Tugas PS 2 Momen

1. Determine the moment about O of the force of magnitude F for the case (a) when the force F is applied at A and for the case (b) when F is applied at B.

Jawab :

rA = LkrB = Di+Lk

FA = -FjFB = -Fj

MA = LFiMB = DFk + Lfi

2. The two forces acting on the handles of the pipe wrenches constitute a couple M. Express the couple as a vector.

Jawab :Kami deskripsikan soal diatas dengan lengan momen. di sepanjang sumbu Y menjadi vector I dan sepanjang sumbu x menjadi vector j dan sepanjang sumbu z manjadi vector z

Diketahui :R1 = (0.250+0.250) jF2 = -150 NR2 = 0.150 iF1 = 150

M total = F2 (R1) + F1 (R2) = -150 x (0.250+0.250) j + 0.150 i x (150) = - 75 j + 22.5 i

3. The bent bar has a mass ρ per unit of length. Determine the moment of the weight of the bar about point O. Assume that the center of the mass is located in the middle of the bar.

Jawab :

Diketahui :

Misalnya massa : ρd Dianggap titik berat berada pada tengah-tengah diameter batang Gaya yang bekerja adalah gaya berat (w)

Ditanya :

Momen di titik O ?

Jawab :

F = W

= massa x gravitasi

= ρd x g

= ρd x g

= ρdg Newton

Karena gaya berat mengarah ke bawah (bekerja pada sumbu z) maka vector F = ρdg k

Vector r = ( ½ d cos 65˚ i + ½ d x sin 65˚ j + h k) meter

Sehingga ,

momen di O = F x r

= (ρdg k) x ( ½ d cos 65˚ i + ½ d x sin 65˚ j + h k)

= (ρdg k) x (0,211 d i + 0,453 d j + h k)

= ( 0,211 ρgd2 j – 0,453 ρgd2 i )N.m

Jika kita asumsikan g = 10 m/s2 maka momen di titik O

MO = (2,11 ρd2 j – 4,53 ρd2 i ) N.m

4. The right-angle pipe OAB of is shown in the figure below. Replace the 750-N tensile force which the cable exerts on point B by a force–couple system at point O.

Jawab :

rOC = 0,7 j + 1,2 k

B = (1,6 ; - 0,8 sin 30 ; 0,8 cos 30)b = 1,6 i - 0,4 j + 0,4 √ 3 k

C = (0 ; 0,7 ; 1,2) c = 0,7 j + 1,2 k

BC = c - b

= (0,7 j + 1,2 k) – (1,6 i - 0,4 j + 0,7 k ¿

= - 1,6 i+1,1 j+¿ 0,5 k

‖BC‖ = √ (−1,6 )2+¿¿

= √2,56+1,21+0,25

= √4,02

= 2,005

nBC = −1,6 i+1,1 j+0,5 k2,005

= - 0,8 i+ 0,55 j+¿ 0,249 k

F = ‖F‖ nBC

= 750 (- 0,8 i+ 0,55 j+¿ 0,249 k)

= -600 i+ 412,5 j+¿ 186,75 k

Mo = rOC X F

= (0,7 j + 1,2 k) X (-600 i+ 412,5 j+¿ 186,75 k ¿

= 420 k + 130,725 i - 720 j - 495 i

= - 364,275 i – 720 j + 420 k

‖Mo‖ = √(Mo )2

= √ (−364,275 )2+¿¿

=√132696,275+518400+176400

= √827496,275

= 909,668 N

5. Determine the moment associated with the pair of 400-N forces applied to the T-shaped structure.

Jawab :

Digambarkan r1 terhadap titik O (sbg titik sumbu) ke B (titik gaya F1) dan r2 terhadap titik

O (sbg titik sumbu) ke A (titik gaya F2) sebagai berikut:

F1 ❑⇒

FX = 0

FY = -400 cos (15⁰)

= -400 (0,96)

= - 386,37

FZ = -400 sin (15⁰)

= -400 (0,26)

= -103,5

Jadi, F1 = -386,37 j – 103,5 k

F2 ❑⇒

FX = 0

FY = 400 cos (15⁰)

