2 kN 4m A 2m 2m MA=2 kN.m O MA=-8 kN.m MA=4 kN.m O None of them
Q: 4 kN RA = 64 kN 30 kN/m 2.0 m 2.5 m MA = 70 kN.m %3D
A:
Q: 3 m 2 m B 4 m F A 30 kN 18 kN
A:
Q: 1 kN 2 kN 2 kN 2 kN 1 kN H 0.46 m F D 2.62 B I A C| E G| 2.4 m 2.4 m 2.4 m 2.4 m Figure 2
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…
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A:
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: 5 m A 30 kN B 5 m 5 m 5 m 5 m 20 KN 5 m 5 m E
A:
Q: F1 = 5 kN F2 = 13 kN 0 = 45° B - 60° C F3 = 4 kN B 2 m 3. 7m F4 = 22 kN E lJa = 30° F5 = 4 kN
A: In case of multiple force on a system, resolve the forces in horizontal and vertical direction. This…
Q: 2 kN 2 kN 2 kN/m 3 kN/m B 3.5 m 15 m 2 m
A: A beam is defined as a structural member subjected to transverse shear load during its…
Q: BM at B is 3 kN 1 m D 1m 1 m B 1 m A
A:
Q: 2 KN/m 2 KN/m 3 m 3 m R R2
A: A beam is defined as a structural member subjected to transverse shear load during its…
Q: Q1: MO =S 400 lb F1 = 400 1 F2 = -600 k F3 = 500 sin 45 i-500 cos 45 k 3" 7" r2=? 600 lb 45 M-…
A: 400 lb force in vector form, F1=400j 600 lb force in vector form, F2=-600k 500 lb force in…
Q: 3 kN 12 kN.m 3 kN / m 5 kN / m 7 3 m 3 m 2 m
A: given:
Q: 20 kN 12 kN A В M = 10 kN-m %3D 4m 4m 4m Figure Q1 (b)
A:
Q: 2 kN 50 mm 5 kN• m D 80 mm B C 1.0 m 1.0 m 1.0 m
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: 3 m 3 m 3 m E C 12 M= 59 N.M 13 3 4 50N D
A:
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: 8 kN 6 kN/m А С 1m 4 m 3 m m E.
A: Calculating the reaction at the support,
Q: h of the following is not a unit of pressure? a) N/m2 b) mbar c) Pascal d)…
A: Which of the following is not a unit of pressure?
Q: 6 m G m- m- 40KN 40KN 40KN
A:
Q: G2 A 16" 13" 3" Answers: NA = 277.3 lb NB: 277.3 lb
A: Draw the free-body diagram of the system.
Q: 6 kN 9 kNm 4 kN/m E 2 m 0.5 m 0.5 m 2 m
A:
Q: 4 m A) 12 kN.m B) 16 kN m C) 20 kN m D) 22 kN.m 4 KN co 2 m 10 kN 2 m 2 m 2 kN/m E
A:
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: 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: 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: k=2 x 10° N/m ITTTTITTT m = 4.2 kg 10 cm 40 cm 60 cm
A: For solution refer below.
Q: 1 m 1 m 30° b. 0.5 m 0.5 m b 1m -75 mm-
A: Given Data The mass of the drum is:m=20 kg The supported diamter of frame is :D=2mm Now draw the…
Q: < 1 of 1 Figure E 6 m D 4 m 4 m 4 m Pi P2
A:
Q: 96 30 lb/ft 180 Ib · ft - 9 ft - 4.5 ft M (lb · ft) 200
A:
Q: 80 kN 40 kN 60 kN -10 kN/m A B 2 mim. 3 m 3 m 2m 1 m (b) 10 k I k/ft B 12 8' (b)
A:
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: 5 panels at 3 m E 3 m G 1.8 kN 15 1.8 kN 3 m 15° 3 m 5 m
A: given data: we consider only calculating side figure: TAB=force in member ABTAE=force in member…
Q: A 1.20m 18 KN 0.75m 12 kN/m 1.20m B
A: The given is as shown below,
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: 3m 6 m -3 m G 20 kN 15 kN H. B -3 m 3 m
A: given; ED=3mCE=4mCH=6mHF=4mFD=9mlets taking tension force in member BC=TBClets taking tension force…
Q: In the Figure 6, the absolute force value in the bar AB is (in kN): Figure 6: 5 m 5 m A 20 KN 5 m B
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: the disc, r = . .mm isc = T r2 s of the disc, h = 10.5 mm %3D he disc, m = 803 g s of the…
A: Given
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: 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: -3 ft- 0.5 ft B 0.5 ft k = 1.20 lb/ft A 1ft
A: first of all convert all quantities in SI units,then apply energy conservation principle.for…
Q: 5 kN 4 m B 5 kN 4 m C D 2 m
A: The free-body diagram of the strut is given as,
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: 2 kN/m 10 kN t. 5 m 3 m 2 m Figure 2
A:
Q: 8 kN/m | 20 kN 4 kN/m 2.0 m 1.5 m
A: For solution refer below images.
Q: 2.6 kN 2.4 kN-m B A 1.0 m 1.0 m 1.0 m
A: To draw the shear force or bending moment diagram, one should first determine the reaction at the…
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- Topic: Couples, Resultant of Force Systems & Equilibrium If F1 = 10N, F2 = 15N, F3 = 8N, F4 = 10N, F5 = 18N, and the length of te side of a red square is 1m. Calculate magnltude (+ for CCW, - for CW) of the resuItant moment (N-m) about point A. Calculate the magnitude of the reItant f0rce (N). (Answer: 50.80Nm, 20.33N)PROBLEM 4: The boom OC is subjected to tension forces at A and C. It is known that the tension in A is 26 kN and at C is 12 kN. Unit vector CE. (ANSWER: -0.38i-0.87j+0.32k) Tension at A. (ANSWER: 13.20i-15.8j+15.8k kN VECTOR COMPONENT FORM) Moment about O due to the given forces. (ANSWER: 78.71i+0j+3.2k kN VECTOR COMPONENT FORM)a = 4 mb = 3.2 mc = 2 mF = {-7i+4j+12k} NSpecified axis: AF Find the magnitude of the moment of F about the specified axis.If the calculated answer is not whole number, express it in 4 significant figures.
