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 2

To find the acceleration of the crate, we start by drawing a free-body diagram.

1. Identify forces acting on the crate:

   • Weight of the crate: $W = mg = 20 \text{kg} \times 9.81 \text{m/s}^2 = 196.2 \text{N}$ downwards.

   • Normal force ($N$) acting upwards.

   • Friction force ($F_f$) opposing the motion, given by:

     $F_f = \mu N$

   • The horizontal component of the tension ($T$):

     $T_x = T \cos(\alpha)$

   • The vertical component of the tension ($T_y$):

     $T_y = T \sin(\alpha)$

2. Set equations of motion:

   • In the vertical direction:

     $$$$

ightarrow N = W - T_y$$

 Substituting for $T_y$ gives:
 
 $$N = 196.2 - 40 \sin(\tan^{-1}(\frac{3}{4}))$$
 

• In the horizontal direction, applying Newton's second law:

  $T_x - F_f = ma$

3. Calculate friction force:

   • First, calculate $N$ using the previous equation, then:

     $F_f = 0.14 N$

4. Combine equations:

   • Substitute $N$ into the equation for $F_f$ and rearrange for $a$:

     $a = \frac{T_x - F_f}{m}$

5. Final calculation:

   • Solve the equations to find the acceleration, yielding:

     $a = \frac{\frac{40 \cdot 4/5}{5/5} - 0.14N}{20} = 1.2 \text{ m/s}^2.$