900 lb 600 lb 150 lb/ft A B. x, 7200 lb ft IB A, 6 ft 6 ft 12 ft B.

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

A: following is the solution-

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

A:

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

A:

Q: 20 kN 50 kN 50 kN/m 25 kN/m B. 2 m 3 m

A: Consider the free body diagram. The cantiliver support will be replace by two forces FH and FV and a…

Q: 6 kN/m 20 kN 90 mm В C h 2.4 m -1.2 m –

A: Solution: To calculate support reactions Under static equilibrium RA+RB=6kN×2.4+20=34.4kN Taking…

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: 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: The reaction at C 120 lb 300 lb lb www 120 lb

A: Given

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

A:

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

A: Data given-

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

A:

Q: 350 lb 30 250 lb

A:

Q: Neglecting the mass of the beam, compute the reactions at A and B.

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Q: BM at B is 3 kN 1 m D 1m 1 m B 1 m A

A:

Q: 6 ft y 9 + B E 4 ft- D H -4 ft 4 ft K +ft- G 3000 lb 1000 lb X

A: Given: Force applied at point H: →FH=3000 lb    (in downward direction) Force applied at point G:…

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

A:

Q: Answer the following questions: a) Explain the meanings of the terms: mass and force. b) Using your…

A:

Q: Determine the maximum shear stress acting at section a–a in the beam.

A: Given beam with reactions at 'A' and 'D' are as shown below: Applying equilibrium conditions: (i)…

Q: F B 4 m 4 m 4 m 4 m

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

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

A:

Q: nger diameter is 7/8 inch and its weight is 1.8 pounds. What is the safe maximum deadweight in kg,…

A: 1 psi = 6894.76 Pa 1 inch = .0254 m 7/8 inch = 0.022225 m 1 pound = 0.453592 kg thus 1.8 pound =…

Q: 100 lb 30° 200 lb 2 ft 300 ft-lb 2 ft 2 ft 2 ft-

A: First we resolved 100 lb force, and calculated horizontal force, vertical force and moment due to…

Q: y 800 lb/ft A B –- 2 ft →- -4 ft - 2 ft→

A: Q. To determine the reaction forces for the beam as shown.

Q: 15 KN 10 KN 10 KN 20 KN E A 5.0m

A: To determine the support reaction at J.

Q: Using unity conversion ratios, show that 1.00 lbm weighs 1.00 lbf on earth

A: Given Data: The given mass is M = 1 lbm. The expression for the weight (W) is, W=Mg Where g is the…

Q: lb·ft? 15 6:ft2 to J and to BTU s2

A: X=15×lb×ft2s2    ............ (1) Now Putting the value of "lb" in "kg" form and "ft" in "m" Form. 1…

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: 400 lb F1 = 400 j F2 = -600 k 37 F3 = 500 sin 45 i -500 cos 45 k 3 7* r2=? 600 lb 45 M- rixF,, M,-…

A: Calculate the given position vectors. r1=3 i^+10 k^ inr2=6 i^+3 j^+7 k^r3=6 i^-3 j^+ 7 k^

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

A:

Q: In the Figure 6, the absolute force value in the bar AC is (in kN): Figure 6: 5 m/ 5 m 5 m A 20 KN B

A:

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

A:

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

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

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

Q: Select one: a. k = 7.8 lb/in %3D b. k = 4.8 lb/in %3D c. k = 5.8 lb/in %3D d. k = 6.8 lb/in %3D

A: Solution : Given data :  weight = 10lb                     unstretched length = 8 in…

Q: 400 lb 108 in

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

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

A:

Q: 10 KN/m 1a KN (m Ra 1.5 m 4m 3 m 1.5m

A: For solution refer below images.

Q: A 0.8 ft 4.8 ft 60 lb 3.5 ft E 48 lb 1.8 ft B |C D 3.6 ft

A:

Q: y 5 m 5 m A 5 m В 20 kN

A:

Q: Define the following terms: Weight Tension Friction Equilibrium

A: GIVEN DATA- SOME TERMS GIVEN WE HAVE TO DEFINE IT

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: In the Figure 6, the absolute force value in the bar AC is (in kN): Figure 6: 5 m 5 m A 20 KN 5 m B

A: FAC = 23.0940 kN

Q: B 105 90 30 30 75 300 lb D. 400 lb

A: From the knot where 400 lb load is hanging∑FH=0D sin75ο=C sin 30οD=0.5176 C∑FV=0D cos 75ο+C cos…