Question

Forces 7kN,10KN,10KN,3KN respectively acting at one of the angular pointsbof a regular pentagon. Towards the other four angular points taken in order. Find there resultant force and its direction
.

Answer 1

respectively one of the angular points to the order

Answer 2

Given: Forces:

7kN , 10kN, 10kN, 3KN

Explanation:

Regular pentagon is a polygon having 5 sides which have equal length.

Taking angles of any regular polygon

= 180 - (360/ Number of sides)

For pentagon angles

= 180 - 360/5 =  108°

= 108 / 3 = 36°

Representing the concurrent forces acting

(i) ∑Fₓ = 7 + 10 cos 36°+ 10 cos 72°- 3 cos 72°

    ∑Fₓ = 7+10cos36° + 10cos72° −3cos72°

    ∑Fₓ = 17.25 kN(→)

(ii) ∑ $F_{y}$ = 10 sin 36°+ 10 sin 72°+ 3 sin 72°

    ∑$F_{y}$ = 10sin36° +10sin72° +3sin72°

  ∑$F_{y}$ = 18.24  kN (↑)

(iii) Magnitude of resultant R

     R = √[(∑Fₓ)² + (∑$F_{y}$)²]

          = √[(17.25)² + (18.24)²]

      R = 25.10 kN

(iv) Inclination of the resultant θ

θ = tan⁻¹(∑$F_{y}$/Fₓ)  :   θ = tan⁻¹( 18.24/17.25)

θ  = 46.59°

(v)  The resulting Position shown in Figure.