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Apply equilibrium conditions in the horizontal and vertical planes: EF, = 0 N-Ncos 30° +0.3N cos 60° = 0 N' = 0.716N.. (1) EF, =0 10* kN) 0.3N" -(5 kN) -| 50×9.81 Nx- 1N +(0.3N sin 60°)+N sin 30° = 0 0.3N" -5.4905+0.7598N =0 0.3N" = 5.4905 – 0.7598N N' =18.30 -2.53N. -(2) Equate equations (1) and (2): 0.716N =18.30-2.53N 3.246N =18.30 N=5.63 kN Substitute the value of N in equation (1): N' =0.716×(5.63 kN) N' = 4.03 kN

Apply equilibrium conditions in the horizontal and vertical planes: EF, = 0 N-Ncos 30° +0.3N cos 60° = 0 N' = 0.716N.. (1) EF, =0 10* kN) 0.3N" -(5 kN) -| 50×9.81 Nx- 1N +(0.3N sin 60°)+N sin 30° = 0 0.3N" -5.4905+0.7598N =0 0.3N" = 5.4905 – 0.7598N N' =18.30 -2.53N. -(2) Equate equations (1) and (2): 0.716N =18.30-2.53N 3.246N =18.30 N=5.63 kN Substitute the value of N in equation (1): N' =0.716×(5.63 kN) N' = 4.03 kN

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Can you explain how to derive to get the values below? and where does the 10^-3/1 came from? (I underlined it with a red ink)
Apply equilibrium conditions in the horizontal and vertical planes: EF, =0 N' – Ncos 30° +0.3N cos 60° =0 N' = 0.716N. (1) EF, =0 10* kN 0.3N" – (5 kN) –| 50×9.81 N×- 1N +(0.3Nsin 60°)+ N sin 30° = 0 0.3N" -5.4905+0.7598N =0 0.3N" = 5.4905 – 0.7598N N' =18.30 -2.53N. -(2) Equate equations (1) and (2): 0.716N =18.30–2.53N 3.246N =18.30 N=5.63 kN Substitute the value of N in equation (1): N' =0.716×(5.63 kN) N" = 4.03 kN
Transcribed Image Text:Apply equilibrium conditions in the horizontal and vertical planes: EF, =0 N' – Ncos 30° +0.3N cos 60° =0 N' = 0.716N. (1) EF, =0 10* kN 0.3N" – (5 kN) –| 50×9.81 N×- 1N +(0.3Nsin 60°)+ N sin 30° = 0 0.3N" -5.4905+0.7598N =0 0.3N" = 5.4905 – 0.7598N N' =18.30 -2.53N. -(2) Equate equations (1) and (2): 0.716N =18.30–2.53N 3.246N =18.30 N=5.63 kN Substitute the value of N in equation (1): N' =0.716×(5.63 kN) N" = 4.03 kN
If block 1 weighs 50 kg, while block 2 weighs 90 kg, determine the external force P2 so that block 2 will maintain its position. Let P1 be equal to 5 kN. Assume 0.3 coefficient of static friction for all surfaces in contact. P1 2 P2 60°
Transcribed Image Text:If block 1 weighs 50 kg, while block 2 weighs 90 kg, determine the external force P2 so that block 2 will maintain its position. Let P1 be equal to 5 kN. Assume 0.3 coefficient of static friction for all surfaces in contact. P1 2 P2 60°
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