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
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
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)

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

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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