A machine part is subjected to forces as shown.Find the resultant of forces in magnitude and in direction.

Also locate the point where resultant cuts the centre line of bar AB.

**Solution**:**Given** : A machine subjected to various forces**To find** : Resultant of forces

Point where the resultant force cuts the bar AB

#### Solution

In △BAM,

∠A=60°

AB=3 m

BM=3sin60

=2.5981 m

In △CAN

AC=1.5m

AN=1.5cos60

=0.75 m

AP=AN+NP

=0.75+0.5

=1.25 m

**Two 20N forces are equal and opposite in direction.Hence,they form a couple**

Perpendicular distance between two 20 N forces = 1m

Moment of couple = 20 x 1

=20 kN-m (Anti clockwise)

Assume R is the resultant of the forces and it is inclined at an angle θ with horizontal

`ΣF_X=5 kN`

`ΣF_Y=-15 kN`

`R=sqrt( F_x^2)+ (F_Y^2)`

`=sqrt(5^2 +(−15)^2)`

=15.8114 kN

`θ = tan^(-1)((R_y)/(R_X))`

`=tan^(-1)((−15)/5)`

=71.5651° (in fourth quadrant)

Assume that the resultant cut the center line of bar AB at point D**Applying Varigon’s theorem**

`ΣM_A=ΣM_A^R`

-5 x BM -15 x AP +20 = R x AQ

-11.7405 = -15.8114 x AQ**AQ = 0.7425 m**

In △AQL, ∠ALQ=θ

∠QAL=90 - θ

∠BAL=60

∠QAD=60 - (90 - θ)

= θ - 30

=71.5651 - 30

=41.5651°

In△DAQ,cos QAD= `(AQ)/(AD)`

`AD=(AQ)/(cosDAQ) = (0.7425)/(cos41.5651) = 0.9924 m`

**Resultant force = 15.8114 kN(at 71.5651° in fourth quadrant)****It cuts the center line of bar AB at point D such that AD=0.9924m**