3. G(s) = 1 4s² + 8s +1
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- 7. Using the formula R = √ (r² + s² /4); solve for r. Page #Let the region R be the area enclosed by the function f (x) = x³ – 1 and g(x) = 2x – 1.Find the volume of the solid generated when the region R is revolved about the line y = -3. You may use a calculator and round to the nearest thousandth. 2 1 -0.5 0.5 1 1.5 -1.5 -1 -2 -4 -5 -6 -7 -8 6- 2.QUESTION 2 Question 2 A cross-section of a beam is shown in Figure Q2. If the shear force in this section is V = 125 KN, determine the value and the location of the maximum shear stress in the section. In Figure Q2, a = 30 mm and the origin of the coordinate system is at centroid of the cross section. 7 y= Z= A a AY S= 20 4a mm; mm; O Figure Q2 Answer The vertical coordinate (y-coordinate; the y-axis serves as the axis of symmetry of the cross- section.) and horizontal coordinate (z-coordinate) of the location where the maximum shear stress occurs in the section are ← a The vertical distance from the location where the maximum shear stress occurs in the section to the bottom side (AB cross section can be calculated as Distance = mm (units: mm) 3a Second moment of area The second moment of area employed in the equation to calculate maximum shear stress can be calculated as I₂ = a (units: mm²) Shear stress The second moment of area employed in the equation to calculate maximum shear…
- Calculate the ?̅ of the center of mass of a wire with the shape of the curve outlined by the graph of ?(?).q1bQ3: Three forces are acting on a steel plate with 4m length by 4m width shown in figure. These three forces are required to cause a horizontal resultant acting through point A. If F = 504N, 1) The value of horizontal resultant R is: E A D 853.814N + O 796.894N → O 711.512N - O 640.361N → O None of these
- Compute a little more by taking the integralinii) terms up to the 5th order ii) taking the terms up to the 7th orderCompare the magnitudes of the equilibrant vectors measured from the experiment with those obtained from the graphical and component methods. Example: A: 200 g 60° above +x axis B: 300 g 45° above -x axis C: 400 g 30°below -x-axis A, A cos a = 1.96 N x cos 60° = 0.98 N B₂B cos b = 2.94 N x cos 135º = -2.08 N C, C cos g = 3.92 N x cos 210°= -3.39 N R₂-A, + B + C₂ = -4.49 N A, A sin a = 1.96 N x sin 60° = 1.70 N By B sin b = 2.94 N x sin 135º = 2.08 N C, C sin g = 3.92 N x sin 210° = -1.96 N Ry= Ay+ By + Cy = 1.82 N Questions: I: (a) A: 200 g along +x axis B: 100 g 45° above -x axis A₂ = A cos a = B₂ =B cos b = R₂-A₂+ B₁₂= A₂ = A sin a = B, = B sin b = R₂ = A + B₂ = R=(R₂. R₂):_ Quadrant R = √ (R₂²+R₂²) = Direction:q=tan [R, /R.] (c) A: 100 g along -y axis B: 200 g along -x axis A = A cos a = B = B cos b = R₂-A₂+ B₂ = A = A sin a = B, B sin b = R=A, +B₂ = R=(R₁, R₂): Quadrant R = √ (R₂²+R₂²) = Direction:q tan¹ [R, /R]Two forces act on the pipe assembly as shown in the figure. Determine the following: 3.1 The magnitude of the resultant FR is Blank 1 lb. 3.2 The coordinate direction angle Alpha of FR is Blank 2°. 3.3 The coordinate direction angle Beta of FR is Blank 3°. 3.4 The coordinate direction angle Gamma of FR is Blank 4°.