- In the figure, the engine of a vehicle is shown as a representation.The crankshaft of the engine around the x-axis It rotates at 4000 rpm and its moment of inertia is 0.5 kgm2 .Moving in the y direction, the vehicle enters the curve with a radius of 70 m at a speed of 120 km/h. In the meantime, find the moment coming to the motor bearings and interpret its effect on the vehicle.In the figure, the engine of a vehicle is shown as a representation.The crankshaft of the engine around the x-axis It rotates at 4000 rpm and its moment of inertia is 0.5 kgm2.Moving in the y direction, the vehicle enters the curve with a radius of 70 m at a speed of 120 km/h. In the meantime, find the moment coming to the motor bearings and interpret its effect on the vehicle.Can anyone help me with this question. The Motor (J1=0.46 kg‐m2) is connected to a gearbox (J2 =0.27 kg‐m2 and J3=0.17 kg‐m2) by a steel shaft. The output of the gearbox is connected to a turbine (J4=0.37 kg‐m2 by a second steel shaft. The shafts are supported by bearings (not shown). The gearing in the gearbox is such that theta3=‐2.7theta2. Use Lagrange methods to determine the equation of motion for the system including the flexibility of the shafts, using variables of theta 1, theta 3, theta 4 based on changing theta 3.
- A mono-cylinder engine has r = 0.3 m, l = 0.9 m, It has: crank mass m2 = 4.5 kg , crank rG2 = 0.4r, conrod mass m3 = 12 kg ,conrod rG3 = 0.3l from A, and piston m4 = 5 kg. The crank is rotating at constant speed ω =3000 rpm. The following equation is an approximation of the gas force over 180° of crank angle with Fgmax = 3000 N and β = 15º.  we need to find the gas force, gas torque, and inertia force . Use approximate expressions in your calculationA flywheel is fitted to a multi cylinder engine, which runs at a mean speed of 360 r.p.m. If the speed varies from 6 % above the mean to 2 % below it and the fluctuation energy is 1200 kN-cm, If the radius of gyration is 20 cm, find (i) mass moment of inertia of the wheel and (ii) mass of flywheel. The mass moment of Inertia in Kgm2 is Mass of the flywheel in kg isA bar is attached to the spring at the point C. The left end of the bar is pin supported and can rotates about the pin at Point A. The mass of the bar is m=20kg. The total length of the bar is LAB=3m and LAC=2m. Point A is 0.6 m below the ceiling. A clockwise constant couple moment M= 30Nm is applied on the bar so that the bar rotates from the horizontal position with θ=0° to the vertical position with θ=90°. The spring always maintains at the vertical position. The spring’s stiffness coefficient is k=30N/m and its unstretched length is 0.5 m. The acceleration due to gravity g=9.81 m/s2. During the process that the bar rotates from the horizontal position to the vertical position, determine the following. (4) the work done by the reaction force of the pin.____________ (J)
- A bar is attached to the spring at the point C. The left end of the bar is pin supported and can rotates about the pin at Point A. The mass of the bar is m=20kg. The total length of the bar is LAB=3m and LAC=2m. Point A is 0.6 m below the ceiling. A clockwise constant couple moment M= 30Nm is applied on the bar so that the bar rotates from the horizontal position with θ=0° to the vertical position with θ=90°. The spring always maintains at the vertical position. The spring’s stiffness coefficient is k=30N/m and its unstretched length is 0.5 m. The acceleration due to gravity g=9.81 m/s2. During the process that the bar rotates from the horizontal position to the vertical position, determine the following. (2) ) the work done by the couple moment. __________(J) (two decimal places)A bar is attached to the spring at the point C. The left end of the bar is pin supported and can rotates about the pin at Point A. The mass of the bar is m=20kg. The total length of the bar is LAB=3m and LAC=2m. Point A is 0.6 m below the ceiling. A clockwise constant couple moment M= 30Nm is applied on the bar so that the bar rotates from the horizontal position with θ=0° to the vertical position with θ=90°. The spring always maintains at the vertical position. The spring’s stiffness coefficient is k=30N/m and its unstretched length is 0.5 m. The acceleration due to gravity g=9.81 m/s2. During the process that the bar rotates from the horizontal position (state 1) to the vertical position (state 2), determine the following. 5) when use T to represent kinetic energy, V potential energy, U work done and if the bar is at rest at state 1, the principle of work-energy in this case could be expressed as____________ . V1+∑U(1-2)=T2+V2 T1+V1=T2+V2…A bar is attached to the spring at the point C. The left end of the bar is pin supported and can rotates about the pin at Point A. The mass of the bar is m=20kg. The total length of the bar is LAB=3m and LAC=2m. Point A is 0.6 m below the ceiling. A clockwise constant couple moment M= 30Nm is applied on the bar so that the bar rotates from the horizontal position with θ=0° to the vertical position with θ=90°. The spring always maintains at the vertical position. The spring’s stiffness coefficient is k=30N/m and its unstretched length is 0.5 m. The acceleration due to gravity g=9.81 m/s2. During the process that the bar rotates from the horizontal position to the vertical position, determine the following. (3) the potential energy of the spring when AB is vertical__________(J)