1. Weight = mass x acceleration due to gravity
Weight, Wt. is the gravitational force acting on an object
Your weight is determined by both your mass and the strength of the gravitational field (the acceleration due to gravity, “g”)
Weight = mass x acceleration due to gravity
Wt. = mg
Since weight is a force, it is measured in Newtons, N
Remember, “g” on Earth is 9.8 m/s2

2. Tension Force
Tension, T, is the force that cables, ropes, and strings pull with.

3. Fnet = ma
A child pulls up on a string that is holding 2 fish of total mass 5 kg. If he is providing a tension of 60 N, what is the net force on the fish?
Fnet = Tension – Weight
Fnet = 60 N – 50 N
Fnet = 10 N
What is the acceleration of the fish?
a = Fnet / m
a = 10 N / 5 kg
a = 2 m/s2

4. The “Normal” Force, N
When an object is pressed against a surface, the surface pushes back.
(That’s Newton’s 3rd Law)
This “push back” from the surface is called the Normal Force, N
The word “normal” in math terminology means “perpendicular”
The surface pushes back in a direction that is perpendicular to the surface.

5. Normal Force, N
Force applied, FA

6. But… what if you were accelerating up or down, like in an elevator?mg N
The Normal force (the floor pushing up on you) would NOT be equal to your weight if you’re accelerating up or down.

7. Free body diagram for pushing a box across a table

8. Normal force, N FA = 20 N f = 4 N m = 2 kg a = ? Fnet = ma
You push on a box with an applied force of 20 N. The frictional force opposing the motion is 4 N. If the mass of the box is 2 kg, what is its acceleration?
Draw the free body diagram.
Write what you know and don’t know.
Write the equation, Fnet = ma
Calculate the value of the net Force and then the acceleration.
Normal force, N
FA = 20 N f = 4 N m = 2 kg
a = ?
Fnet = ma
20 N – 4 N = ma
a = Fnet / m
a = 8 m/s2
f = 4 N
FA = 20 N
Weight = mg

10. Normal force, N
Weight = mg
If the box is not accelerating up or down, then the support force from the table, the “Normal force”, must by equal to the Weight

11. But… what if you were accelerating up or down?
The Normal force would NOT be equal to your weight if you’re accelerating up or down. And… you feel heavier or lighter! This feeling is called your “apparent weight”.

12. Apparent Weight
When you ride an elevator, you “feel” heavier or lighter than you actually weigh because of the acceleration of the elevator.
Your own sense of your weight is called your “apparent weight”. That sensation comes from the support underneath us.
If the floor underneath you quickly starts pushing up hard upon your feet, you feel heavier.
If the floor underneath your feet quickly starts moving downward, you feel lighter.
If the floor underneath your feet falls away completely, you feel weightless!

13. A 50 kg woman steps on a scale in an elevator that accelerates upward at 1.5 m/s2.
What is her weight (use g = 10 m/s2) ?
Weight = mg = 500 N
How heavy does she feel- in other words,
what is her APPARENT weight?
SF = ma
N – mg = ma
Her APPARENT weight is what she feels like she weighs, which is determined by how hard the floor is pushing up against her- the “Normal” force.
N = mg + ma
N = 500 N kg x 1.5 m/s2
Apparent weight = 575 N mg N

14. Do falling objects REALLY accelerate toward the Earth at 9.8 m/s2?
No, because of air resistance.
Air resistance is a force that pushes up on an object as it falls.
The faster you fall…
The greater the air resistance.
Eventually, the air resistance pushing up on you is just as large as your weight that is pulling down on you!!

16. The faster the man falls, the more air resistance pushes up on him
The faster the man falls, the more air resistance pushes up on him. Eventually, there will be just as much air resistance pushing up on him as his weight pulling him down. What will be the NET force acting then?
What will be his acceleration?

17. Air resistance = your weight
Once the air resistance pushing up is as large as the weight pushing down, the NET force acting on you is ZERO!
If the net force is zero, what is your acceleration?
ZERO!
This doesn’t mean you stop in mid air. But it does mean that you stop accelerating!
You still continue to fall towards the Earth, but you don’t pick up any more velocity- you continue to fall towards the Earth at the same velocity.
This speed is called your “terminal velocity”
You will reach your terminal velocity when
Air resistance = your weight

18. Which one will have a faster terminal velocity?
You don’t reach terminal velocity until the air resistance grows to as large a force as your weight.
The more massive skydiver will have a faster terminal velocity and hit the ground at a faster speed

19. Terminal Velocity and Cats

21. Terminal Velocity and Cats

23. Apparent weight = mg + ma
When you ride an elevator, you “feel” heavier or lighter than you actually weigh because of the acceleration of the elevator.
Your “apparent weight” comes from your support force and is found by taking your REAL weight, mg, and adding the term ma, where “a” is your acceleration in the elevator.
Apparent weight = mg + ma

24. A man of mass 75 kg is standing inside an elevator.
What is his weight (use g = 10 m/s2)?
750 N
How heavy does he feel when the elevator has an acceleration of -2 m/s2?
Apparent weight = mg ma
Apparent weight = 75 kg x 10 m/s kg x (-2 m/s2) = 600 N
How heavy does he feel when the elevator is moving upward with a constant velocity?
Since a = 0, he only feels his real weight- 750 N.

