2021-09-21T17:13:52-04:00
2.4 A force F is applied at point A of the block with edge lengths of a, b, c Determine (1) the projections of F on the x, y and z axes; (2) the moments of F about x, y and z axes; (3) the moment of F about OB axis.
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2021-09-22T00:22:41-0400
(1)
F x = − ∣ F ⃗ ∣ sin β cos α F_x=-|\vec F|\sin \beta\cos \alpha F x = − ∣ F ∣ sin β cos α
F y = − ∣ F ⃗ ∣ cos β cos α F_y=-|\vec F|\cos \beta\cos \alpha F y = − ∣ F ∣ cos β cos α
F z = ∣ F ⃗ ∣ sin α F_z=|\vec F|\sin \alpha F z = ∣ F ∣ sin α
(2)
F ⃗ = ( − ∣ F ⃗ ∣ sin β sin α , − ∣ F ⃗ ∣ sin β cos α , ∣ F ⃗ ∣ cos β ) \vec F=(-|\vec F|\sin \beta\sin \alpha, -|\vec F|\sin \beta\cos \alpha, |\vec F|\cos \beta) F = ( − ∣ F ∣ sin β sin α , − ∣ F ∣ sin β cos α , ∣ F ∣ cos β )
r ⃗ = ( a , b , c ) \vec r=(a, b, c) r = ( a , b , c )
M ⃗ O = r ⃗ × F ⃗ \vec M_O=\vec r \times\vec F M O = r × F
= ∣ i ⃗ j ⃗ k ⃗ a b c − ∣ F ⃗ ∣ sin β cos α − ∣ F ⃗ ∣ cos β cos α ∣ F ⃗ ∣ sin α ∣ =\begin{vmatrix}
\vec i & \vec j & \vec k \\
a & b & c\\
-|\vec F|\sin \beta\cos \alpha & -|\vec F|\cos \beta\cos \alpha & |\vec F|\sin \alpha\\
\end{vmatrix} = ∣ ∣ i a − ∣ F ∣ sin β cos α j b − ∣ F ∣ cos β cos α k c ∣ F ∣ sin α ∣ ∣
= ( ∣ F ⃗ ∣ b sin α + ∣ F ⃗ ∣ c cos β cos α ) i ⃗ =( |\vec F|b\sin \alpha+|\vec F|c\cos \beta\cos \alpha)\vec i = ( ∣ F ∣ b sin α + ∣ F ∣ c cos β cos α ) i
+ ( − ∣ F ⃗ ∣ a sin α − ∣ F ⃗ ∣ c sin β cos α ) j ⃗ +(- |\vec F|a\sin \alpha-|\vec F|c\sin \beta\cos \alpha)\vec j + ( − ∣ F ∣ a sin α − ∣ F ∣ c sin β cos α ) j
+ ( − ∣ F ⃗ ∣ a cos α cos β + ∣ F ⃗ ∣ b sin β cos α ) k ⃗ +(- |\vec F|a\cos \alpha\cos \beta+|\vec F|b\sin \beta\cos \alpha)\vec k + ( − ∣ F ∣ a cos α cos β + ∣ F ∣ b sin β cos α ) k
M x = i ⃗ ⋅ ( r ⃗ × F ⃗ ) = M_x=\vec i\cdot(\vec r\times\vec F)= M x = i ⋅ ( r × F ) =
= ∣ F ⃗ ∣ b sin α + ∣ F ⃗ ∣ c cos β cos α = |\vec F|b\sin \alpha+|\vec F|c\cos \beta\cos \alpha = ∣ F ∣ b sin α + ∣ F ∣ c cos β cos α
M y = j ⃗ ⋅ ( r ⃗ × F ⃗ ) = M_y=\vec j\cdot(\vec r\times\vec F)= M y = j ⋅ ( r × F ) =
= − ∣ F ⃗ ∣ a sin α − ∣ F ⃗ ∣ c sin β cos α =- |\vec F|a\sin \alpha-|\vec F|c\sin \beta\cos \alpha = − ∣ F ∣ a sin α − ∣ F ∣ c sin β cos α
M z = k ⃗ ⋅ ( r ⃗ × F ⃗ ) = M_z=\vec k\cdot(\vec r\times\vec F)= M z = k ⋅ ( r × F ) =
− ∣ F ⃗ ∣ a cos α cos β + ∣ F ⃗ ∣ b sin β cos α - |\vec F|a\cos \alpha\cos \beta+|\vec F|b\sin \beta\cos \alpha − ∣ F ∣ a cos α cos β + ∣ F ∣ b sin β cos α
(3)
O B → = ( a , 0 , c ) ) \overrightarrow{OB}=(a, 0, c)) OB = ( a , 0 , c ))
u ⃗ = ( a a 2 + b 2 , 0 , c a 2 + c 2 ) \vec u=(\dfrac{a}{\sqrt{a^2+b^2}}, 0, \dfrac{c}{\sqrt{a^2+c^2}}) u = ( a 2 + b 2 a , 0 , a 2 + c 2 c )
M O B = u ⃗ ⋅ ( r ⃗ × F ⃗ ) = M_{OB}=\vec u\cdot(\vec r\times\vec F)= M OB = u ⋅ ( r × F ) =
= ∣ F ⃗ ∣ a b sin α + ∣ F ⃗ ∣ a c cos β cos α a 2 + b 2 =\dfrac{ |\vec F|ab\sin \alpha+|\vec F|ac\cos \beta\cos \alpha}{\sqrt{a^2+b^2}} = a 2 + b 2 ∣ F ∣ ab sin α + ∣ F ∣ a c cos β cos α
+ − ∣ F ⃗ ∣ a c cos α cos β + ∣ F ⃗ ∣ b c sin β cos α a 2 + b 2 +\dfrac{- |\vec F|ac\cos \alpha\cos \beta+|\vec F|bc\sin \beta\cos \alpha}{\sqrt{a^2+b^2}} + a 2 + b 2 − ∣ F ∣ a c cos α cos β + ∣ F ∣ b c sin β cos α
= ∣ F ⃗ ∣ a b sin α + ∣ F ⃗ ∣ b c sin β cos α a 2 + b 2 =\dfrac{ |\vec F|ab\sin \alpha+|\vec F|bc\sin \beta\cos \alpha}{\sqrt{a^2+b^2}} = a 2 + b 2 ∣ F ∣ ab sin α + ∣ F ∣ b c sin β cos α
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