300 lb/ft 1500 lb-ft B A 16'- 6' 1200 lb

Q: 900 lb 60 lb/ft 400 lb D A |B - 8 ft 6 ft→ 4 ft

A: following is the solution-

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

A:

Q: 3 ft- Wo G 4 ft 60% OD MG 2 ft FG (NG B G E B.

A:

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

A:

Q: B 4a E M A L AM = 10a 8a D L

A:

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

A:

Q: E 4 ft G B D 12 kip 4 ft A 4 ft 4 ft 4 ft- -

A: A truss is a structural member, used to create rigid structures. The main function of the truss is…

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: -5 ft 3 ft- 2 ft 4 ft B В 5 3 12 13 4 5 500 lb 200 lb 260 lb

A:

Q: 600 lb-ft B 350 lb-ft 3ft 1 ft 500 lb-ft 2 ft

A: For solution refer below images.

Q: 300 lb B. Ao 20 in. 48 in.- 15 in.

A: Given dataLoad P = 300lb (vertically downwards)

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: 2000 Ib 300 lb/ft 6 in, TI A 4 in. -3 ft- 3 ft-

A: Calculate the reaction at the support, Total load on beam is=300×6+2000…

Q: 500 lb 600 lb · ft A B -5 ft - -5 ft -5 ft

A: Forces and moments generated in any member depend upon the type of support and type of loading…

Q: 100 lb 800 lb · ft A |C B 5 ft 5 ft

A:

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.

A:

Q: 350 lb 30 250 lb

A:

Q: F 1000 lb y 500 lb 45° 30°

A:

Q: KIN 14 kN / m 10 kN / m A B ID 4 m 2.5 m 2.5 m 3 m

A: Finding support reactions:RA+RB=12*4*14+14*5+20+30RA+RB=148…

Q: F-1200 +o 3t N A В 1.4 m

A: First of all take A+B as a system, Now apply force equation along and traverse to the inclined…

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

A:

Q: Figure 1 of 1 |2P A -2 ft- -2 ft- d 2 in.

A: find the maximum magnitude of load P.

Q: 12KN 4 kN /m 12m 24 KN 12m 36m 一- 10

A: Dear student, we are suppose to solve only one question. Please post other question as a separate…

Q: F = 250 lb - 30°, 45 F2 = 375 lb

A:

Q: 13378.2898 lbm to lbf

A:

Q: 120° F= 400 lb F = 600 lb

A: Here for force F1, Angle with x-Axis (alpha) = 60o Angle with Y-Axis (beta) = ? Angle with Z-Axis…

Q: 18 kN 3 B 25 kN 8m DE 6m D 10 kN Figure 1

A:

Q: 5 kip/ft 15 kip It 15 kip · ft B- 10 ft 6 ft 19-

A: lets take x-x in section BD at distance x

Q: A 2 ft F, 4 ft M B. - 3 ft- 30 -x F,

A: Given that, F1=150 lb,F2=50 lbM=500 lb.ft Consider the given frame, Resolve the given force in two…

Q: In each case, draw the shear and moment diagrams for the beam.

A: The free body diagram of the first beam is as follows: Calculate the force in the vertical…

Q: 150 Ib/ft 100 lb

A: Consider the member AB and drawing FBD as shown below, Balancing vertical…

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

A: For solution refer images.

Q: 125 kN 50 kN/m VIIE A C 3 m 12 m

A:

Q: 180 lb · ft В. C 9 ft 4.5 ft

A: Now draw the free body diagram for the above problem. Convert the triangular varying load into the…

Q: 600 N 600 N -F 40° 405

A:

Q: 3.5 ft 18 ft

A:

Q: RB 50KN bm 100 KN 4m B 4m

A:

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: 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: kG w(b)=500 %3D b-3 4m

A:

Q: 30 2f- 6 ft-

A:

Q: A steel bar of c

A: 1.2125mm

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

A:

Q: G. 620mm 1.8m 1.2m

A: As per honor code and answering guidelines I can answer 1st 3 sub parts at a time.So i answered…

Q: 3.kN 30 kN/m B.

A: given; upper left beam; ⇒lets take reaction force at A=RA⇒lets take reaction force at B=RB⇒load…

Q: 15 ft 150 Ib 35 5ft 2 ft 250 Ib 25

A:

Properties of Fluids

The term fluid represents a type of substance that can easily change its shape under any external pressure/force and can flow. Fluids can acquire any shape in which it is stored and can resist normal stresses but cannot resist shear stress in rest condition. Whenever external shear stress is applied to fluids, it starts to flow. The fluids may be liquid or gas. Example- water, air, etc.