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:

   • The weight of the crate ( W = mg = 20 \text{kg} \times 9.81 \text{m/s}^2 = 196.2 \text{N} )
   • The vertical component of the tension in the handle is ( T \sin \alpha ).

   Using equilibrium in the vertical direction:

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

   where ( R ) is the normal reaction force from the ground.

   Substituting the known values:

   ( R + 40\sin \left( \tan^{-1}\left(\frac{3}{4}\right) \right) = 196.2 \text{N} )

2. Horizontal Forces:

   • The horizontal component of the tension is ( T \cos \alpha ).
   • The frictional force is given by ( F = \mu R = 0.14R ).

   Again, using Newton's second law:

   $40\cos \alpha - F = ma \implies 40\cos \alpha - 0.14R = 20a$

3. Substituting Values:

   Solving for ( R ) from the first equation and substituting into the second, we can calculate:

   From these equations, after substituting all the known values and resolving, we find:

   ( a = 0.396 \text{m/s}^2 ) or ( a = 0.40 \text{m/s}^2 ).