A wooden crate of mass 20kg is pulled in a straight line along a rough horizontal floor using a handle attached to the crate - Edexcel - A-Level Maths Mechanics - Question 7 - 2018 - Paper 1

To find the acceleration of the crate, we will resolve the forces acting on it.

1. Vertical Forces: We apply the condition for vertical forces:

   $R + 40 \sin \alpha = 20g$

   where:

   • $R$ is the normal reaction force,
   • $g$ is the acceleration due to gravity (approximately 9.81 m/s²).

   Given that $\tan \alpha = \frac{3}{4}$, we can find $\sin \alpha$ and $\cos \alpha$:

   $\sin \alpha = \frac{3}{5} \quad (hypotenuse = 5)$ $\cos \alpha = \frac{4}{5}$

   Substituting these into the equation gives:

   $R + 40 \times \frac{3}{5} = 20 \times 9.81$

   Therefore:

   $R + 24 = 196.2$

   Solving for $R$ gives:

   $R = 172.2$

2. Horizontal Forces: Now, considering the horizontal direction, we have:

   $40 \cos \alpha - F = 20a$

   where $F$ is the frictional force, given by:

   $F = \mu R = 0.14 \times 172.2 = 24.084$

   Substituting back into the horizontal forces equation:

   $40 \times \frac{4}{5} - 24.084 = 20a$

   This simplifies to:

   $32 - 24.084 = 20a$

   Giving:

   $7.916 = 20a$

   Solving for $a$ yields:

   $a = 0.3968 \text{ m/s}^2 \approx 0.40 \text{ m/s}^2$.