# Chapter 6 Practice Problems

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### Transcript of Chapter 6 Practice Problems

Chapter 6 Practice Problems

Equations

• Sin θ = opposite side• hypotenuse• Cos θ = adjacent side• hypotenuse• Tan θ = opposite side• adjacent side

Equations

• vR2 = vp

2 + vw2

• R2 = A2 + B2

• Fv = F sin θ

• Fh = F cos θ

• Fnet2 = Fnetx

2 + Fnety2

• A + B + W = 0

• F ll = W Sin θ

• Fl = W cos θ

• Ff = μFN

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

Problem 1

• Basic equation: R2 = A2 + B2 or• vR

2 = vp2 + vw

2

Problem 1

• Answer: vR = 103 km/ h at 29o

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

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

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

Problem 2

• From 0o cos 30o =vo/v

• V0 = 34.6 km/h at 0.0o

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

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

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

Problem 3

• Find the angle of the force• Tan θ = 1.14• Θ = 49o

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?

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

Problem 4

• Basic equations:• In equilibrium net force is 0• A + B + W = 0• Fh = F Cos θ

• Fv = F Sin θ

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

Problem 4

• Thus• Av = Bv = ½ (168 N )• = 84 N• A = Av

• sin 22.5o = 220 N• B = A = 220 N

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.

Problem 5

• Given: • W = 562 N• Θ = 30o

• Unknown:• Fl

• F ll

Problem 5

• Solution:• Resolve the weight into components

perpendicular and parallel to plane • F ll = W sin θ F l = W Cos θ

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

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?

Problem 7

• Finding the Coefficient of Static Friction