WJEC Physics for AS Level Student Book 2nd Edition

22 WJEC Physics for AS Level Applying condition 2: The resultant moment = 0 (about any point). Let’s take moments about end A : The 3 , 1 and 2 N forces each have a clockwise moment about A ; F (= 6 N ) has an anticlockwise moment. Taking CM as positive: \ 3.0 N × 10 cm – 6.0 N × 40 cm + 1.0 N × 50 cm + 2.0 N × d = 0 \ (simplifying) 2 d = 160 cm . \ The 2.0 N weight must be at the 80 cm mark. The last problem we are going to look at is how to find an unknown force if there are forces at different angles. For example, what force F must we apply in Fig. 1.1.27 so that the forces are in equilibrium? We don’t need to worry about rotations because all the forces pass through the same point. We have a choice of three techniques – but two of them are essentially the same! a) Add the 10 N and 6 N . Then F must be equal and opposite to the resultant. b) Add the 10 N , the 6 N and F as in Fig. 1.1.7, so that the resultant is 0. c) Resolve in two directions – horizontally and vertically are the obvious ones: Vertically: F cos θ = 10 N [1] Remember: through the angle l cos Horizontally: F sin θ = 6 N [2] Through 90° – θ l sin Dividing [2] by [1] and remembering that sin θ cos θ = tan θ l tan θ = 6 10 . This lets us calculate θ and then we can use [1] or [2] to calculate F . We finish off with two more difficult examples. The first shows that the principle of moments applies even when there isn’t a pivot: if something doesn’t start to rotate, the moments of the forces about any point must add to zero, even for a bridge! The second is more difficult because the forces are not parallel. – F (a) 10 N 6 N F θ (b) 10 N 6 N Fig. 1.1.27 Find the equilibrant θ 10 N 6 N F Knowledge check Find d and F in Fig. 1.1.26 by taking moments, about A , then about B and solving the simultaneous equations. 1.1.12 Knowledge check Find F and θ in Fig. 1.1.27 using methods (a), (b) and finish off (c). 1.1.13