A force of F = 160 N is applied to the rope that causes the angle 8₁ = 60 degrees to keep the system at N equilibrium. The spring constant is = 80- m B. a b FAB FAC a S 0₁ Values for the figure are given in the following table. Note the figure may not be to scale. Variable Value 2m 2m A 2m F a. Determine the magnitude of the force in the cord AB, FAB b. Determine the magnitude of the force in the spring AC, FAC c. Determine the stretch of the spring, s d. Determine the original length (unstretched length) of the spring, Lo Round your final answers to 3 significant digits/figures. m b wwwww N @080 2013 Michael Swanbom

A force of F = 160 N is applied to the rope that causes the angle 8₁ = 60 degrees to keep the system at N equilibrium. The spring constant is = 80- m B. a b FAB FAC a S 0₁ Values for the figure are given in the following table. Note the figure may not be to scale. Variable Value 2m 2m A 2m F a. Determine the magnitude of the force in the cord AB, FAB b. Determine the magnitude of the force in the spring AC, FAC c. Determine the stretch of the spring, s d. Determine the original length (unstretched length) of the spring, Lo Round your final answers to 3 significant digits/figures. m b wwwww N @080 2013 Michael Swanbom

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Equilibrium of Concurrent Force System. The piston of a reciprocating engine exerts a force of 175 kN on the crosshead when the crank is 45° (angle B) past TDC. If the stroke of the piston is 800 mm and the length of the connecting rod is 1.80 m, find the guide force and the force in the connecting rod. Hint: 1=½ stroke; Angle o may be solved using Sine Law in AACO. K= Compressive Force in the connecting Rod E= Piston Effort 70 120 F, = Guide Force Fig. P-1 Fig. P-2 a Tools: EF cos0=0; F sin .=0. R = Ev)* + (E#): ; R 771 =- : tane = sin A sin B sin C
3) 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.)
Q2/scale the Rhombus defined by the points p1=( 3,1), p2=(5,3), p3=(3,5) & p43(1,3) by a scale factor of 1/2 keeping the first point in the same location
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Consider the four bar suspension linkage shown in the figure. Use the given dimensions from the table below to write the → position vector rße describing the lowest link and the position →>> vector TGH describing the shock. Express each vector in Cartesian components and also as a magnitude and an angle (measured counterclockwise from an x-axis defined parallel to the ground and pointing right). h₂ Variable Value h₁ 10.3 in h₂ 17 in h3 18 in h4 12.6 in h5 5.7 in ← "BE = W₂ → rGH = Values for dimensions on the figure are given in the following table. Note the figure may not be to scale. → rGH at an angle of WANAN → IBE has magnitude BE = at an angle of Variable Value 8.5 in 5.1 in 9.1 in 6.8 in 12 in W1 W2 W3 W4 W5 î has magnitude TGH = W4 î+ h₂ h₂ cc 30 BY NC SA 2013 Michael Swanbom 3) inches inches and points degrees CCW from horizontal. 3) inches inches and points degrees CCW from horizontal.