Question

If the resultant force is directed along the boom from point A towards O, determine the values of x and z for the coordinates of point C. Set FB = 1420 N and FC = 2470 N ?

Answer 1

z = 1.547725 m

x = -1.0318 m

Step-by-step 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