Question 4.8: Three coplanar forces (forces that act within the same plane...
Three coplanar forces (forces that act within the same plane) –A, B and C – are all applied to a pin joint (Figure 4.16(a)). Determine the magnitude and direction of the equilibrant for the system.
Each force needs to be resolved into its two orthogonal (at right-angles) components, which act along the vertical and horizontal axes, respectively. Using the normal algebraic sign convention with our axes, V is positive above the origin and negative below it. Similarly, H is positive to the right of the origin and negative to the left. Using this convention we need only consider acute angles for the sine and cosine functions. These are tabulated below:
| Magnitude (KN) |
Horizontal component (kN) | Vertical component (kN) |
| 10 | + 10 (→) | 0 |
| 14 | + 14 cos 60 (→) | + 14 sin 60 (↑) |
| 8 | −8 cos 45 (←) | − 8 sin 45 (↓) |
Then total horizontal component = (10 + 7 − 5.66) kN = 11.34 kN (→) and total vertical component = (0 + 12.22 − 5.66) kN = 6.46 kN (↑).
Since both the horizontal and vertical components are positive, the resultant force will act upwards to the right of the origin. The three original forces have now been reduced to two which act orthogonally. The magnitude of the resultant R, or the equilibrant, may now be obtained using Pythagoras’ theorem on the right-angle triangle obtained from the orthogonal vectors, as shown in Figure 4.16(b).
From Pythagoras we get:
R² = 6.46² + 11.34² = 170.33
and so resultant
R = 13.05 kN
So the magnitude of the equilibrant also = 13.05 kN.
From the right-angled triangle shown in Figure 4.16(b), the angle θ that the resultant R makes with the given axes may be calculated using trigonometric ratios:
\tan \theta=\frac{6.46}{11.34} = 0.5697 and θ = 29.67°
Therefore, the resultant R = 13.05 kN ∠29.67°.
The equilibrant will act in the opposite sense and therefore = 13.05 kN ∠209.67°.