= 400 (0,96)

= 386,37

Fz = 400 sin (15⁰)

= 400 (0,26)

= 103,5

Jadi, F2 = 386,37 j+103,5 k

r1 = 0,45 k+0,25 i

r2 = 0,45 k−0,25 i

M1 = r x F = ( 0,25 i + 0,45 k ) x ( -386,37 j - 103,5 k )

= -96,6 k + 25,9 j + 173, 9 i

M3 = r2 x F2 = ( -0,25 i + 0,45 k ) x ( 386,37 j + 103,5 k )

= -96,6 k + 25,9 j−173,9 i

Jadi, MO = ( M 1 + M 2 )

= -96,6 k + 25,9 j + 173,9 i−¿ 96,6 k + 25,9 j−¿ 173,9 i

= 51,8 j−¿ 193,2 k

6. If the magnitude of the moment of F about line C Dis 50 N 1⁄7 m, determine the magnitude of F.

Jawab :

A = (0;0,2;0)B = (0,4;0,4;0,2)

AB=0,4 i+0,2 j+0,2 k

AB=0,2(2 i+ j+ k )

nAB=1

√0,24(0,4 i+0,2 j+0,2 k )

F= F

√0,24( 0,4 i+0,2 j+0,2 k )

C = (0;0;0,2)D = (0,4;0;0)

CD=0,4 i−0,2 k

nCD=1

√0,2(0,4 i−0,2 k )

Ambil titik Cr CA = 0,2 j+0,2 k

‖M CD‖=(rC A × F ∙nCD)

50 = [(( 0,2 j+0,2 k ) × F

√0,24( 0,4 i+0,2 j+0,2 k ))∙ 1

√0,2(0,4 i−0,2 k )]

50 = F

√0,24( 0,08 i−0,08 j−0,08 k ) ∙ 1

√0,2(0,4 i−0,2 k )

50 = F

√0,048(0,032+0,016)

F = 228 N7. A mechanic applies the horizontal 100-N force perpendicular to the wrench as indicated in

the figure. Determine the moment of this force about the bolt center O. The wrench centerline lies in the x-y plane.

Jawab :

Diketahui:R = 0,185 mF = 100Nθ = 15o

Ditanyakan:a. Moment O

Jawab:

F = 100 kR = 0,185 cos 150 i + 0, 185 sin 150 jR = (0,185x0,17869)i+(0,2588x0,185)jMomen= R x F= (0,1786 i + 0,04788 j) x 100k)= -17,85 j + 4,788 i Nm

Bila dijadikan scalar maka = √(−17,85)2+(4,788)2

= √318,62+22,92= 18,4 Nm

8. Two 4-N thrusters on the nonrotating satellite are simultaneously fired as shown. Compute the moment associated with this couple and state about which satellite axes rotations will begin to occur.

Jawab :

Momen A = τ A=r × F

= (−0,625 k−0,5 i ) ×−4 j

=−2,5 i−2 k

Momen B =τ B=r × F

=(−0,625 k−0,5 i ) ×−4 j

=−2,5 i−2 k

τ total= τ A+ τB

= (−2 , 5 i++2 k )+(−2 , 5 i++2 k ) = −5 i+4 k

9. A space shuttle orbiter is subjected to thrusts from five of the engines of its reaction

control system. Four of the thrusts are shown in the figure; the fifth is an 850-N upward

thrust at the right rear, symmetric to the 850-N thrust shown on the left rear. Compute the

moment of these forces about point G and show that the forces have the same moment

about all points.

Jawab :

r1=18 i F1=−1700 j M 1=−30600 k

r2=18 i F2=1700 k M 2=−30600 j

r3=−12 i−3,2 j −2 k F3=1700 j M 3=−20400 k+3400 i

r 4=−12 i−3,2 j−2 k F4=−850 k M 4=−10200 j+2720 i

r5=−12 i+3,2 j−2 k F5=−850 k M 5=−10200 j−2720 i

M=M 1+M 2+ M3+M 4+M 5

M=(−30600 k )+(−30600 j)+ (−20400 k+3400 i )+(−10200 j+2720 i )+¿

(−10200 j−2720 i)

M=3400 i−51000 j−51000 k

−1700 j → F1

1700 j → F3

1700 k →F2

−1700 k {−850 k → F4

−850 k → F5

That is shown that each forces has same moment about all point 9898

10. In picking up a load from position B, a cable tension T of magnitude 24 kN is developed. Calculate the moment which T produces about the base O of the construction crane.

