causing object to change motion = F – Fk
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Coefficient of friction (µ) expresses the dependence of frictional forces on the particular surface they are in contact with Coefficient of kinetic friction: µk = Fk/Fn
Fn = mgcosΘ
Coefficient of static friction: µs = Fsmax/Fn
Force of friction: Ff = µ Fn
Try another one: part of homework
Practice 4C page 145 #2 (use sample 4C)
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Net Force Forces up should equal forces down if no
change in motion occurs up or down 69.3N right minus 40N left = 29.3N right If the object weighs 100N then its mass is weight/gravity = 100N/9.8m/s/s =about
10kgAcceleration = F/m = 29.3N/10kg =
2.93m/s/s to the right
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Let’s Practice
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Practice: Try this one onThree forces are acting on an object. 1. 15N at 90degrees 2. 25N at 220degrees 3. 20N at 300degrees Find the x and y components of each Find the object’s mass assuming no up or
down motion Calculate the net force acting on the object
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Fnet = ΣF = ma Practice: given a net force of 5N to the left
acting on a 20kg force find the objects rate of acceleration
a= F/m
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Try yet another: part of homework Given: forces acting on an object
40N at 60 degrees
35N at 170 degrees
Object weighs 100kg, Ff = 2N
Find: Fg, Fn, Fnet, µ, and the acceleration of the object
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Quiz: Forces & frictionGiven Fapplied = 125N 30º above the horizontal
East, mass = 250 kg, and Ff = 25N
Find: Fg, Fn, µ, and a
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Let’s mix it up Given m=1.5x107 kg, F=7.5x105 N,
Vi = 0, Vf = 85 km/h
Find time to increase speed from Vi to Vf
Use a = F/m, solve a = Vf – Vi /t for time
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Given: m = 3.00 kg, Δy = -176.4m,
Fw = 12.0N, g = 9.8 m/s/s
Find: time to hit ground, Δx, V
Use: Δy = -½ g Δt2, solve for t
ax = Fw/m, plug into Δx = ½ ax Δt2
Vy = -gΔt
Vx = axΔt
V = (Vx2 + Vy
2)½
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Given: m = 40.0 kg, Θ = 18.5º,
Fapplied,x = 1.40 x 102 N, Δx = 30.0 m,
g = 9.8m/s/s, Vi = 0 m/s
Find Vf = (Vi2 + 2ax Δx)1/2
Use Fgx = mg(sin Θ), Fgy = mg(cos Θ)
Fx,net = max = Fapplied,x –Fg,x
ax- = Fx,net / m
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Quiz: Putting it togetherA 250 kg box is pulled along a horizontal
surface. A rope attached to the box pulls the box east with a force of 125N at an angle of 25degrees above the horizontal. The force of friction acts with a force of 30N opposite the direction of the box. What is the coefficient of friction and acceleration of the box?
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Buoyant forceFluids: matter that flows (liquids or gases)
-liquid: has volume and takes the shape of its container.
-gas: has no shape and fills its container
Mass density (ρ) = mass (kg)/volume(m3)
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Buoyant force: upward force exerted on an object buy the fluid it is immersed in.
Apparent weight: the weight of an object immersed in a fluid.
Magnitude of buoyant force: (Archimedes principle) any object partially or completely immersed in a liquid experiences an upward buoyant force equal in magnitude to the weight of the fluid displaced by the object.
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Buoyant force (FB) = Fg (displaced fluid) = mfg
Buoyant force on floating object = weight of floating object