Linear motions and Newton's law

Breaking down forces

Physics G10 Class 1

Free body diagrams & vectors

​Vectors

​Arrows that indicate the direction a force is applied to. A vector consists of two main components: directions and magnitudes.

​Free body diagram case 1:

​An object does not move(The laws of inertia)

f = F(Static friction)

​mg = N

​Free body diagram case 2:

​An object moves in one fixed direction.(The law of forces)

f < F(Kinetic friction)

​mg = N

​Free body diagram case 3:

​An object moves in multiple direction simultaneously.

f < F(Kinetic friction)

​F is broken down into it's horizontal and vertical direction.

​mg = N

​Free body diagram case 4:

Free fall incline planes

mg is broken down into it's parallel and perpendicular direction

​N = mgcos(theta)

​g = mgsin(theta)

​Accelerations does not depend on mass but the angle of incline

​Free body diagram case 5:

Incline planes with friction

mg is broken down into it's parallel and perpendicular direction

​N = mgcos(theta)

​Net force = Fg - mgsin(theta)-uN

​Accelerations does not depend on mass but the angle of incline

Multiple Select

If the object is moving to the left what can be concluded?

• The force we apply(F) is less or equal to the friction force.

• We need to apply less force to make the object move slower or stop moving

• The friction force is lesser than the force we apply(F)

• The normal force is lesser than the gravitational force

Multiple Choice

If the object is floating and hovering above the ground what can we conclude?

• The force we apply(F) is less or equal to the friction force.

• The normal force is more than the gravitational force

• The friction force is lesser than the force we apply(F)

• The normal force is lesser than the gravitational force

If F=20N. Angle = 30 degrees

​

​Calculate the horizontal and vertical forces applied.

​Horizontal force = Fcos(30) = 20(sqr(3)/2) = 10sqr(3)N

​Vertical force = Fsin(30) = 20(1/2) = 10N

Multiple Choice

Calculate the horizontal and vertical force that is applied.

Angle = 45 degrees

• Horizontal = 20sqr(3)N

  Vertical = 20N

• Horizontal = 40N

  Vertical = 40sqr(3)N

• Horizontal = 40/sqr(2)

  Vertical = 40/sqr(2)

• Horizontal = 40sqr(3)N

  Vertical = 40N

• Horizontal = 20/sqr(2)

  Vertical = 20/sqr(2)

Multiple Choice

Calculate the horizontal and vertical force that is applied.

Angle = 30 degrees

• Horizontal = 20sqr(3)N

  Vertical = 20N

• Horizontal = 40N

  Vertical = 40sqr(3)N

• Horizontal = 20N

  Vertical = 20sqr(3)N

• Horizontal = 40sqr(3)N

  Vertical = 40N

• Horizontal = 40/sqr(2)

  Vertical = 40/sqr(2)

Fill in the Blank

Calculate the horizontal force that is applied.

Angles = 30 degrees

Newton's first law of motion

F = 0(The laws of inertia)

Newton's second law of motion

F = ma(The laws of forces)

Frictions

f = uN

Multiple Select

John is walking on a slippery floor, how can he increases his friction to make him not slip?

• Wear socks

• Wear rubber shoes

• Make the floor more wet

• Put a towel or carpet over the floor

Multiple Choice

The friction has it's coefficient(u) equal to 0.5 and the normal forces acting upon the applied force is 100N. We then apply a force of 75N to the object, will it move?

• Yes

  f = 50N

• No

  f = 100N

• Yes

  f = 100N

• No

  f = 50N

Newton's second law of motion's related formulas

F = ma

​f = uN

​N = mg-Fsin(theta)

F-f=ma​

​

​​​F = applied forces

​f = friction forces

​u = friction coefficient

​N = normal force

​mg = the force of gravity

​a = acceleration

Multiple Choice

If we apply the same force to a ball that weights 1 kg and a lawn mower that weights 12 kg. What can we conclude?

