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 need to resolve the forces acting on it.

1. Vertical Forces: Using equilibrium in the vertical direction:

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

   Where:

   • R is the normal reaction force
   • g = 9.81 m/s² (acceleration due to gravity)
   • (\sin \alpha = \frac{3}{5}) (from tanα = 3/4)

   This gives us: $R + 40 \cdot \frac{3}{5} = 20 \cdot 9.81$

   Thus: $R + 24 = 196.2 \implies R = 196.2 - 24 = 172.2 \, \text{N}$

2. Horizontal Forces: For the horizontal direction, we apply:

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

   Where F is the friction force given by: $F = 0.14R$

   Substituting R:

   $F = 0.14 \cdot 172.2 = 24.068 \, \text{N}$

   And using (\cos \alpha = \frac{4}{5}):

   $40 \cdot \frac{4}{5} - 24.068 = 20a$

   This simplifies to:

   $32 - 24.068 = 20a \implies 7.932 = 20a \implies a = \frac{7.932}{20} = 0.3966 \, \text{m/s²}$

   Thus, the acceleration of the crate is approximately 0.40 m/s².