Jawab :

T = 24 KN

Titik O = (0i,0j,0k)

Titik A = (0i,18j,30k)

Titik B = (6i,13j,0k)

r = Titik A – Titik O

= (0i,18j,30k) - (0i,0j,0k)

= 18j + 30k

AB = Titik B - Titik A

= (6i,13j,0k) - (0i,18j,30k)

= 6i -5j -30k

Vektor Satuan AB =1

√62+ (−6 )2+(−30 )2(6 i−5 j−30 k )

=(6 i−5 j−30 k )

√62+ (−6 )2+(−30 )2

=(6 i−5 j−30 k )

√961

=(6 i−5 j−30 k )

31

F=24 KN(6 i−5 j−30 k )

31

=(144 i−120 j−720 k )

31KN

= (4,65i−3,87 j−23,23 k ) KN

MO= r× F

= (18j + 30k) x (4,65i -3,87j -23,23k) KN

= (83,7 (−k )+0−418,14 (i)+139,5( j)−116,1(−i)+0)KN

= (−302,04 i+139,5 j−83,7k )KN

11. The specialty wrench shown is used for difficult-toaccess bolts which have a low torque specification. Determine the moment about O of the 40-N force applied at point A of the wrench handle.

Jawab :

12. In order to decrease the undesirable moment about the x-axis, a mechanic uses his left hand to support the wrench handle at point B. What upward force F must he exert in order that there be no moment about the x-axis? What then is the net moment about O?

Jawab :

Ditanya :Berapakah Sigma Momen apabila momen di sumbu x = 0 ?

Penyelesaian :a. Menghitung M1

Diketahui : F1 = −40 j rOa =225 i+300 k ,maka dapat dihitung sebagai berikut :

M 1=(225 i+300 k ) x (−40 j )

M 1 =−9000 k+12000 i

b. Menghitung lokasi M2

Diketahui : F2 = F j r2 =100 i+150 k , maka dapat dihitung sebagai berikut :

M 2 =(100 i+150 k ) x ( F j )

M 2 =100 F k−150 F i

Maka ∑ M =(−9000 k+12000 i )+(100 F k−150 F i)∑ M =(12000−150 F )i+(100 F−9000) k

Apabila sb. X = 0, maka momen adalah :12000−150 F=0

F=12000150

=80

∑ M =(12000−12000)i+(15000−9000) k

∑ M =6000 k

13. Determine the moment of the 400-N force about point A by (a) using the vector cross-product relation and (b) resolving the force into its components and finding their respective moments.

Jawab :

14. The specialty wrench shown in the figure is designed for access to the hold-down bolt on certain automobile distributors. For the configuration shown where the wrench lies in a vertical plane and a horizontal 200-N force is applied at A perpendicular to the handle, calculate the moment MO applied to the bolt at O. For what value of the distance d would the z-component of MO be zero?

Jawab :

Dik : F = 200 N

r = 0 i - 47,5 j + 242,5 k

Ditanya :

a. Menghitung saat MO diterapkan pada baut di O

125 mm x

200

Cos θ = depanmiring

Cos 200 = 70+x125

(0.94 x 125 ) - 70 = XX = 47,5

MO= r× F

= (−47,5 j+242,5 k )x –200 i ¿ = −9500 k – 48500 j

jika MO = 0 maka gaya harus kita geser tegak lurus dengan sumbu Z dan pada kordinat (0,0,k) sehingga jarak d berubah

x

200

70

Cos 200 = 70x

0,94 = 70x

X = 70

0.94X = 74,47

Sin 200 = y

74,47

0.3 = y

74,47Y = 25,32

Maka jarak d = 74,47 pada titik ( 0 ; 0 ; 225,32 )

15. The moment M applied to the pulley wheel causes a tension T 80 N in the cable which is secured to the wheel at D and to the ground at E. determine the moment about O of this 80-N force as applied at C.

