1. Refresher:
*Acceleration is only caused by an unbalanced net force acting on an object.
F = ma
F = F1 + F2 + …
*The weight of an object is referred to as a Gravitational Force
Fg = mg
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2. Refresher:
*Friction is a force that always opposes motion. “” is the coefficient of friction which represents the roughness or smoothness of the surfaces in contact.
Ff = FN
The equation is typically found in a “net force” equation.
*The Normal Force is a supporting force caused by a surface. The normal force is always perpendicular to the surface. Sometimes called the “apparent weight” of an object.
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3. 4.8 The Normal Force
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4. 4.8 The Normal Force
Apparent Weight
The apparent weight of an object is the reading of the scale.
It is equal to the normal force the scale exerts on the man.
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5. 4.8 The Normal Force
true
weight
apparent
weight
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6. FA *An Applied Force is any push or pull
Refresher:
*An Applied Force is any push or pull FA
The equation is typically found in a “net force” equation.
*A Tension Force (FT) is a supporting force caused by a string or cable.
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7. Tension and Normal forces help “support” the weight
Grocery Bag FT Fg FN Fg
Tension and Normal forces help “support” the weight
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8. The coefficient of kinetic friction between the red block and the floor is In order to accelerate this system at a rate of 2 m/s2, what force must be applied? Assume the top block does not slip.
Use g = 10 m/s2
4 kg
8 kg F
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9. Which Block will accelerate?
a.) Consider the frictionless pulley that has two masses hanging over each side. What will happen to the apparatus if the blocks are released from rest?
Which Block will accelerate?
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10. a.) Consider the frictionless pulley that has two masses hanging over each side. What will happen to the apparatus if the blocks are released from rest?
If the green block is 3 kg and the blue block is 1 kg, calculate that acceleration.
3 kg
1 kg
Use g = 10 m/s2
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11. Hint: set up a Net Force problem. Label the direction of your forces.
a.) Consider the frictionless pulley that has two masses hanging over each side. What will happen to the apparatus if the blocks are released from rest?
Hint: set up a Net Force problem. Label the direction of your forces.
Are these forces acting with or against each other?
3 kg
1 kg
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12. A 20. 0 kg mass sits on a tabletop
A 20.0 kg mass sits on a tabletop. The coefficient of kinetic friction between the tabletop and the mass is 0.25.
20.0 kg
It is attached to a 12.5 kg mass that is hung over a massless pulley by a string.
12.5 kg
Use g = 10 m/s2
What is the acceleration of the system?
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13. What is the magnitude and direction of the Net Force (Resultant)?
Two children push on a 5 kg toy box at the same time. One pushes with a force of 125 N West and the other pushes 165 N South.
What is the magnitude and direction of the Net Force (Resultant)?
R2 = x2 + y2
125 N
165 N Ө
Ө= Tan-1(o/a)
207 N
Ө = 53º S of W
Recall Pythagorean Thm.
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14. What is the magnitude and direction of the Acceleration?
Two children push on a 5 kg toy box at the same time. One pushes with a force of 125 N West and the other pushes 165 N South.
What is the magnitude and direction of the Acceleration?
F = ma
125 N
165 N Ө
207 N = (5kg)a
41.4 m/s2 = a
207 N
Ө = 53º S of W
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15. Two children push on a 5 kg toy box at the same time
Two children push on a 5 kg toy box at the same time. One pushes with a force of 125 N West and the other pushes 165 N South.
If a third child were to apply a force to this box to suddenly make it maintain a constant velocity (stop accelerating), how much force would be needed and in what direction?
207 N
Ө = 53º N of E
125 N
165 N Ө
207 N
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16. m F Fy = F sin  Fx = F cos Ff = 0 Fx = 100N cos30 = 86.6 N = ma
A 40.0 kg crate is pulled across the ice with a rope. A force of N is applied at an angle of 30 above horizontal. Neglect friction. A) Calculate the acceleration of the crate. F
Fy = F sin  m
Fx = F cos
Ff = 0
Fx = 100N cos30 = 86.6 N = ma
Fx = 86.6 N = (40kg)a
a = 2.17 m/s2
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17. A 40. 0 kg crate is pulled across the ice with a rope. A force of 100
A 40.0 kg crate is pulled across the ice with a rope. A force of N is applied at an angle of 30 above horizontal. Neglect friction. B) Calculate the upward force the ice exerts on the crate as it is pulled (FN). F
Fy = F sin
= 50 N  m
Since the box is not accelerating upward or downward, the sum of the vertical forces = 0 mg
Fnorm
Fy = Fy + Fnorm – mg = 0
Fy + Fnorm = mg
(Upward = Downward)
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18. 4.8 The Normal Force
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19. m = 342 N F Fy = F sin = 50 N  Fy + Fnorm = mg mg Fnorm
A 40.0 kg crate is pulled across the ice with a rope. A force of N is applied at an angle of 30 above horizontal. Neglect friction. B) Calculate the upward force the ice exerts on the crate as it is pulled. F
Fy = F sin
= 50 N  m
Fy + Fnorm = mg mg
Fnorm
50 N + Fnorm = (40kg)(9.8 m/s2)
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50 N + Fnorm = 392 N
= 342 N
Fnorm = 392 N – 50 N

20. Inclined Surface  Fnorm Fg Fnorm Fg = Fnorm = Fg cos
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This is called the perpendicular component

21. Inclined Surface F  Fnorm Fg F = Fg sin
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There is also a parallel component of Fg

22. What is the coefficient of friction? Ff Fg = Ff Fg sin = (Fg cos)
mg sin = (mg cos) 
sin = 
Suppose this person is not moving… Label the forces
cos
tan = 
Suppose  = 30°
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23. What is the coefficient of friction? Ff Fg = Ff
Same answer as before…except this would represent the Kinetic Friction 
Suppose this person slides downhill at a constant velocity
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24. What is the acceleration?
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What is the acceleration? Fg
= F
Fg sin = ma Fg
mg sin = ma 
g sin = a
Suppose this person is accelerating downhill because there is zero friction.
Suppose  = 30°
Remember this for the lab

25. Pg. 25

26. Fa Fg Ff Fa = Ff + mg Fa = (Fg cos) + Fg sin
Because it slides at const velocity
Fa = (Fg cos) + Fg sin
Suppose this block gets pushed uphill at a constant velocity…label the forces
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27. FT mg FT Ff mg 
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