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Transcript
Forces and the Laws of Motion
Newton’s Laws: Chapter 4
Buoyant Force: Chapter 9.1
Quiz
Given: Force of 30N, east
Force of 40N, north
Find: resultant force: magnitude and
direction

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
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.
Classes of Forces


Contact force: results from physical
contact between two objects
Field force: does not require physical
contact: ex: gravity, electrical charges
Force: a push or a pull
Force Diagrams
Try this one, part of homework
1. 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.
If Forces are
balanced
Head to Tail Method
Parallelogram method
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
Everyday Forces: Chapter 4-4
1.
2.
Weight = Fg = mg = kg x 9.8m/s2 on earth
Normal force = Fnorm = Fn = opposite in
direction to contact surface = mgcosΘ
where Θ is the angle between the normal
force and the vertical line.
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.
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

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)

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
10kg
Acceleration = F/m = 29.3N/10kg =
2.93m/s/s to the right
Let’s Practice
Practice: Try this one on
Three 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
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
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
Quiz: Forces & friction
Given Fapplied = 125N 30º above the horizontal
East, mass = 250 kg, and Ff = 25N
Find: Fg, Fn, µ, and a
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
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 + Vy2)½
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
Quiz: Putting it together
A 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?
Buoyant force
Fluids: 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)
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.
Buoyant force (FB) = Fg (displaced fluid) = mfg
Buoyant force on floating object = weight of
floating object