In this question use $g = 9.8 \, \text{ms}^{-2}$\n\nA boy attempts to move a wooden crate of mass 20 kg along horizontal ground - AQA - A-Level Maths Pure - Question 13 - 2018 - Paper 2

In this case, we need to resolve the applied force into its horizontal and vertical components. The vertical component is given by:\n\n$F_{y} = 150 \sin(15°)$\n\nCalculating this gives:\n\n$F_{y} = 150 \times 0.2588 \approx 38.82 \text{ N}$\n\nNext, we determine the normal force (R):\n\n$R = mg - F_{y}$\n\nSubstituting for mass and gravity:\n\n$R = 20 \times 9.8 - 38.82 \approx 157.18 \text{ N}$\n\nNow, we calculate the static friction force using the coefficient of friction:\n\n$F_{static} = \mu R = 0.85 \times 157.18 \approx 133.6 \text{ N}$\n\nNext, we need the horizontal component of the applied force:\n\n$F_{x} = 150 \cos(15°) \approx 150 \times 0.9659 \approx 144.9 \text{ N}$\n\nFinally, we compare the horizontal force to the maximum static friction force:\n\n- Horizontal force = 144.9 N\n- Maximum static friction = 133.6 N\n\nSince the horizontal force exceeds the static friction, the crate begins to move.