Compute the resultant of three forces acting on the plate shown in the figure. Locate it’s intersection with AB and BC.

#### Solution

Solution :

Given : Various forces acting on a body

To find : Resultant of the forces and intersection of resultant with AB and BC

Solution :

In △ AFG ,

tanα = `(AG)/(AF)` =`(DE)/(BH)` = `3/2` = 1.5

α = tan^{-1}(1.5) = 56.31°

In △DAE,

tan θ = `(DE)/(AD)`=`(DE)/(BC)` =`3/4` = 0.75

θ = tan^{-1}0.75 = 36.87°

In △DHC

tanβ =`(DC)/(HC)` =`6/2` = 3

β = tan^{-1}(3)

β = 71.565°

Assume R be the resultant of the forces

ΣFx = -722cos α + 1000cos θ + 632cos β

= 599.3624 N

ΣFy = -722sin α - 1000sin θ + 632sin β

= -601.1725 N

R=`sqrt((ΣFx)^2+(ΣFy)^2)`

R=`sqrt((599.3624)^2+(−601.1725)^2`

R=848.9073 N

ϕ = tan^{-1}`(Σ Fy)/(ΣFx)`

= tan^{-1} `((-601.1725)/(599.3624))`

= 45.0863° (in fourth quadrant)

Let R cut AB and BC at points M and N respectively

Draw AL ⊥ R

**Taking moments about point A**

M_{A} = 632 sin β x AD -722cos α x AG

= 632 x sin71.565o x 4 – 722cos56.31° x 3

=1196.7908 Nm**Applying Varigon’s theorem**MA = R x AL

1196.7908 = 848.9073 x AL

AL=1.4098 m

In △AML,

cos Ф = `(AL)/(AM)`

cos 45.0863 =`(1.4098)/(AM)`

AM = 1.9967 m

MB = AB - AM

= 6 - 1.9967

= 4.0033 m

In △BMN

tan Ф =`(BM)/(BN)`

tan 45.0863=`(4.0033)/(BN)`

R=848.9073 N (45.0863° in fourth quadrant)

Resultant force intersects AB and BC at M and N such that AM=1.9967 m and BN=3.9912 m