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- Q3) Determine the magnitude of the resultant force and its direction (0) measured .counterclockwise from the positive x axis y F2 = 61 N F1 = 80 N F2 = 20 N %3D %3DThe bar shown below is inclined at 45 degrees with the horizontal and at equilibrium. The force at C is applied perpendicular to the beam. The force at E is nearest to: a. 40kN b. 50kN c. 55kN d. 45kNCalculate the vertical deflection of point B and the horizontal movement at D shown in the figure. Consider that all the elements of the truss are linearly elastic and have sectional areas of 1800mm2. Also consider an E of 200,000N/mm2. Add the calculation process of solving the problem step by step and complete the table accordingly. L F FB ∂FB / ∂PB FD ∂FD/∂PD AE - 2√2/3 PB -2/3 √2 EB EF AB BC CD BF FD FC
- >10KN 4m A 3m 3m Using Virtual Work: Compute for the vertical component of the deflection at B and the horizontal component for the deflection at B. E=30GPa and 500sqmm cross sectional area for the membersThe two forces are equivalent to a force R that has a line of action passin through point A. Determine R and the distance X. 200mm-f JOON 490 N 90mmAnalyze the beam
- 2. How far is the centroid of the cross scction above its bottom edge? What is the area moment of inertia, I, about the centroid?* Q2: The effect or two forces FA,FB is (5 KN) and acting along u axis, determine the value of FB by parallelogram method (nun perpendicular coordinate) FA=3 kN 00 00 0 30 00 ************* ************ ス☆☆☆☆☆☆☆☆. . .- العهبه۱قحم المدني ) « University of Babylon College of Engineering Engineering Mechanics I Dept, of Civil Eng. Homework 1 Q1: Two forces act on the hook shown in figure. Determine the magnitude and direction of the 600 N resultant using Parallelogram Law. (Ans. R=910.241 N, angle=50.328° horizontal) with 50° 400 N 20° Q2: Determine the moment of the (50 N) force 50 N shown in figure with respect to: 30° 1-Point A, 2-Point O (Ans. Ma=50 N.m c.w., Mo=25 N.m c.c.w) 60° Q3: Replace the force and couple system 400 N shown in figure by an equivalent force and couple moment acting at: 1-Point A, 2-Point B 3 m 2 m 1.5 m- (Ans. MA=465 N.m c.c.w., Mp=575 N.m c.w) 150 N 150 N إضافة وصف. . .
- The access door (rectangle of sides 1430 mm by 770 mm) is held in the 36° open position by the chain AB. If the tension in the chain is 360 N, determine the scalar projection of the tension force (which acts on point A) onto the diagonal axis CD of the door. 770 mm 170 mm 1430 mm D. 36 Answer: TcD= i 249.524View Policies Current Attempt in Progress Replace the three forces shown by an equivalent force-couple system at point A. If the forces are replaced by a single resultant force, determine the distance d below point A to its line of action. 200 N Answers: R= 285 N i d = M = i Force-couple system at A. The force is positive if to the right, and the couple is positive if counterclockwise. Single resultant force. 150 N 690 mm 690 mm 690 mm N N-m mmdetermine: a) the differential area b) centroid from x and y axis