FRICTION
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Transcript of FRICTION
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FRICTION
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Friction- a force that resists the motion of an object
• Acts parallel to the surface and opposite the direction of the motion.
• Dependent on the type of contact surfaces.• Independent of contact area.• Equal to applied force when object is at rest or
traveling at a constant velocity.
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TYPES OF FRICTION
• Static friction(starting friction)- Friction that is produced by surface projections and bonding.
• Kinetic friction(sliding friction)- friction an object has while in motion.
• Rolling friction-one object rolls over another• Air resistance- friction produced by the air
pushing against something in free fall.
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FORMULA
Ff = µ FNFf = frictional force (N)
FN = normal force (N)
µ-Coefficient of friction*(µ) Has No Units*
*See reference table*
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ExA. A block of wood rests on a wood desktop. If it has a mass of 10kg, what is the minimum force required to move it?
FN = Fg = mg
FN = 10kg(9.81m/s2)
FN = 98.1N
Ff = µ FN = .42(98.1N) = 41.2N
• All forces are balanced if at rest or traveling at a constant velocity.
• The value for static friction is used because its at rest.
Ff= 41.2N FN= 98.1N
F= 41.2N
Fg= 98.1N
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exB-If a 60N force is applied to the block, what will be its acceleration?
Fnet = 60N – 41.2N
Fnet = 18.8N
a = Fnet/m = 18.8N/10kg
a = 1.88 m/s2
Ff= 41.2N F = 60N
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FORCE VECTOR DIAGRAMSSTATIONARY OBJECT:
Fg- weight Fg = mg
Fn – normal force- force
pushing surfaces together Fn = Fg
Fg
Fn
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Applied Parallel force
Fp - Parallel force
Ff – frictional force
Fn – normal force
Fg- weight- Fn
At constant velocity Fp = Ff
FpFf
Fg
Fn
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Applied force at an angle
Fa – applied force
Fp – parallel force
Fp = Fa cosΘ = Ff
Fy = FasinΘ
Fn = Fg - Fy
Fa
Fp
Fg = mg
Ff
Fy
Fn
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A worker pulls a 40kg crate across the floor at a constant velocity. If he applies 214N of force along a rope held at 30 degrees with the ground, find Fg , Fp, Fy , Fn , and μ.
Fg = mgF = 40kg(9.81m/s2)Fg =392.4N
Fp = Fa cosθ = Ff
Fp = 214N cos 30Fp = 185.3N
Fy = Fa sin θFy = 214N sin 30Fy = 107N
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Fn = Fg – Fy
Fn = 392.4N – 107NFn = 285.4N
μ = Ff / Fn
μ = 185.3N/ 285.4Nμ = 0.65
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Sliding object on an incline
Ѳ
Fn
Fn
Ff
Fg
Fp
Fg = mg (always perpendicular to the ground)
Fn = Fg cosθ (perpendicular to the surface)
Fp = Ff ( both are parallel to the surface)
Fp = Fg sinθ
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Ex: A 20kg wooden box is sliding down an incline of 30 degrees at a constant velocity. Find
Fg , Fp , Fn and μ.
Fg = m gFg = 20kg(9.81m/s2)Fg = 196N
Fp = Fg sin ѲFp = 196N sin 30Fp= 98N
Fn = Fg cosѲFn= 196N cos 30Fn = 170N
µ = Ff / Fn = 98N/170N = .576
FnFg
Fp
θ
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COOL MATH TRICK
Sin Ѳ = opp/hyp cos Ѳ= adj/hyp
µ = Ff = Fgsin Ѳ = opp/hyp = opp/adj
Fn Fg cos Ѳ adj/hyp
µ = tan Ѳ
*Only works for objects sliding down an incline*