Can you futher explain how the to get the coeeffiecnt (angle at the end) The last equation is kind of confusing, i dont know evactly how the 738sin @ -1755cos@=0 to get the angle @=67.1
Can you futher explain how the to get the coeeffiecnt (angle at the end) The last equation is kind of confusing, i dont know evactly how the 738sin @ -1755cos@=0 to get the angle @=67.1
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Can you futher explain how the to get the coeeffiecnt (angle at the end) The last equation is kind of confusing, i dont know evactly how the 738sin @ -1755cos@=0 to get the angle @=67.1
Transcribed Image Text:Using free body diagram equating using equilibrium conditions.
Calculation:
Free hody diagram of ladder.
P= 175 lb
3 -0"
Ng
3' -0"
D
6' -0"
W= 30 lb
Friction force at the finor.
Fa - NA
From the triangle OAB from given figure,
OA - AB coso
OA - 12 cos
OB - AB sino
OB - 12 sino
AC - ft
Equilibrium equation in X direction.
la - Na
Na - Na(1)
Equilibrium equation in Y direction.
EFy-0
NA - (. + mp) - 0
NA - (30 + 175) - 0
NA - 205b. (2)
Consider moments about A
EMA -0
Ng (OB) – 175 (AD cas ) – 30AC cas0 -0. (3)
Suhstitute NA - 205 in equation I. we get
Na - 0.3205)
Ny - 61.S
Suhstituting Na - 615in equation 3, we get,
Ng (OB) – 175 (AD cas ) - 30AC caso -0
61.5(12 sin - 175 (9 cos )- 30 x 6 cos0 -0
738 sin o- 1575 cos - 180 coso-0
738 sin o-1755 cnso-0
0-67.19
Conclusion:
The angle required for the ladder without slipping to the left is o- 67.19.
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