60 lb A 15 14 65 1.2

Q: 73.12 lb/in^2 to kg/m^2

A: Answer: 73.12 lb/in2 = 51400 kg/m2.

Q: lin = 401. %3D Vc =? Given Oz BE 1.25" Be= 2.15" CD=2. S2" W2= 400 Las 02 4 400 AD 3 2.

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Q: ensity is 1000 kg/m³), determin

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Q: 2m A * 4m 4m E 4M 4200 kN

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Q: A B 60° 3 kN 40° 2 kN

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Q: (M=450 lb-ft (M)s = 300 lb-ft 3.5 ft 1.5 ft 2 ft 2 ft (Mh = 250 lb-ft

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Q: 120 kg 60 A B 50 kg

A: Free body diagram of block B , Weight of block B ,WB=mBg      =50×9.81      =490.5NUsing…

Q: F2 = 81 lb F = 100 lb 4 ft 3 ft 7 ft 40° 4 ft

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Q: 25 lb 40 lb 24 in. 0.5 in. 4 in. a B Section a – a

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Q: A = 20 in.² Oil SG=0.86 A = 50 in.² 5 ft 0.8 ft SG = 2.5

A: Calculated discharge Q.

Q: 60 Ib 90 lb 30 45 150 lb

A: GIVEN DATA- A 150lb BOX GIVEN ACTED BY TWO FORCES WE HAVE TO FIND ACCELERATION OF IT

Q: kj/kg, ochange in eafrey ky/My k solve coorte

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Q: 10 kN 0 r=2m 10 kN

A: Given data: Magnitude of the force = 10 kN Need to determine the magnitude of the couple about the…

Q: H G 60 E D 7 kN 7 kN 9 kN 6m 6m 6m 6m

A:

Q: Given b- Bm * 3m a 60 degrees R6m C 10 m Eind

A: Solution:Given,b=8mr=3m @ 60 degreesR=6mC=10m Scale(1m=1cm)

Q: 166 kN HOKM 2 bokN 4m 4m 2m 4mi 3m 3m

A: In step 1 we draw given diagram and calculate reaction forces:-

Q: 1= S in. A B C ww k = 250 lb/in. r=3 in. = 400 lb

A: Drawing the free diagram of the lever,   Following forces acting on the lever, (a) The reaction at…

Q: A 90 N 40° L L = Im 80 N 37° 60 N L 70 N B 30° SON

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Q: 2m 2m 2m 2m he=70° 0=

A:

Q: 4. If 6.5 m³ of oil weighs 46,800 N, calculate its density in kg/m³. (a) 285 (b) 258 (c) 82 (d) 734

A: Given : volume of oil , V = 6.5 m3             weight of oil , W = 46800 N To find : density

Q: 60 lb A 15 14 65 1.2"

A:

Q: 1.75 Fg 1S00 FB = 857.14 |lbs Fr =Fe Pase- ISoo lbs Fo %3D Fr=642.86 lbs (4) F. 3 F. 유 Fs +…

A: given  equation are = 3/4 FB +FB  -  1500 =0 TO FIND =THE VALUE OF FB

Q: The length of a ribbon is 1.28 m. The length of a rope is 2.74 m longer than the ribbon. What is the…

A: Given that length of a ribbon = 1.28m length of rope =2.74 + length of ribbon=2.74+1.28=4.02m

Q: 200 lb/ft 800 lb 37 A Bカ 1 5' 4' 21 21

A: I HAVE FULL EXPLAINED BUT IF YOU WANT  ONLY BX DIRECTLY GO FOR SECOND STEP

Q: 80 lb/ft В - 5 ft – 2-5 5 ft A 60 lb/ft 12 ft

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Q: =150 le 120 F = 80 lb.

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Q: 2000 lb M = 4800 lb-ft В C A D 3 ft RA 6 ft 3 ft Ro 2.

A: Given DataP=2000lbMx=4800lb-ftTo find SFD and BMD **As question demanding for solving Question no 2…

Q: 3000 lb 90 Ib /ft 9 ft 200 lb/ft

A: For solution refer below images.

Q: PA 2 V

A: For solution refer below images.

Q: 200 Ib 30

A:

Q: C 3 ft B 0.8 ft A

A:

Q: Magnitude of AL = 275 pounds Find: TAG= 0 = 5 0 Pounds A Jø B D = 450 pounds

A:

Q: convertion 400lb·ft to kN·m

A: converting 400lb·ft to kN·m we know that 1 lb-ft to kN-m = 0.00136 kN-m

Q: ROBLEM 2.3 Y Y 3 kips 425 lb 4. 2 3 3 X X X 12 Y 110 lb Y Y (а) (b) (c)

A: find the vertical and horizontal component of forces.

Q: B A 10m 6m C 8m 6m

A: Given:

Q: F 105° 60° 45° 300 lb 60° 400 lb 200 lb

A:

Q: 2 KN 20 20 5 KN X 45° D F 2 KN

A:

Q: figure. 400 lb 30 Note: 1" x 1" 228 lb

A: To find resultant

Q: F = 52 lb 13/ 12 F = 80 lb 25°

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Q: 3 h -3t 1.5R It 2.

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Q: З KN 8 kN A 25 kN m 2 m 2 m 2 m 5 kN

A: Note: As per bartleby guidelines for more than one question asked only one question to be solved…

Q: 60 21.65ft 30 2000lbf 50ft 50ft

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Q: For each of the plates and loadings shown, determine the reactions at A and B.

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Q: OA 6 m 2 m 1.5 m 3 m

A: Given V=6m/s

Q: 2 kN 2 kN/m A B (a) 4 m 2 m RA Re

A: A beam is defined as a structural member subjected to transverse shear load during its…

Q: Joul In English unit Ft. lbf None of the these lbm.ft/lbf lbm.ft N.m

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Q: 2107 hy volu

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Q: 2.5 kN 1.5 kN 3 kN 30 A B. 2m 4m 2m

A: Resolve the two forces along the beam and perpendicular to the beam as shown. Write the x component…