8000 lb 150 lb/ft -4 ft 4 ft 1.5 ftT1.5 ft 2 in. 0.75 in. 6 in. 0.5 in.- B 4 4 in.' 0.75 in.

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

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

Q: 2 ft F D E -1.5 ft 600 lb 300 lb -3 ft -3 ft-

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Q: ensity is 1000 kg/m³), determin

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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: 3cm 3 Cum 4 Cun

A:

Q: 150 lb/ft A В -6 ft -8 ft

A: Let us draw the reaction at the supports  Weight of the rectangular load = 150 lb/ft(8ft)=1200 lb…

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

A:

Q: 18 ft 24 ft 10 ft 6ft

A: Given that, Weight of each light, W= 65 lb, Drawing the FBD of the light, The angle θ…

Q: 1. For the shown hydraulic jack the gage pressure indicates 140 kpa how much kilograms should be…

A:

Q: 11.2 kN/m A B 2.8 m 3.1 m Fig. Q1 Note: not drawn to scale

A: To determine the reactions at A and B and to draw the hear force diagram.

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

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Q: 50 KN/m 120 KN.m B- 9m 4m

A: For solution refer below images.

Q: 2T/m ΑΔ 3 m 6m 4T/m BA

A: Given problem consists of an overhanging beam supported by two hinge supports and acted upon by Two…

Q: A-20 in Oil SG 0.86 51t 0.8 t A-50 in. SG 2.5

A:

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: 2 ft 2.5 ft 1.5 t

A: Centroid, also called the center area of anybody is the point at which the whole area of the body…

Q: free-body diagrams. 6 ft 3 kip/ft B. 2 kip/ft -3 ft C

A: As per bartleby guidelines we are allowed to solve only 1st question if multiple questions are…

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: 200 lb - 3 ft 3 ft - 50 Ib 30 -3 ft 100 lb

A: Data given-

Q: 100 lb B 12а++ - 4 ft - 80 lb - 3 1 -

A: To find : The shear stress in the beam. Free body diagram : The free body diagram is shown below:…

Q: D 50 lb-ft B 4 in. 30 6 in. A.

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Q: 5000 lb 25 000 lb • ft A D - 3 ft 3 ft

A: It is required to draw SFD and BMD

Q: 1000 Ib -12 ft 12 ft- B 2000 lb 8 ft D. ton -12 ft- 6 ft 6 ft

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: W=4 kN/m 5 kN A 2.5 m + 1.5 m

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

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Q: 1 ft 3 ft 3 ft (-900î + 800j 3 ft

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Q: 4 (a) L-1 9. (b) L-1 s²+4s+20. s² +3s)

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Q: norma Bulk Modulus is defined as the ratio of nor stress to.

A: Given data : Direct stress=σVolumetric strain=∆VV

Q: 7.5 ft A. Smooth wall 15 ft 9 ft

A:

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

A: Given:

Q: 29.20 kg, 4220.29 N

A:

Q: او جد القوه ونقطه تأثیر 1 w=2KN/m 4m

A: Equivalent force and its location  for given beams

Q: 6 kip/ft 8 kip AB A 9 ft 4.5 ft 5 ft -

A:

Q: gure |2P Pro A |B -2 ft2 ft - –2 ft d 2 in.

A: For solution refer images.

Q: 12 KN 2.5 KN/M 介介企 A 12 m B 4m

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

Q: 17 4-m 180 lbs 3-m 2-m 5-m 160 lbs

A:

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

A: To find resultant

Q: в A y 10m 6m C 8m 6m

A: Given: If the used turbine converts only three-quarters of the energy that the fluid will lose after…

Q: 12 kN/m 40 kN C A B 2 m 1m-

A: Let us draw the reaction force at A and C

Q: 400 lb 108 in

A: Calculate the reaction of the support, RA+RB=400   .............. (1)Taking moment about…

Q: Determine the x and y components of each of the forces shown.

A: To find out the x component of the forces Resolve all the force in the x-axis…

Q: 40cm 4000 N Determine pressure? Diameter force —

A: To Find : The pressure applied on the piston. Given :  The Force applied to the piston is F=2000 N.…

Q: X -2 ft- F By 2 ft F P 2 ft

A: As per honor code and answering guidelines I can answer 1st 3 subparts at a time.If you want other…

Q: F2-13 KN C A4 kN A 2m 4- 22 KN 30 FS-4 KN

A: Since on the system, some of applied forces are acting at some angle, so it is better to resolve…

Q: 400 lb F1 = 400 F2 = -600 k F3 = 500 sin 45 i-s00 cos 45k 7t Y600 lb 45 M- rxF,. M,- r2x , M,- rgxf,…

A: "Since you have asked multiple questions, we will solve the first question for you. If you want any…

Q: 1500 lb 300 lb/ft 300 lb/ft A B 3 ft- 3 ft- HINGE ROLLER (2)

A:

Q: 600 lb · ft -5 ft – -5 ft -5 ft

A: The beam can said to be in static equilibrium, when the vector sum of all the forces on the beam is…