Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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- 3arrow_forwardThe figures show a frame and its FBDS. If an additional couple moment is applied at A, then how will you change the ?FBD of member BC at B 2000 N B. 3 m 2 m 2 m 60° .Will have two forces and a moment at B .a O Will have two forces, Bx and By, at B .b O .No change, still just one force (FAB) at B .c O Will add one moment at B .d Oarrow_forwardQ.2) An exerciser shown in Figure 2 begins with his arm in the relaxed vertical position OA, at which the elastic band is unstretched. He then rotates his arm to the horizontal position OB. The elastic modulus of the band is k 150 N/m that is, 150 N of force is required to stretch the band each additional meter of elongation. Determine the moment about 0 of the force which the band exerts on the hand B. -680 800mmarrow_forward
- I need quick solution, attempt only if you can solve it fast. Answer is around 166.7lb (don't know where on the beam does the resultant acts) Show work with FBDarrow_forwardExample 3- Home work Determine the couple moment acting on the pipe. Segment AB is directed 30°below the x-y plane. 250 N 0.8 m 300 250 N 0.6 m Barrow_forwardand For problem above, Replace the loading on the system below by a single resultant force. Find the points on the x- and y – axes through which the resultant force must pass.arrow_forward
- Version: b Determine the length b of the triangular load and its position a on the beam such that the equivalent resultant force is zero and the resultant couple moment is MR Let M₂ = -30 kN - m R Assume the intensities of the distributed loading be w = 12 kN/m w = 5 kN/m r Let the length of the beam be d = 8 m A b W₁ Wr d Note that the sign of MR indicates whether the couple moment is positive (CCW) or negative (CW). Also, position a may be negative, which indicates that the length of the rectangular load is shorter than the triangular load.arrow_forwardQ2 Draw the S.F.and B.M. diagrams for the overhanging beam carrying uniformly distribu 30KN/m overthe entire length and a point load of (Yourroll no.kN) as shown in figure, Locat of contraflexure. Your roll no. kN 30kN/m 2 ofarrow_forwardPlease give detailed solution with free body diagram. Strictly don't copy from othersite ,I will report such answer.arrow_forward
- The rigid pipe is subject to the following loads. Neglecting the weight of the pipe, determine the moment at O. SOLVE THE FF 1. Vector component form of 50-lb force (ANSWER: 25.000i +0j -43.30k lb) and the Moment at point O, in vector component form. (ANSWER: 1153.01i -829.52j +250.00k lb-in) 2. Magnitude of the resultant force (ANSWER: 78.10 lb) 3. Position vector of the 50-lb force from point O to its point of application on the pipe. (ANSWER: -15i -10j +12k inch)arrow_forwardPlease don't provide handwritten solution .....arrow_forward
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