The value of T1 is 100 kN.
The value of T2 is 80 kN.
The value of T3 is 50 kN.
Consider the figure
The formula to calculate the unit vector is
u = r ∣ r ∣ u= \frac{r}{|r|} u = ∣ r ∣ r
Here, r is the position vector.
The formula to calculate the force vector is
T=T u
Here, T is the force.
The formula to calculate the resultant force vector is
T R = T 1 + T 2 + T 3 T_R=T_1+T_2+T_3 T R = T 1 + T 2 + T 3
The coordinates of the points A, B, C, and P are
A(1,-2,0)
B(2,3,0)
C(-2,3,0)
P(0,0,6)
The position vector for AP, BP and CP are
r A P = − 1 i + 2 j + 6 k r B P = − 2 i − 3 j + 6 k r C P = 2 i − 3 j + 6 k r_{AP}=-1i+2j+6k \\
r_{BP}=-2i-3j+6k \\
r_{CP}=2i-3j+6k r A P = − 1 i + 2 j + 6 k r BP = − 2 i − 3 j + 6 k r CP = 2 i − 3 j + 6 k
Calculate the force vector at AP.
T 1 = T 1 u A P = 100 × r A P ∣ r A P ∣ = 100 × − 1 i + 2 j + 6 k ( − 1 ) 2 + 2 2 + 6 2 = − 100 41 i + 200 41 j + 600 41 k = − 15.62 i + 31.23 j + 93.7 k k N T_1=T_1u_{AP} \\
= 100 \times \frac{r_{AP}}{|r_{AP}|} \\
= 100 \times \frac{-1i+2j+6k}{\sqrt{(-1)^2+2^2+6^2}} \\
= -\frac{100}{\sqrt{41}}i + \frac{200}{\sqrt{41}}j+ \frac{600}{\sqrt{41}}k \\
= {-15.62i +31.23j+93.7k}\;kN T 1 = T 1 u A P = 100 × ∣ r A P ∣ r A P = 100 × ( − 1 ) 2 + 2 2 + 6 2 − 1 i + 2 j + 6 k = − 41 100 i + 41 200 j + 41 600 k = − 15.62 i + 31.23 j + 93.7 k k N
Calculate the force vector at BP.
T 2 = T 2 u B P = 80 × r B P ∣ r B P ∣ = 80 × − 2 i − 3 j + 6 k ( − 2 ) 2 + ( − 3 ) 2 + 6 2 = − 160 7 i − 240 7 j + 480 7 k = − 22.85 i − 34.28 j + 68.57 k k N T_2=T_2u_{BP} \\
= 80 \times \frac{r_{BP}}{|r_{BP}|} \\
= 80 \times \frac{-2i-3j+6k}{\sqrt{(-2)^2+(-3)^2+6^2}} \\
= -\frac{160}{7}i - \frac{240}{7}j+ \frac{480}{7}k \\
= {-22.85i -34.28j+68.57k} \;kN T 2 = T 2 u BP = 80 × ∣ r BP ∣ r BP = 80 × ( − 2 ) 2 + ( − 3 ) 2 + 6 2 − 2 i − 3 j + 6 k = − 7 160 i − 7 240 j + 7 480 k = − 22.85 i − 34.28 j + 68.57 k k N
Calculate the force vector at CP.
T 3 = T 3 u C P = 50 × r C P ∣ r C P ∣ = 50 × 2 i − 3 j + 6 k 2 2 + ( − 3 ) 2 + 6 2 = 100 7 i − 150 7 j + 300 7 k = 14.28 i − 21.43 j + 42.85 k k N T_3=T_3u_{CP} \\
= 50 \times \frac{r_{CP}}{|r_{CP}|} \\
= 50 \times \frac{2i-3j+6k}{\sqrt{2^2+(-3)^2+6^2}} \\
= \frac{100}{7}i - \frac{150}{7}j+ \frac{300}{7}k \\
= {14.28i -21.43j+42.85k}\;kN T 3 = T 3 u CP = 50 × ∣ r CP ∣ r CP = 50 × 2 2 + ( − 3 ) 2 + 6 2 2 i − 3 j + 6 k = 7 100 i − 7 150 j + 7 300 k = 14.28 i − 21.43 j + 42.85 k k N
Calculate the resultant force vector.
T R = T − 1 + T 2 + T 3 = [ ( − 15.62 i + 31.23 j + 93.7 k ) + ( − 22.85 i − 34.28 j + 68.57 k ) + ( 14.28 i − 21.43 j + 42.85 k ) ] = − 24.19 i − 24.48 j + 205.12 k k N T_R=T-1+T_2+T_3 \\
= [(-15.62i +31.23j+93.7k) + (-22.85i -34.28j+68.57k) + (14.28i -21.43j+42.85k)] \\
= {-24.19i-24.48j + 205.12k} \;kN T R = T − 1 + T 2 + T 3 = [( − 15.62 i + 31.23 j + 93.7 k ) + ( − 22.85 i − 34.28 j + 68.57 k ) + ( 14.28 i − 21.43 j + 42.85 k )] = − 24.19 i − 24.48 j + 205.12 k k N
Calculate the magnitude of the resultant force.
∣ T R ∣ = ( − 24.19 ) 2 + ( − 24.48 ) 2 + ( 205.12 ) 2 = 208 k N |T_R| = \sqrt{(-24.19)^2 +(-24.48)^2 +(205.12)^2} \\
= 208 \;kN ∣ T R ∣ = ( − 24.19 ) 2 + ( − 24.48 ) 2 + ( 205.12 ) 2 = 208 k N
Calculate the unit vector of the resultant force.
u R = T R ∣ T R ∣ = − 24.19 i − 24.48 j + 205.12 k 208 = − 0.116 i − 0.118 j + 0.986 k u_R= \frac{T_R}{|T_R|} \\
= \frac{-24.19i-24.48j + 205.12k}{208} \\
= -0.116i -0.118j +0.986k u R = ∣ T R ∣ T R = 208 − 24.19 i − 24.48 j + 205.12 k = − 0.116 i − 0.118 j + 0.986 k
The coordinate is (-0.116,-0.118,0.986).
The equivalent force is {-24.19i - 24.48j + 205.12k} kN and the coordinate is (-0.116,-0.118,0.986).
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