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Determine the horizontal and vertical components of reaction on the beam caused by the pin at B and the rocker at A. Neglect the weight of the beam. A 600 N 45% -2 m 0.2 m -3 m D 100 N 2 m 200 N B

Determine the horizontal and vertical components of reaction on the beam caused by the pin at B and the rocker at A. Neglect the weight of the beam. A 600 N 45% -2 m 0.2 m -3 m D 100 N 2 m 200 N B

Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter11: Columns
Section: Chapter Questions
Problem 11.5.10P: Solve the preceding problem (W 250 × 44.8) if the resultant force P equals 110 kN and E = 200 GPa.
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the +38 EXAMPLE NO. 1 Determine the horizontal and vertical components of reaction on the beam caused by the pin at B and the rocker at A. Neglect the weight of the beam. Free-Body Diagram A 600 sin 45° N 600 cos 45° N 2 m A 0.2 m 3 m 600 N 100 N -2 m + ΣF = 0; (+ΣΜ» = 0; +1 ΣF, = 0; 45° 0.2 m 2 m | 200 N SOLUTION B 3 m B, 600 cos 45° N D 100 N B₁ = 0 2 m 200 N B₁ = 424 N 100 N (2 m) + (600 sin 45° N)(5 m) B Equations of Equilibrium 319 N 1319 N 319 N (600 cos 45° N)(0.2 m) - A,(7 m) = 0 Ay = 319 N Ans. Ans. 319 N-600 sin 45° N 100 N 200 N + By = 0 By = 405 N Ans. 319 N
Transcribed Image Text:the +38 EXAMPLE NO. 1 Determine the horizontal and vertical components of reaction on the beam caused by the pin at B and the rocker at A. Neglect the weight of the beam. Free-Body Diagram A 600 sin 45° N 600 cos 45° N 2 m A 0.2 m 3 m 600 N 100 N -2 m + ΣF = 0; (+ΣΜ» = 0; +1 ΣF, = 0; 45° 0.2 m 2 m | 200 N SOLUTION B 3 m B, 600 cos 45° N D 100 N B₁ = 0 2 m 200 N B₁ = 424 N 100 N (2 m) + (600 sin 45° N)(5 m) B Equations of Equilibrium 319 N 1319 N 319 N (600 cos 45° N)(0.2 m) - A,(7 m) = 0 Ay = 319 N Ans. Ans. 319 N-600 sin 45° N 100 N 200 N + By = 0 By = 405 N Ans. 319 N
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