Question

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

Answer 1

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^-10.75 = 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