Problems Set 02

BME-131

Th f h it dThe force has a magnitude of 500N Express F asof 500N. Express F as vector in terms of the unit vectors i and j. Identify h d lthe x and y scalar

components of Fcomponents of F.

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x(a) F = F cos θx( )

xF = 500 cos40383 N= 383 N

yF = F sinθy

yF = -500 sin 40

-321 N=

(b) x yF F i F j= +

383 321 NF i j= −

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Th it d f th fThe magnitude of the force F is 400 lb Express F as aF is 400 lb. Express F as a vector in terms of the unit vectors i and j. Identify b h h l dboth the scalar and vector components of Fcomponents of F.

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x(a) F = F cos θxF = -400 cos30 = -346 N

yF = F sinθyF = 400 sin 30

S l C= 200 N

Scalar Components:xF = -346 N

F = 200 NVector Components:

xF = -346i NyF 200 N

yF = 200j N

346 200 NF i j= − +(b) x yF F i F j= +

Th l f th 5 2 kNThe slope of the 5.2 kNforce F is specified asforce F is specified as shown in the figure. gExpress F as a vector in

f h iterms of the unit vectors iand jand j.

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2 2( ) (12) 135

2 2(5) (12) 13+ =

125Opposite 5

sin13

OppositeHypotenuse

θ = =

Instead of using angle θ, the12

cos13

adjacentHypotenuse

θ = = Instead of using angle θ, thedirection of force can also bedefined by using a small slope

13Hypotenuse

defined by using a small slopetriangle.

5

F F i θ

5sin

13θ =

12cosθ

F = F cos θyF = F sinθ

52 ( )

cos13

θ =

xF F cos θ

12F 5 2 ( )

y

5F = -5.2 ( )

13×

xF = -5.2 ( )13

×

4 8 kN= -2 kN

= -4.8 kN

F F i F j= +

4 8 2 kNF i j=

x yF F i F j= +

4.8 2 kNF i j= − − of 14

Th li f ti f thThe line of action of the 3000 lb force runs through3000 lb force runs through the points A and B as pshown in the figure.

i h dDetermine the x and yscalar components of Fscalar components of F.

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1 2( , ) ( 7, 2)A a a = − −1 2( , ) ( , )

1 2( , ) (8,6)B b b =

1 1 2 2( ) ( )AB b a i b a j= − + −

(8 7) (6 2)AB i j= + + +

15 8AB i j= +2 2( ) ( )AB b b2 2

1 1 2 2( ) ( )AB b a b a= − + −

AB

ABn

AB=

2 2

15 8

(15) (8)

i j+=+AB (15) (8)+

3000 lbF =

ABF Fn=

15 83000 lb

i j⎡ ⎤+= ⎢ ⎥2 23000 lb

(15) (8)⎢ ⎥

+⎣ ⎦2647 1412F i j= +

F F i F j

S l C t F 2647 lb

x yF F i F j= +

Scalar Components:xF = 2647 lb

F = 1412 lbyF 1412 lb of 14

Th 1800 N f F iThe 1800-N force F is li d t th d f thapplied to the end of the

I b E FI-beam. Express F as a t i th itvector using the unit t i d jvectors i and j.

2 2(4) (3)4

2 2(4) (3) 5+ =

34Opposite 4

sin5

OppositeHypotenuse

θ = =

3cos

5adjacent

Hypotenuseθ = =

5Hypotenuse

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4i θ

3θ sinθ=

5cosθ=

5

xF = F cos θ x

3F = -1800 ( )

5× = -1080 N

5

F = F sinθ4

F = -1800 ( )× = -1440 Ny s θyF 1800 ( )

5

F F F

1080 1440 NF i j

x yF F i F j= +

1080 1440 NF i j= − −

Th l b fThe two structural members, one of which is in tension and the other inwhich is in tension and the other in compression, exert the indicated pforces on joint O. Determine the

it d f th lt t R f thmagnitude of the resultant R of the two forces and the angle θ which Rtwo forces and the angle θ which Rmakes with the positive x-axis.

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1 16 kN, 15F θ= =1 1

2 14 kN, 45F θ= =

x xR F= ∑ y yR F= ∑

1 1 2 2cos cosF Fθ θ= + 1 1 2 2sin sinF Fθ θ= +

6 15 4 45 6sin15 4sin 45+6cos15 4cos45= − − 6sin15 4sin 45= − +8.62 kN= − 1.276 kN=

2 2

x yR R R= + 2 2( 8.62) (1.276)= − + 8.72 kN=x y ( ) ( ) 8.7 N

1tan yRθ − ⎛ ⎞= ⎜ ⎟

1 1.276tan− ⎛ ⎞= ⎜ ⎟

⎝ ⎠171 6=tan

xRθ ⎜ ⎟

⎝ ⎠tan

8.62⎜ ⎟−⎝ ⎠

171.6= of 14

Th f h f FThe y-component of the force Fhich a person e erts on thewhich a person exerts on the

handle of the box wrench ishandle of the box wrench is known to be 70 lb. Determineknown to be 70 lb. Determine the x-component and the pmagnitude of F.

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5

125 122 2(5) (12) 13+ =

12sin

13θ =

5cos

13θ =

12

yF =70 lb

F = F sinθ12

70 = F ( )× F = 75 83 lb⇒

θ5

83 ( )

yF = F sinθ 70 F ( )13

× F = 75.83 lb⇒

xF = F cos θ x

5F = 75.83 ( )

13×

xF = 29.16 lb⇒

D i h l R f hDetermine the resultant R of the t o forces sho n btwo forces shown by

( ) applying the parallelogram(a) applying the parallelogram rule for vector addition andrule for vector addition and

(b) (b) summing scalar (b) (b) summing scalar components.p

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Law of Cosines:A2 2 2 2 cosc a b ab C= + −A

Ccb

Cc

aLaw of Sines:

Ba

a b c= =sin sin sinA B C

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(a) Using Law of cosines:2 2 2

(a) Using Law of cosines:

R = (600) +(400) -2(600)(400)cos60( ) ( ) ( )( )

R = 529 NUsing Laws of sines:

529 600529 600sin 60 sinθ

= oθ = 79.1⇒sin 60 sinθ

1 1( ) 600N, 60b F θ= =400N 0F θ

R F= ∑ R F= ∑2 1400N, 0F θ= =

x xR F= ∑ y yR F∑

1 1 2 2cos cosF Fθ θ= + 1 1 2 2sin sinF Fθ θ= +1 1 2 2cos cosF Fθ θ+ 1 1 2 2

600cos60 400cos0= − 600sin 60 400sin 0= +100 N= − 520 N=

2 2

x yR R R= + 2 2( 100) (520)= − + 529.52 N=R⎛ ⎞ 520⎛ ⎞1tan y

x

R

Rθ − ⎛ ⎞= ⎜ ⎟

⎝ ⎠1 520

tan100

− ⎛ ⎞= ⎜ ⎟−⎝ ⎠79.11= −

x⎝ ⎠ ⎝ ⎠100 520 NR i j= − + of 14

Problems Set 02

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