120 mm 60 mm 90 mn D 80 mm C 45 mim 480 N

Q: 4 kN RA = 64 kN 30 kN/m 2.0 m 2.5 m MA = 70 kN.m %3D

A:

Q: 10 mile/h = ……… m/s (1mile = 1852m) Convert from FPS to SI.

A: Velocity, V = 10 miles/h 1 mile = 1852 m

Q: 0.8 1 0.8 m 0.8 m D 0.6 m E 1.2 m 1.2 m y F 1.2 m

A:

Q: N 20 mm 200 mm 20 mm 300 mm pmri 20 mm A 1.5 m C N N 站 B D TTTTTTT 1.5 m 5m

A: given; load at point c (Pc)=430Nload at point D(PD)=634NAC=1.5mBD=1.5mAB=5mCD=2mlets take reaction…

Q: 15 ft 350 Ib 55° 6 9 ft 6 B 4 ft 550 Ib 35° 3.

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Q: 20 KN/m 10 KN/m 學 2.00 1.70 1.30

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Q: E 600 mm B Rigid P. 80 mm 160 mm 80 mm

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Q: 120 mm 600 mm 10-N 4N Fig. 5 600 mm 4N

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Q: 5 kips 30 kip • ft I = 90 in.4. 6 in. B NA 2 in. 4 ft 4 ft

A: For solution refer below images.

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: 10 kN 5 kN 3 kN/m F C D E 3 m 1 m 4 m 3 m 1 m

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Q: 330 mm 220 100 mm mm 800 N 75 mm 200 N 10 1200 N

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Q: 125 mm 175 mm D -225 mm- 75 mm 150 N Fig.3

A: Answer: (a) The tension in cable ABD is 195 N. (b) The reaction at support C is 216.33 N at an angle…

Q: 50 KN 100 KN/m 60 KN/m 40 KN/in 10 KN.m 2.0 m 6.0 m 2.0 m 4.0 m

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Q: 300 N -1.35 m. -1.35 m 1.2m 1.5m 300 N 552 N 500 N

A: For solution refer below images.

Q: R(A)sin30 R(B)si R(A) 400lb R(B) DN 45 R(B)cos45 45 B 30 60 R(A)cos30

A: Given Data→ Beam

Q: 400 N 3 m A, 8 m 4 m D 1200 N

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Q: 20 sq mm 40 sq mm 10° W W= 40 kg stress AB = Stress AC

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: 0.25 m 25° B 0.5 m 1.5 m - 0.12 m 10 kN 5 m

A: There are two reaction force acting at point A, ( RAX & RAY). The distance  between the…

Q: Im 35 N/m 2m 30 N Im в 85 N 100 N.m 60 2m 2m 0.5m 1.5 m

A: Mechanical Equilibrium: A body is said to be in mechanical equilibrium if it satisfies the following…

Q: mN)(6 GN). 0 mm)(0.6 MN)²

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Q: 150 mm y30 mm 120 mm Area 2 X2 30 + (2b/3) 100 mm X3 = 150- (4r/3n) Area1 y2 TArens y3 120 mm 130 mm

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Q: 3 m 3 m 3 m E C 12 M= 59 N.M 13 3 4 50N D

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Q: 2Nm 15KN L. 12m 0.6 12m 18m 12m

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Q: 8 kN 6 kN/m А С 1m 4 m 3 m m E.

A: Calculating the reaction at the support,

Q: 200N 500N 600 N.m 3 m 4m 2m 45 2 m + 150 N 400N 5m

A: Resultant force A resultant force is a single force and associated torque obtained by combining a…

Q: H G 0° 60 6 kN 6 kN 8 kN Sm Sm Sm Sm

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Q: Water 10.0 in 666 in 30.0 in 222 in 40.3 Mercury (8- 13.6)

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Q: 6 kN 9 kNm 4 kN/m E 2 m 0.5 m 0.5 m 2 m

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Q: 2 tór/m 8 ton F 4 ton-m E 3 ton/m 2 m 4 m Im1m 2 m 2 m

A: According to the principal of engineering mechanics  Reaction and moment at support is equal to zero

Q: Which of the following nearly gives the W = 200 kN W = 400 kN 1m DI 2 m B. 5.6 m

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Q: C cord 2m 1.5m 10 N B. 56 N.m 2.5m E- 0.25m F 3m 1.5m 3m

A: Find the reactions at point B on member ABC

Q: 10 m 10 m 10 m 10 m 180 KN 8 150 KN 11 4 7 10 5 9. 12 12 m 1 3 6 14 20 m 13 200 KN 2 A

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Q: 100 kN/m 150 KN E (m) 5 7 10

A: "Since you have posted a question with multiple sub-parts, we will solve the first three sub-parts…

Q: E= 103 GPa E = 75 GPa E= 190 GPa 0.8 mm 150 mm 100 mm 300 mm Bronze Al Steel

A: Stress produces in the material due to changes in temperature is known as thermal stress. Condition…

Q: A steel pipe has an inside diameter of d = 350 mm and an outside diameter of D = 158 mm. The length…

A:

Q: 1.0 m 2.0 m 05 m 05 m 20 kN 4 KN 10 kN 4.0 m

A: Given Data: Force (F) = 10 kN

Q: |20 kN 15 kN/m A 2 m 4 m B.

A: The reactions on the supports of beam can be drawn as follows,

Q: 20 kN D 20 kN 20 kN 3 m 10 kN 3 m 3 m A 4 m 4 m E 4 m 4 m B,

A:

Q: 40 mn 40 mm 40 mm 40 mm N.A N.A 80 mm 80 mm 40 mm 40 mm 100 mm 100 mm 100 mm 100 mm case 2 case 1

A: For case 1:let:b=40  d=160b1=100    d1=40y=40+20 =60mmy'=80…

Q: 4.5KM/M 225 mm 8m 125mm 35mm

A: Given Data: The magnitude of load, w=4.5 kN/m

Q: M N.M A150 N 300 mm 9o N 400 mm

A: Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and…

Q: 24 mm 5 mm 30 mm 8 mm 5 mm 30 mm Section AA 12,000 N

A: The correction factor is rc=3015=2 The correction factor for 2 is k=1.41. The moment of inertia is…

Q: F126KN 180KN 26 -3m 1.5m 48 65° 145KN V165 KN

A: Drawing the system of force as shown below – Let us take point A as an origin of the coordinate…

Q: 8 kN/m | 20 kN 4 kN/m 2.0 m 1.5 m

A: For solution refer below images.

Q: Determine the reactions at the supports. E is constant.

A: Uniformly distributed load, P=3K/ftLength of the beam , L=10 ft

Q: 1471b 243 Ib "L. 2 ft 61 10 ft 3 ft 2 ft

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

A:

Conservation of Energy

In physics, the rule of the conservation of energy states that energy doesn’t disappear from the atmosphere; instead, it changes its form from one type to another. Energy is one of the most important physical quantities in nature, and it cannot be created or destroyed — only transformed.