Engineering Mechanics Statics - Intructor Solutions manual 2.45 - 245 If the resultant force acting on the bracket is to be directed along the positive

Engineering Mechanics Statics - Intructor Solutions manual 2.45

This preview shows page 1 out of 1 page.

Image of page 1

Unformatted text preview: 2—45. If the resultant force acting on the bracket is to be directed along the positive u axis, and the magnitude of F1 is required to be minimum, determine the magnitudes of the resultant force and F1. SOLUTION Rectangular Components: By referring to Figs. a and b, the x and y components of F1, F2, F3, and FR can be written as (F1)x=F1 COS¢ (F1)y=Flsin¢ (F2)x = 650G) = 390N (F2), = 650 (g) = 520N (F3), = 500 cos 45° = 353.55 N (F3), = 500 sin 45° = 353.55 N (FR)x = FR cos 45° = 0.7071FR (FR)y = FR sin 45° = 0.7071FR Resultant Force: Summing the force components algebraically along the x and y axes, we have i, 2(FR)X = 213,; 0.7071FR = F1 cos ¢ — 390 + 353.55 (1) +12%)y = 213,; 0.7071FR = F1 sin ¢ + 520 — 353.55 (2) Eliminating F R from Eqs. (1) and (2), yields 202.89 F = — 3 1 cos¢ — sin¢ ( ) The first derivative of Eq. (3) is dF1 sin (1) + cos (1) = — 4 dd) (cos d) — sin ¢)2 ( ) The second derivative of Eq. (3) is W 2 ' + 2 1 = (smqfi cos qb) + 1 (5) d¢2 (cos (I) — sin (1))3 cos d) — sin ¢ For F1 to be minimum, —1 = 0. Thus, from Eq. (4) 01¢ sin (15 + cos ()5 = 0 tan (b = —1 ¢ = —45° Substituting 45 = —45° into Eq. (5), yields d2F1 2 = 0.7071 > 0 d¢ This shows that 45 = —45° indeed produces minimum F1.Thus, from Eq. (3) (F3)? 202.89 C“) F1 = 143.47 N = 143 N Ans. = cos (—45°) — sin (—45°) Substituting ()5 = —45° and F1 = 143.47 N into either Eq. (1) or Eq. (2), yields FR = 919 N Ans. ...
View Full Document

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Ask Expert Tutors You can ask 0 bonus questions You can ask 0 questions (0 expire soon) You can ask 0 questions (will expire )
Answers in as fast as 15 minutes