25. and Newton’s second law
FRICTION
and Newton’s second law

26. The “Normal” Force, N
When an object is pressed against a surface, the surface pushes back.
(That’s Newton’s 3rd Law)
This “push back” from the surface is called the Normal Force, N
The word “normal” in math terminology means “perpendicular”
The surface pushes back in a direction that is perpendicular to the surface.

27. Normal force, N
Weight = mg
If the box is not accelerating up or down, then the net force is zero and the Normal force must balance the Weight

28. Normal force, N FA = 20 N f = 4 N m = 2 kg a = ? Fnet = ma
You pull on a box with an applied force of 20 N. The frictional force opposing the motion is 4 N. If the mass of the box is 2 kg, what is its acceleration?
Draw the free body diagram.
Write what you know and don’t know.
Write the equation, Fnet = ma
Calculate the value of the net Force and then the acceleration.
Normal force, N
FA = 20 N f = 4 N m = 2 kg
a = ?
Fnet = ma
20 N – 4 N = ma
a = Fnet / m
a = 8 m/s2
f = 4 N
FA = 20 N
Weight = mg

29. Friction, f f = mN A force that always opposes motion
Depends on two things: the roughness of the surfaces and how hard they are pressed together.
f = mN
, mu- the “coefficient of friction” tells how rough the surfaces are.
N, the Normal force tells how hard the surfaces are pressed together

30. Example: How large is the frictional force between 2 surfaces if the coefficient of friction is 0.2 and the Normal force is 80 N?
f = mN
f = 0.2 x 80 N
f = 16 N

31. Static friction > Kinetic friction
There are two kinds of friction:
“static friction” (not moving) must be overcome to initiate motion.
“kinetic friction” must be overcome while an object is moving
Static friction > Kinetic friction

32. Applied Force
Friction = mN
IF you pull just hard enough to make the object move, you overcome static friction: Your applied force = static friction force Your applied force = mstaticN Once an object is moving, if you pull in such a way that the object is moving with constant velocity, then Your applied force = kinetic friction force Your applied force = mkineticN And… what is the normal force, N? N = mg

33. A 75 kg firefighter is sliding down the pole at the fire station
A 75 kg firefighter is sliding down the pole at the fire station. If he is moving at a constant velocity, what is the friction force? Fnet = ma friction – weight = ma = 0 Therefore, friction = weight friction = mg = 75 kg x 10 m/s2 friction = 750 N
friction
weight

34. Normal force, N FA = 20 N f = 4 N m = 2 kg a = ? Fnet = ma
You pull on a box with an applied force of 20 N. The frictional force opposing the motion is 4 N. If the mass of the box is 2 kg, what is its acceleration?
Draw the free body diagram.
Write what you know and don’t know.
Write the equation, Fnet = ma
Calculate the value of the net Force and then the acceleration.
Normal force, N
FA = 20 N f = 4 N m = 2 kg
a = ?
Fnet = ma
20 N – 4 N = ma
a = Fnet / m
a = 8 m/s2
f = 4 N
FA = 20 N
Weight = mg

35. Horizontal : Fnet = FA - f Fnet = 30 N – 8 N = 22 N a = Fnet / m
You pull on a box with an applied force of 30 N. The coefficient of friction is m = If the mass of the box is 2 kg, what is its acceleration?
Draw the free body diagram.
Write what you know and don’t know.
Write the equations, Fnet = ma and f = mN
Calculate the Normal force and the friction force.
Calculate the value of the net Force and then the acceleration.
Normal force, N
FA = 30 N m = 2 kg
m = 0.4 a = ?
Fnet = ma f = mN
N = mg = 2 kg x 10 m/s2 = 20 N
f = mN = 0.4(20 N) = 8 N
Horizontal : Fnet = FA - f
Fnet = 30 N – 8 N = 22 N
a = Fnet / m
a = 11 m/s2
f = mN FA
= 30 N
Weight = mg

36. Pre-AP only…

37. FA = 25N m = 2 kg q = 43 m = 0.4 a = ? SFx = ma f = mN
Fx = - mN + FA cos q = ma
(-0.4(36.65) + 25cos 43 ) / 2= a
a = 1.81 m/s2
N = mg + Fsin q
= N
Normal force, N f q FA
Weight = mg

38. FA = 15 N m = 2 kg q = 43 m = 0.4 a = ? SFx = ma f = mN
S Fx = - mN + FA cos q = ma
(-0.4(9.77) + 15cos 43) / 2 = a
a = 3.53 m/s2
N = mg – Fsin q
= 9.77 N
Normal force, N FA q f
Weight = mg

39. If the box is moving at constant velocity, there is no acceleration,
Therefore the net force must be zero…
so the horizontal forces must cancel each other.
mN = Fcos q
If you push hard enough to just get the box moving, the acceleration is zero in that case also, but the friction is static, not kinetic.
Normal force, N FA q f
Weight = mg

