chapter 6 practice problems
DESCRIPTION
Chapter 6 Practice Problems. Equations. Sin θ = opposite side hypotenuse Cos θ = adjacent side hypotenuse Tan θ = opposite side adjacent side. Equations. v R 2 = v p 2 + v w 2 R 2 = A 2 + B 2 F v = F sin θ F h = F cos θ - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 6 Practice Problems
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Equations
• Sin θ = opposite side• hypotenuse• Cos θ = adjacent side• hypotenuse• Tan θ = opposite side• adjacent side
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Equations
• vR2 = vp
2 + vw2
• R2 = A2 + B2
• Fv = F sin θ
• Fh = F cos θ
• Fnet2 = Fnetx
2 + Fnety2
• A + B + W = 0
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• F ll = W Sin θ
• Fl = W cos θ
• Ff = μFN
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Problem 1
• Finding a resultant velocity.• An airplane flying toward 0o at 90 km /h is being
blown toward 90o at 50 km/h. What is the resultant velocity of the plane.
• Given: vp = 90 km /h at 0o vw = 50 km/h at 90o tan Θ = side opposite /side adjacent
• Unknown ; resultant velocity vr
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Problem 1
• Basic equation: R2 = A2 + B2 or• vR
2 = vp2 + vw
2
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Problem 1
• Answer: vR = 103 km/ h at 29o
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Problem 1
• Solution: • VR
2 = (90 km/h)2 + (50 km/h)2 =• sq root 1.06 X 104 (km/h)2
• vR2 = 103 km /h• Tan θ = 50 km/h• 90 km/h• Tan θ -1 = .556 = 29o
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Problem 2
– Resolving a Velocity Vector into its Components– A wind with a velocity of 40 km/h blows toward
30o.
– A. What is the component of the wind’s velocity toward 90o
– B. What is the component of the wind’s velocity toward 0o
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Problem 2
• Given: v= 40.0 km/h, 30.0o
• Unknown: v90, v0
• Solution: Toward 0o and 90o are positive. Angles are measured from 0o. To find the component toward 90o use the following:
• sin 30o = v90/v then v90 = 20 km/h at 90o
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Problem 2
• From 0o cos 30o =vo/v
• V0 = 34.6 km/h at 0.0o
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Problem 3
• Adding Non perpendicular Vectors• Two ropes(F1 = 12.0 N at 10.0o) (F2= 8.0 N at
120o ) are pulling on a log. What is the net force on the log?
• Given: F1 = 12.0 N at 10.0o
• F2 = 8.0 N at 120o
• Unknown: Fnet
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Problem 3
• Find the perpendicular components of each force.(figure 6-12)
• F1x =(12 N) cos 30o = 11.8 N
• F1y = (12 N) sin 30o = 2. N
• F2x = ( 8.0 N) cos 120o = -4.0 N
• F2y = (8.0 N) sin 120o = 6.9 N
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Problem 3
• Sum x and y components• Fnet x = F1x + F2x = 11.8 N + -4.0 N = 7.8 N
• F nety = F1y + F2y= 2.0 N + 6 .9 N = 8.9 N• Find the magnitude of the net force• F net = sq root Fx
2 +Fy2 = sq root of (7.8 N)2 +
(8.9 N )2 = 11 N
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Problem 3
• Find the angle of the force• Tan θ = 1.14• Θ = 49o
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Problem 4
• Finding Forces when Components are Known• A sign that weighs 168 N is supported by ropes
a and b (figure6-16)that make 22.5o angles with the horizontal. The sign is not moving. What forces do the ropes exert on the sign?
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Problem 4
• Given: • The sign is in
equilibrium• Weight = W = 168 N
(down)• Angles ropes make with
horizontal is 22.5o
• Unknowns:• Force of rope a is A• Force of rope b is B
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Problem 4
• Basic equations:• In equilibrium net force is 0• A + B + W = 0• Fh = F Cos θ
• Fv = F Sin θ
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Problem 4
• Since W is down the direction of A + B is up• Since the sum of A + B has no horizontal
components . Therefore Ah and Bh have equal magnitude
• Now, Ah = A Cos 22.5o and Bh = B cos 22.5o
• Av + Bv = 168 N• Since A = B then Av + Bv
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Problem 4
• Thus• Av = Bv = ½ (168 N )• = 84 N• A = Av
• sin 22.5o = 220 N• B = A = 220 N
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Problem 5
• Finding F l and Fll
• A trunk weighing 562 N is resting on a plane inclined at 30o from the horizontal (Figure 6-18 ) Find the components of the weight parallel and perpendicular to the plane.
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Problem 5
• Given: • W = 562 N• Θ = 30o
• Unknown:• Fl
• F ll
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Problem 5
• Solution:• Resolve the weight into components
perpendicular and parallel to plane • F ll = W sin θ F l = W Cos θ
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Problem 5
• F ll = (+562 N)( sin 30.0o)
• = +(562 N) (0.500)• = +281 N
• F l = + (562 N)(cos 30o)• = + (562 N) (.866 ) = + 487 N
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Problem 6
• Finding acceleration down a plane• The 562 N trunk is on a frictionless plane
inclined at 30o from the horizontal . Find the acceleration of the trunk. What is its direction?
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Problem 7
• Finding the Coefficient of Static Friction