A block is resting on a horizontal surface - Scottish Highers Physics - Question 2 - 2017

The applied force of 24 N is at an angle of 60° with the horizontal. The horizontal component of this force is given by:

$F_{applied,x} = 24 \text{ N} \cdot \cos(60°) = 24 \text{ N} \cdot 0.5 = 12 \text{ N}.$

The vertical component is:

$$F_{applied,y} = 24 \text{ N} \cdot \sin(60°) = 24 \text{ N} \cdot \frac{\sqrt{3}}{2} \approx 20.78 \text{ N}.$

The normal force ($F_{N}$) is affected by the weight of the block and the vertical component of the applied force:

$F_N = m \cdot g - F_{applied,y},$ where $g$ (acceleration due to gravity) is approximately 9.81 m/s².

Calculating weight: $Weight = m \cdot g = 20 \text{ kg} \cdot 9.81 \text{ m/s}^2 = 196.2 \text{ N}.$

Thus, $$F_N = 196.2 \text{ N} - 20.78 \text{ N} \approx 175.42 \text{ N}.$

Using the net force calculated earlier and the components of the applied force, the force of friction can be derived from:

$F_{friction} = F_{applied,x} - F_{net}$

Substituting the known values: $F_{friction} = 12 \text{ N} - 4 \text{ N} = 8 \text{ N}.$

Thus, the force of friction acting on the block is 8 N.