Jawab :

DiketahuiTorsi = 80N AB = 0.2 m BE = 0.4 mBM = 0.1 m

Ditanya momen pada titik O ?

Sin α ¿ BCBE

=100400

=14,48 °

T = TnCE

= 80(- cos α k – sin α k)= - 20k – 75,5i N

MO = rOE ×T , dimana rOE= 0,2j m= 0,2j x (20k – 75,5i)

= - 15,41i + 4k Nm

16. A 5-N vertical force is applied to the knob of the window-opener mechanism when the crank BC is horizontal. Determine the moment of the force about point A and about line AB.

Jawab :

r=75 cos30 ° i+75 j+75 sin 30 k

¿75( 12√3)i+75 j+75( 1

2) k

¿64,95 i+75 j+37,5 k

Arah F = -k

F= 5

√12(−k )

¿51(−k )

¿−5 k

17. Determine the vector expression for the moment MO of the 600-N force about point O. The design specification for the bolt at O would require this result.

Jawab :

Penyelesaian:

M = r x F

Mencari nilai r

vector i = 50 + 130 sin 60 0i= 162,583 i

Vektor j = - (140 + 130 cos 60 0)

= -205 j Vektor k = 150 k

r = i - j + k

=162,583 i – 205 j + 150 k

Mencari nilai F

Fx = F. cos 450 . sin 600

= 600 cos 450 . sin 600 i

= 367,42 i

Fy = F. cos 450 . cos 600

= 600 - cos 450 . cos600

= 212,13 j

Fz = F. sin 450

= 424,64 k

Maka:

F = Fx + Fy+ Fx

= 367,42 i - 212,13 j + 424,64 k

M 0 = r x F

= (162,583 i – 205 j + 150 k ) x (367,42 i - 212,13 j + 424,64 k )

= - 34488,73 k – 68911,853 j + 75321,1 k – 87051,2 i + 55113 j + 31819,5 i

= 40832,37 k – 13798,853 j - 55.231 i

18. A force of 400 N is applied at A to the handle of the control lever which is attached to

the fixed shaft OB. In determining the effect of the force on the shaft at a cross section

such as that at O, determine the moment about point O.

Jawab :

19. The turnbuckle is tightened until the tension in the cable AB equals 2.4 KN. Determine the vector expression for the tension T as a force acting on member AD. Determine also the moment about point O.

Jawab :

20. The tension in the supporting cable AB is 10 kN. Write the force which the cable exerts

on the boom BC as a vector T. Determine the moment about point C.

Jawab :

║ F║ = 10 kN

A = 4 i + 5 k B = 7.5 j

AB = - 4i + 7.5 j - 5 k

BA = 4i - 7.5 j + 5 k

ȠBA = 1

√42+¿¿¿ ( 4i - 7.5 j + 5 k )

= 1

√97.25 ( 4i - 7.5 j + 5 k )

= 1

9.86 ( 4i - 7.5 j + 5 k )

F =║F║ x ƞBA

= 10

9.86 ( 4i - 7.5 j + 5 k )

= 4.06 i - 7.6 j + 5.07 k

Cara 1 :

r = 4 i + 5 k

MC = r x F

= (4 i + 5 k) x (4.06 i - 7.6 j + 5.07 k )

= - 30.4 k - 20.28 j + 20.3 j + 38 i

= 38 i + 0.02 j - 30.4 k

Cara 2 :

Transmibilitas gaya, maka r = 7.5 j

MC = r x F

= (7.5 j) x ( 4.06 i - 7.6 j + 5.07 k )

= - 30.45 k + 38.025 i

Kesimpulan :

Dengan cara jaraknya diukur seperti biasa ataupun di transmibilitaskan akan

menghasilkan besar momen gaya yang sama saja besarnya dan hanya beda

pembulatan pada koma.