A particle of weight W newtons is held in equilibrium on a rough inclined plane by a horizontal force of magnitude 4 N - Edexcel - A-Level Maths Mechanics - Question 3 - 2011 - Paper 1

To find W, we continue from the previous result:

From the first part, we established that the normal force is 20 N. Now using equilibrium conditions for forces:

1. The frictional force at the point of sliding is: $F_{ ext{friction}} = 0.5 imes R = 0.5 imes 20 = 10 ext{ N}$

2. The weight component acting down the incline can be rewritten: W imes rac{3}{5} = 10 + 2.4 Simplifying gives: $W imes 0.6 = 12.4$

3. To find W, we rearrange: W = rac{12.4}{0.6} = rac{124}{6} = 20.67 ext{ N}

Thus, the weight of the particle is approximately 22 N.