• The acceleration of the lawn mover will be 1/12 times the acceleration of the ball.

• The acceleration of the lawn mover will be 12 times the acceleration of the ball.

• The acceleration of the lawn mover will be 1/24 times the acceleration of the ball.

• The acceleration of the lawn mover will be 24 times the acceleration of the ball.

Multiple Choice

A table 12kg is pushed across a room with an acceleration of 4m/s^2. How much forces has to be apply to push this table successfully?

• 2N

• 36N

• 60N

• 48N

Multiple Choice

Tom pushed the piano with a force of 100N to the left and Annie pulled the piano in the opposite direction with a force of 50N. The piano weights 25kg. What will the acceleration of the piano be? Assuming there is no friction.

• 6m/s^2

• 2m/s^2

• 4m/s^2

• 6m/s^2 to the right

• 2m/s^2 to the left

Multiple Choice

Avery pushed the dresser to the other side of her room. She applies a force of 200N with a kinetic friction counteracting her force at 100N. The dresser accelerated at a rate of 2m/s^2. How much does the dresser weight?

• 20kg

• 50kg

• 25kg

• 100kg

Multiple Choice

วัตถุนึงมีมวล 10 kg เคลื่อนที่ด้วยนำประสิทธิ์แรงเสียดทานเท่ากับ 0.5 ถ้าเราใช้แรง 100N กระทำในทิศ 60 องศาบนพื้นที่ราบ ความเร่งจะเท่ากับเท่าใด g=10n/s^2

sqr(3) = 1.5

• ประมาณ 2m/s^2

• ประมาณ 1.25m/s^2

• ประมาณ 0.375m/s^2

• ประมาณ 0.5m/s^2

Multiple Choice

วัตถุนึงเคลื่อนที่ด้วยนำประสิทธิ์แรงเสียดทานเท่ากับ 1 ถ้าวัตถุมีมวล 5 kg และใช้แรง 100N กระทำในทิศ 53 องศาบนพื้นที่ราบ ความเร่งจะเท่ากับเท่าใด g=10n/s^2

sin(53) = 4/5

cos(53) = 3/5

• 18m/s^2

• 16m/s^2

• 20m/s^2

• 24m/s^2

Newton's third law of motion

Action = Reaction(The law of equilibrium)

Multiple Choice

If James hit the table with a force of 30N, what would be the appropriate reactive force that the table will do on James?

• -15N

• 30N

• -30N

• 15N

Calculate the vertical and horizontal force of a 10kg object on an incline plane with an angle of 30 degrees.

​Vertical = mgcos(30) = 100sqr(3)/2 = 50sqr(3)

​Horizontal = mgsin(30) = 100/2 = 50

Multiple Choice

Mass of the object is 5 kg.

Angle = 60 degrees

Find the vertical and horizontal force that is applied.

• Vertical = 50N

  Horizontal = 50sqr(3)N

• Vertical = 25N

  Horizontal = 25sqr(3)N

• Vertical = 50sqr(3)N

  Horizontal = 50N

• Vertical = 25sqr(3)N

  Horizontal = 25N

Multiple Choice

Mass of the object is 8 kg.

Angle = 30 degrees

Find the vertical and horizontal force that is applied.

• Vertical = 80N

  Horizontal = 80sqr(3)N

• Vertical = 40N

  Horizontal = 40sqr(3)N

• Vertical = 40sqr(3)N

  Horizontal = 40N

• Vertical = 80sqr(3)N

  Horizontal = 80N

Horizontal force = mgsin(theta)-uN

​Net horizontal force = Fg = mgsin(theta)-uN

​Net horizontal force = ma(Newton's second law)

​

Multiple Choice

A block slides down a 30 degree incline starting from resting. Find the acceleration.

g = 9.8m/s^2

• 4.9m/s^2

• 5m/s^2

• 4.8m/s^2

• 5.8m/s^2