Acceleration a=F/m

Velocityv= v0 + at

Positionx = x0 + v0t + ½ at2

FORCES

NET FORCE

FREE BODY

DIAGRAM

FBD

From Simple to Complex

• 1) Horizontal plane

• 2) Inclined plane

Drawing a FBD of forces on an object (on, not by)

1. Choose the object to analyze. Draw it as a dot.2. What forces physically touch this object?

This object, not some other3. What “action at a distance” forces act on the object?

Gravity is the only one for this PHYS2053

4. Draw these forces as arrows with tails at the dot (object).5. Forces only! No accelerations, velocities, …

Get components of Newton’s 2nd Law

Choose a convenient xy coordinate systemFind the x and y components of each force in the FBDAdd the x and y components separately

x

y

x positivey positive

x negativey positive

x

y

x

y

x negativey negative

x

y

x positivey negative

x

y

??????

x positivey positive

x

y

Horizontal plane

• 1) Horizontal force

A horizontal force Fa of 12 N is appliedTo a block with mass m=6kg, on a frictionlesstable. The block was originally at rest when the force was applied. Draw a FBD and find the acceleration of the block and its velocity after it travels 0.4m from the origin

Fa

FBD

mFa

N

W

Normal force

weight

applied force

x

y

Normal force: mg = 6kg* 9.81 m/s2 (along y)Weight: -mg (along y)Applied force = 12 N along x

The net force F : the y component is zero because the normalForce and the weight cancel. The x component is the appliedForce. Hence Fx = 12 N and Fy = 0 N

The net force F : the y component is zero because the normal

Force and the weight cancel. The x component is the appliedForce. Hence Fx = 12 N and Fy = 0 N

Applying the second Newton/s law

F = m a

Fx = max ax = Fx/m

ax = 12N/6kg ax = 2 m/s2

Velocity after it travels 0.4 m from the origin:

The block was originally at rest: vx0 = 0m/svx

2 = vx02 + 2ax(x-x0)

vx = √2ax(x-x0) vx = 1.26 m/s

Horizontal plane• 2) Force at an angle

A force Fa of 15 N making an angle of 35o from the horizontal is applied to a block with mass m=6kg, on a frictionless table. The block was originally at rest when the force was applied. Draw a FBD and find the acceleration of the block and its velocity after it travels for 5 seconds from the origin

FBD

m

Fa

N

W

Normal force

weight

applied force

x

y

Fa,y

Fa,x

There is no motion in the yDirection (the block does not jump !!!)Fy = 0 NHence:Normal force N = w+Fsin(

Motion along x:

Fa,x = m ax

ax = 15N cos (35o)/6kg

ax = 2.05 m/s2

vx = vo + axtvx = 0m/s + (2.05m/s2)(5s) vx= 10.25 m/s

Fx = Fcos(Fy = Fsin(

Horizontal plane• 3) Force at an angle + friction

A force Fa of 15 N making an angle of 35o from the horizontal is applied to a block with mass m=6kg, on a table with friction force Ff opposing the motion of the block of 5.2 N magnitude. The block was originally at rest when the force was applied. Draw a FBD and find the acceleration of the block and its position after it travels for 5 seconds from the origin

FBD

m

Fa

N

W

Normal force

weight

applied force

x

y

Fa,y

Fa,x

There is no motion in the yDirection (the block does not jump !!!)Fy = 0 NHence:Normal force N = w+Fsin(

Motion along x:

Fa,x – Ff = m ax

ax = (15N cos (35o) -5.2N)/6kg

ax = 1.18 m/s2

x = xo + voxt + (1/2) axt2

x = (1/2)(1.85m/s2)(5s)2 x = 14.77 m

Fx = Fcos(Fy = Fsin(

Ff

x

y

α

β ???

β = 90o - αβ = 90o - (90o - )

β =

α = 90o -

Inclined plane1) Only gravitational force

M =25 kg θ= 25o h= 2 m

M

h

Question: What is the velocity of the block at the bottom of the frictionless incline?

d

d= h/sin(

FBD

Wx = W sin θ Wy = W cos θ N =-Wy

y

W

N

x

Wx

Wyθ

y-component of the net force is zero!

M

h

Fa

Ff

A box with mass 45 kg is at rest when a force Fa (20N)making an angle of 370 with the inclined plane. The inclined plane makes an angle of 220 with the horizontal plane. When the box is moving on the inclinedPlane, there is a friction force Ff of 5 N opposing the motion. The box wasOriginally at a height h from the ground (h=4 m).Draw the free body diagram for the box. Determine the components x, y for all the forces acting on the box. Find the net force and accelerationOf the box.

The conditions for a particle to be in equilibrium

• Necessary conditions for an object to settle into equilibrium F = 0 Fx = 0 and Fy = 0

•Both x and y forces must be considered separately.

Homework from chapter 4

• 3, 9, 14, 20, 21, 23, 30, 34, 41, 52