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

To find the vertical component ($F_{y}$) of the force, we use the equation:

$F_{y} = F imes ext{sin}( heta)$

Substituting the values:

$F_{y} = 60 imes ext{sin}(30°) = 60 imes 0.5 = 30 ext{ N}$

The frictional force ($F_{f}$) can be calculated using:

$F_{f} = ext{μ}_k imes F_{N}$

To find the normal force ($F_{N}$), we apply the formula:

$F_{N} = mg - F_{y}$

Where:

• $m = 6$ kg
• $g = 9.8$ m/s²
• Using the value of $F_{y}$ calculated previously:

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

Now substituting into the frictional force formula:

$F_{f} = 0.13 imes 58.8 = 7.644 ext{ N}$

To find the horizontal component ($F_{x}$) of the force, we use:

$F_{x} = F imes ext{cos}( heta)$

Calculating:

F_{x} = 60 imes ext{cos}(30°) = 60 imes rac{ ext{√}3}{2} hickapprox 51.96 ext{ N}

The net force can be calculated by subtracting the frictional force from the horizontal component:

$F_{net} = F_{x} - F_{f}$

So,

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

Using Newton's second law ($F = ma$) to find acceleration ($a$):

a = rac{F_{net}}{m} = rac{44.316}{6} hickapprox 7.39 ext{ m/s}^2

The friction will DECREASE. This is because a larger angle leads to a reduced vertical component of the force, which in turn decreases the normal force acting on the object. Since friction is proportional to the normal force, a decrease in the normal force results in less friction.

Newton’s Third Law states that for every action, there is an equal and opposite reaction. This means that forces always occur in pairs; if object A exerts a force on object B, then object B simultaneously exerts a force of equal magnitude and opposite direction on object A.

In the free-body diagram, illustrate the following forces:

1. Weight ($F_{g}$) acting downwards due to gravity.
2. Normal force ($F_{N}$) acting upwards perpendicular to the surface.
3. Frictional force ($F_{f}$) acting opposite to the direction of motion.
4. Tension ($F_{T}$) from the car pulling the caravan.

Each force should be represented with arrows showing their direction and labelled accordingly.