Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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- Why does this guy always provide incorrect solutions. :( I will reportarrow_forward3) A point mass is attached to one end of a beam levered against a spike which acts as the pivot. The point mass weighs 250N and the beam has a uniform mass of 10kg and is 8m long. Parts a and b Part c Fapplied 3 of 6mov 1m 8m Krist 0 = 60° Scale 250 N Scale a. If a scale placed beneath the point mass reads 250N, what is the magnitude and direction for the force at the pivot point? Direction: North, 90° from x-axis Magnitude to k* 250N = 2500 N b. For the same scenario, how far must the pivot point be located relative to the left edge of the beam? (Hint, both the boxes weight and the force from the scale act at the farthest right edge of the beam.) www.acta 0672nos onl c. The pivot is then moved to be 6m from the left edge of the beam which now makes a 60 angle with the floor. How much downward force would be needed at the left-most edge of the beam to reduce the reading on the scale to zero? (Hint: the beam is still in static eq. when the scale reads zero.)arrow_forwardReview Learning Goal: To use the vector cross product to calculate the moment produced by a force, or forces, about a specified point on a member. Part A - Moment due to a force specified by magnitude and endpoints The moment of a force F about the moment axis passing through O and perpendicular to the plane containing O and F can be expressed using the vector cross product, Mo =r x F. In a properly constructed Cartesian coordinate system, the vector cross product can be calculated using a matrix determinant: As shown, a member is fixed at the origin, point O, and has an applied force F, the tension in the rope, applied at the free end, point B. (Figure 1) The force has magnitude F = 180 N and is directed as shown. The dimensions are ¤1 = 0.350 m, x2 = 1.90 m, y1 = 2.30 m, and z1 = 1.20 m. What is the moment about the origin due to the applied force F? i j M =rx F =|rz k Express the individual components of the Cartesian vector to three significant figures, separated by commas. ry F F,…arrow_forward
- L=9m W=5kN/m 360 UB 50.7arrow_forward15 kN/m 50 kN 150 kN m A 3 m E B 6 m 3 m -3 m a) Determine the reactions at supports A (pin) and D (roller). b) Determine the internal loading at x= 3 m (point E) measured from point A. c) Write functions for shear force (V) in terms of x d) Write functions for bending moment (M) in terms of x.arrow_forwardFarrow_forward
- A bar with varying cross section is subjected to various forces as shown below. Compute: a) Stresses in each section b) strain in each section c) total extension of the bar. Take E = 2.1x105 N/mm² 400 mm² P1=10kN A 1000 mm B 800 mm mm² P₂ P3=45KN+ 1500 mm 600 mm² 800 mm → P4-35kNarrow_forward5. Determine the value of the distributed load w and the concentrated loads x, y and z in the load diagram. [x=5.75kN, y=15kN, w= 5kN/m, z=34.25kN] 80 kN-m 5.75 X V (kN) 50N A 5m V (N) M (Nm) B 5 -9.25 6. Determine the value of A, B in the shear force diagram and C in the bending moment diagram. [A =50N; B = -50N; C = 125Nm] 10N/m 10m -5m- C W Z -.x(m) -34.25 50N -x (m) B -x (m)arrow_forwardA composite beam is under external loads as shown below. a) Draw the internal shear force and bending moment diagrams under the graph for the beam. Label the shear force and bending moment with values for each controlling point on the diagrams. b) Find the maximum tensile and compressive stresses. 12 mm 75 kN 40 kN/m 48 mm В 12 mm A 96 mm 1.8 m Cross section 0.9 m (not drawn to scale) 3.6 marrow_forward
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