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 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.

If Forces are balanced

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 4-41. 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.

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

10kgAcceleration = F/m = 29.3N/10kg =

2.93m/s/s to the right

Let’s Practice

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

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 & frictionGiven 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 + Vy

2)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 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?

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)

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