41. Friction Lab frictional force = (m)Normal force f = mN
Normal force, N = Wt1
frictional force, f = Wt2
Weight = mg N
Frictional force, f
Wt1

42. Pre-AP Only…

43. More examples…
Xavier, who was stopped at a light, accelerated forward at 6 m/s2 when the light changed. He had dice hanging from his rear view mirror. What angle did they make with the vertical during this acceleration? 1- draw free body diagram 2- write Newton’s 2nd law for BOTH directions SFx = max Tsinq = max Tcosq = mg tanq = a/g q = degrees q T
SFy = may
Tcosq – mg = may = 0 mg

44. An “Atwood’s Machine” + a SFext = mtotala
Two masses of 5 kg and 2 kg are suspended from a massless, frictionless pulley, When released from rest, what is their acceleration? What is the Tension in the string?
1- Draw a free body diagram
2- Write Newton’s Second Law for EXTERNAL forces acting on the whole system.
SFext = mtotala
The Tension in the string is an “internal” force. The only “external” forces are the gravitational forces of weight, which oppose each other. So…
SFext = m1g – m2g = mtotala
Solving for the acceleration yields
a = 4.2 m/s2
Now, to find the Tension, we must “zoom in” on mass 2:
SF= T – m2g = m2a
T = m2a + m2g
T = 28 N a +

45. Objects on Inclines- sliding down, no friction
The free-body diagram ALWAYS comes first:
Draw the weight vector, mg
Draw the Normal force vector.
Are there any other forces???
Since the motion is parallel to the plane, ROTATE the axis from horizontal and vertical to “parallel” and “perpendicular”. Then draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle.
Does the Normal force balance with the force of weight, mg?
NO! What force must balance the Normal force?
N = mgcosq
Does all of the weight, mg, pull the box down the incline?
Write Newton’s Second Law for the forces on box parallel to the plane with “down” being negative.
S F = ma
-mgsin q = ma N
No! Only mgsinq q mg q

46. Inclines: pushed upward, no friction
Draw the weight vector, mg
Draw the Normal force vector.
Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle.
Write Newton’s Second Law for the box.
S F = ma
FA - mgsin q = ma N FA q mg q

47. Inclines: pushed downward, no friction
Draw the weight vector, mg
Draw the Normal force vector.
Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle.
Write Newton’s Second Law for the box.
S F = ma
- FA - mgsin q = ma N q FA mg q

49. Inclines: pushed upward, no friction
Draw the weight vector, mg
Draw the Normal force vector.
Draw the components of the weight vector AND the applied force, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle.
Write Newton’s Second Law for all forces acting parallel to the incline.
S F = ma
FA cos q - mgsin q = ma N q FA q mg q

50. Inclines: With friction, case 1: at rest or sliding down
Draw the weight vector, mg
Draw the Normal force vector.
Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle.
Write Newton’s Second Law for the box.
S F = ma
f - mgsin q = ma
mmgcos q - mgsinq = ma
mgcos q - gsinq = a
What is friction?
f = mN
What is N?
N = mgcos q
f = mmgcos q N f q mg q

51. Inclines: With friction, case 2: pushed downward
Draw the weight vector, mg
Draw the Normal force vector.
Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle.
Write Newton’s Second Law for the box.
S F = ma
f - mgsin q – FA = ma
mmgcos q - mgsinq – FA = ma
What is friction?
f = mN
f = mmgcos q N f FA q mg q

52. Inclines: With friction, case 3: pushed upward
Draw the weight vector, mg
Draw the Normal force vector.
Draw the components of the weight vector, both perpendicular and parallel to the incline. The components help form a right triangle. Label the angle.
Write Newton’s Second Law for the box.
S F = ma
FA - mgsin q – f = ma
FA - mgsinq - mmgcos q = ma
What is friction?
f = mN
f = mmgcos q N FA q f mg q

53. Friction along an incline
An object placed along an incline will eventually slide down if the incline is elevated high enough. The angle at which it slides depends on how rough the incline surface is. To find the angle where it slides:
Since it doesn’t move:
mgsinq = mmgcosq
Therefore:
Tan qmax = mmax, the coefficient of static friction Or
The angle qmax = tan -1 mmax q

54. Spring Forces

55. Spring Force Fs = kx, If a mass is suspended from a spring,
two forces act on the mass:
its weight and the spring force.
The spring force for many springs
is given by
Fs = kx,
Where x is the distance the spring is stretched (or compressed) from its normal length and “k” is called the spring constant, which tells the stiffness of the spring. The stiffer the spring, the larger the spring constant k.
This relationship is known as “Hooke’s Law”. Fs mg

56. Spring Force
If the mass is at rest, then the two forces are balanced, so that kx = mg These balanced forces provide a quick way to determine the spring constant of a spring, or anything springy, like a rubber band: Hang a known mass from the spring and measure how much it stretches, then solve for “k” ! k = mg/x
Fs= kx mg