Question

Abcd is a square, each side being 20 cm and e is the middle point of ab. Forces of 7, 8, 12, 5, 9 and 6 kn act on the lines of directions ab, ec, bc, bd, ca and de respectively. Find the magnitude, direction and position of the resultant force.

Answer 1

Answer:

Step-by-step explanation:

Step 1:

Magnitude of resultant forceLet∠BEC = α

We know that

Since tanα = 20/10 = 2

sinα = 2/5 = 0.894

and

cosα = 1/5 = 0.447

Resolving all the forces horizontally,

ΣH= 8 sin α+ 12 + 5 sin 45° – 9 sin 45° – 6 sin α

= (8 × 0.894) + (12) + (5 × 0.707) – (9 × 0.707) – (6 × 0.894) kN

= 10.96 kN

Step 2:

and now resolving all the forces vertically,

ΣV=7 + 8 cos α– 5 cos 45° – 9 cos 45° + 6 cos α

=7 + (8 × 0.447) – (5 × 0.707) – (9 × 0.707) + (6 × 0.447) kN

=3.36 kN...(ii)

We know that magnitude of the resultant force,

R = (ΣH)^2 + (ΣV)^2

= (10.96)^2 + (3.36)^2 = 11.46 kN

Direction of the resultant force Let

θ=Angle, which the resultant force makes with BC i.e., with the horizontal

tanθ =  ΣV /ΣH  

       = 3.36 /10.96 = 0.3066  or θ= 17.05°

Step 3:

Since both the values of ΣHand ΣV are + ve, therefore the resultant actual angle of the resultant force lies between 0° and 90°.

Position of the resultant force

Letx= Perpendicular distance between point E and the line of action of the resultant force.

Taking moments about *E and equating the same,

11.46 x=(7 × 0) + (8 × 0) + (12 × 10) + (5 × 0.707) + (9 × 0.707) + (6 × 0)= 129.9

x = 129.9 / 11.46 = 11.33 cm