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 6kN act on the lines of directions AB, EC, BC, BD, CA and DE respectively. Find the magnitude, direction and position of the resultant force.
Answers
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
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Step 1:
Magnitude of resultant force Let ∠BEC = A
We know that
Since tanA = 20/10 = 2
sinA = 2/5 = 0.894
and
cosA= 1/5 = 0.447
Resolving all the forces horizontally,
ΣH= 8 sin A+ 12 + 5 sin 45° – 9 sin 45° – 6 sin A
= (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 A– 5 cos 45° – 9 cos 45° + 6 cos A
=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