Question

The moment of the 30 N force passing

through the coordinates (4, 0) and (0, 3)

about the origin

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Answer 1

Answer:

This is concurrent system of force system of 3 forces passing through origin

Let F1 be the force in OA

F1=F1.rOA=40[(−i+2j+4k)/(12+22+42‾‾‾‾‾‾‾‾‾‾‾‾√)]=−8.73i+17.4j+34.915kN

Let F2 be the force in OB

F2=F2.rOB=10[(3i−3k)/(32+32‾‾‾‾‾‾‾√)]=7.071i−7.071kN

Let F3 be the force in OC

F3=F3.rOC=30[(2i−2j+4k)/(22+22+42‾‾‾‾‾‾‾‾‾‾‾‾√)]=12.247i−12.247j−24.5kN

Now, resultant force

R=F1+F2+F3R=−10.588i−5.153j−3.344kN

The resultant of force system is R=−10.588i−5.153j−3.344kN

Explanation:

Answer 2

Given:

Force= 30 N passes through the coordinates (4,0) and (0,3)

To Find:

Moment of force about the origin.

Solution:

equation of a line passing throgh (4,0) and (0,3)

x/4+y/3=1

3x+4y=12

3x+4y-12=0

Now find the perpendicular distance from origin (0,0)  to this line along which the force is acting.

= I 3×0+4×0-12I/$(\sqrt{3^2+4^2} )$

= 12/$\sqrt{25}$

=12/5

We know, Moment = Force × Perpendicular distance

                               = 30 × 12/5

                               =72 N-m

Hence the value of the moment is 72N-m