In this question use g = 9.8 m.s^{-2} A boy attempts to move a wooden crate of mass 20 kg along horizontal ground - AQA - A-Level Maths Mechanics - Question 13 - 2018 - Paper 2

To determine if the crate remains stationary, we resolve the applied force into its horizontal and vertical components:

The horizontal component of the force is:

$F_{horizontal} = 150 \cos(15^\circ)$

And the vertical component is:

$F_{vertical} = 150 \sin(15^\circ)$

Calculating these:

1. Horizontal component:

   • $F_{horizontal} ≈ 150 \times 0.9659 ≈ 144.89 \text{ N}$

2. Vertical component:

   • $F_{vertical} ≈ 150 \times 0.2588 ≈ 38.82 \text{ N}$

Next, we calculate the normal force (R). The weight of the crate is:

$W = mg = 20 \times 9.8 = 196 \text{ N}$

Thus, the normal force is:

$R = W - F_{vertical} = 196 - 38.82 ≈ 157.18 \text{ N}$

Now, we calculate the maximum static friction:

$F_{max} = ext{μR} = 0.85 \times 157.18 ≈ 133.6 \text{ N}$

Finally, we compare the horizontal force with the maximum friction. Since:

$144.89 > 133.6$

The crate will begin to move, which suggests that it does not remain stationary. Therefore, the applied force is greater than the maximum static friction, and the crate begins to move.