Respuesta :
Answer:
z = 1.547725 m
x = -1.0318 m
Explanation:
Given:
- FB = 1420 N
- FC = 2470 N
- load P = 1500 N (diagram)
- OA = < 0 , 6 , 0 > (diagram)
- OB = < -2 , 0 , 3 > (diagram)
- vec(R) = < 0 , -1 , 0 >
Find:
The x and z components of point C
Solution:
- Compute unit vectors AB and AC
vec (AB) = -OA + OB = -<0 , 6 , 0> + <-2 , 0 , 3 > = < -2 , -6 , 3 >
vec (AC) = -OA + OC = -<0 , 6 , 0> + <x , 0 , z > = < x , 6 , z >
mag (AB) = sqrt(2^2 + 6^2 + 3^2) = 7
mag (AC) = sqrt(x^2 + 6^2 + z^2) = sqrt(x^2 + 36 + z^2)
unit (AB) = < -2 / 7 , -6 / 7 , 3 / 7 >
unit (AC) = (1 / sqrt(x^2 + 36 + z^2)) < x , -6 , z >
- Compute Force vectors FB, FC, R and Fg:
FB = FB . < -2 / 7 , -6 / 7 , 3 / 7 >
FC = FC . (1 / sqrt(x^2 + 36 + z^2))< x , -6 , z >
R = R. < 0 , -1 , 0 >
FG = P . < 0 , 0 , -1 >
- Compute sum of forces in x and z directions:
In x- direction:
- FB*2 / 7 + FC * ( x / sqrt(x^2 + 36 + z^2)) = 0 ... 1
In z- direction:
FB*3 / 7 + FC * ( z / sqrt(x^2 + 36 + z^2)) = 0 .... 2
- Dividing two equations:
x / z = -2 / 3
x = -2*z / 3
- Substitute x into 2:
FC*( z / sqrt(36 + 13z^2/9)) = - (3*FB/7)
( z / sqrt(36 + 13z^2/9)) = - (3*FB/7FC)
- Square both sides:
z^2 / (36 + 13z^2/9) = (9/49)*(FB/FC)^2
z^2 = z^2*(13/49)*(FB/FC)^2 + (324/49)*(FB/FC)^2
z^2*(1 - (13/49)*(FB/FC)^2) = (324/49)*(FB/FC)^2
z = sqrt( (324/49)*(FB/FC)^2 / (1 - (13/49)*(FB/FC)^2) )
- Compute z and x:
z = sqrt( (324/49)*(1420/2470)^2 / (1 - (13/49)*(1420/2470)^2) )
z = 1.547725 m
x = -1.0318 m
