A 6 kg object is pulled with a force of 60 N at an angle of 30° across a rough surface, as shown in the diagram below - NSC Technical Sciences - Question 2 - 2021 - Paper 1

The vertical component (F_y) of the force can be calculated using the sine function:

$F_y = 60 imes ext{sin}(30°)$

Calculating this gives:

$F_y = 60 imes 0.5 = 30 ext{ N}$

The frictional force (f_k) can be calculated using the equation:

$f_k = ext{μ}_k imes F_n$

Where:

• Normal force (F_n) is defined as:

$F_n = mg - F_y$

Given the mass (m) of the object is 6 kg and weight (g) is approximately 9.8 m/s², we find:

$F_n = (6 imes 9.8) - 30 = 58.8 ext{ N}$

Thus,

$f_k = 0.13 imes 58.8 = 7.644 ext{ N}$

The horizontal component (F_x) of the force can be calculated using the cosine function:

$F_x = 60 imes ext{cos}(30°)$

Calculating this gives:

F_x = 60 imes rac{ ext{√3}}{2} = 51.96 ext{ N}

Using Newton's second law, we can find the acceleration (a) of the object:

$F_{ ext{net}} = m imes a$

Where the net force is:

$F_{ ext{net}} = F_x - f_k = 51.96 - 7.644 = 44.316 ext{ N}$

Thus,

a = rac{F_{ ext{net}}}{m} = rac{44.316}{6} = 7.386 ext{ m/s}^2

The friction will DECREASE because as the angle increases, the vertical component of the applied force will increase, reducing the normal force and thus the frictional force.

Newton's Third Law states that for every action, there is an equal and opposite reaction.

A free-body diagram for the caravan should include:

• Weight force acting downwards (F_w = mg)
• Normal force acting upwards (F_n)
• Tension force from the car pulling the caravan (F_T)
• Frictional force acting opposite to the direction of movement (F_f)

Each force should be represented by arrows indicating their direction and labelled accordingly.