Question

Q.6) Three forces Fi = 100 lb, F2 = 90 lb, and F3 = 120 lb, and the moment C= 200 lb.ft are applied to the box below. Find the wrench that is equivalent to the force system and the coordinates of the point of the intersection of the wrench and x-y plane. 6 ft 3 ft 4 ft

Answer 1

First, we write vectors for given forces and the couple:

$\small F_1 = 100\hat {AB} = 100*\frac{-4i-6j+3k}{\sqrt{4^2+6^2+3^2}}$

$\small \Rightarrow F_1 = -51.12i-76.82j+38.41k (lb)$

F2901b)

$\small F_3 = 120 j (lb)$

$\small C = 200\hat {BG} = 100*\frac{4i-3k}{\sqrt{4^2+3^2}}$

$\small \Rightarrow C = 160i-120k(lb-ft)$

Now, Resultant force is given by:

$\small R = F_1+F_2+F_3 = -51.12i-76.82j+38.41k+90i+120 j$

$\small \Rightarrow R =38.78i+43.18j+38.41 k$

Couple due to F1:

$\small C_1 = OB \times F_1$

$\small C_1 = (3k) \times -51.12i-76.82j+38.41k$

$\small C_1 = 230.46i-153.36j$

Couple due to F2:

$\small C_2 = OB \times F_2$

$\small C_2 = 3k\times 90i$

$\small C_2 = 270j$

Couple due to F3:

$\small C_3 = OG \times F_3$

$\small C_3 = 4i \times 120j$

$\small C_3 = 480 k$

total Couple:

$\small C_{eq} = C+C_1+C_2+C_3$

$\small C_{eq} = 160i-120k+230.46i-153.36j+270j+480k$

$\small C_{eq} = 390.46i+116.64j+360k (lb.ft)$

Let the point P in xy plane where the equivalent system intersects be:

$\small OP = xi+yj$

$\small OP \times R = C_{eq}$

$\small \Rightarrow (xi+yj)\times(38.78i+43.18j+38.41 k) = 390.46i+116.64j+360k$

$\small \Rightarrow 38.41y i - 38.41x j + (43.18x-38.78y) k= 390.46i+116.64j+360k$

$\small \Rightarrow 38.41y =390.46; -38.41x = 116.64; (43.18x-38.78y) =360$

r-3.3067; y= 10.166;