Force F 1860 N acts from A towards B C 175 mm 40 mm 350 mm D

Force F 1,860 N acts from A towards B. C/ 175 mm ·40 mm 350 mm Determine the moment of the force about point C in N·m. (The line of action due to the isometric drawing. Express your answer in vector form.) for F does not pass through O. That illusion occurs Mc = N·m Determine the shortest distance from point C to the line of action of the force in mm.

Solution

coordinates of A=(0.14,0.35,0)m

coordinates of B=(0.14,0,0.175)m

coordinates of C=(0,0.35,0.175)m

vector along AB = -0.35j+0.175k m

Unit vector along AB = (-0.35j+0.175k)/sqrt(0.35²+0.175²)=-0.894i+0.447j

Force vector = 1860*(-0.894i+0.447j)=-1663.6j+831.8k N

Vector CA = 0.14i-0.175k

Moment about C=(0.14i-0.175k)x(-1663.6j+831.8k) = -291.13i-116.452j-232.904k

Magnitude = sqrt(291.13²+116.452²+232.904²)=390.6 Nm

Shortest distnace = 390.6/1860=0.21 m