If force F rightarrow is perpendicular to the inclined roof

If force F ^ rightarrow is perpendicular to the inclined roof plane ABC, Find the moment along CB.^ rightarrow

Ans:

r_AB = r_B - r_A = (0,0, 2) - (2, 2, 0) = (-2, -2, 2)

r_AC =r_C - r_A = (0, 3, 0) - (2, 2, 0) = (-2, 1, 0)

To find the direction of force, we need to determine unit vector along the force

cross product of any two vectors on the roof would generate the vector along the force, so from that vector we can determine unit vectors

Let u is the vector perpedicular to roof

u = r_AB x r_AC = (-2, -2, 2) x (-2, 1, 0) = (-2, -4, -2)

Unit vector of u = (-2, -4, -2)/ (24)^1/2 = (-1, -2, -1)/ sqrt(6) This is direction of force

Now we have to calculale perpendicular distance from line BC

Mid point of BC = (0, -3/2, 1)

Distance between Mid point and A is (2, 0, 0) - (0, -3/2, 1) = (2, 3/2, -1 )

Now, Moment = 200 [ (-1, -2, -1) ] /(6)^1/2 x (2, 3/2, -1)

= 100 [(7 i -10j + 4k)]/ / (6)^1/2