Crate P of mass 1,25 kg is connected to another crate, Q, of mass 2 kg by a light inextensible string - NSC Physical Sciences - Question 2 - 2022 - Paper 1

To illustrate the forces acting on crate P, the free-body diagram should include:

• The gravitational force, $F_g = mg = 1.25 ext{ kg} imes 9.81 ext{ m/s}^2 = 12.25 ext{ N}$, acting downward.
• The normal force, $F_N$, acting upward, equal in magnitude to the gravitational force (12.25 N).
• The tension in the string, $T$, acting horizontally to the right.
• The friction force, $F_f = 1.8 ext{ N}$, acting horizontally to the left.

The net forces can be displayed as follows:

1. Weight (gravitational force - downward): $12.25 ext{ N}$
2. Normal force (upward): $F_N = 12.25 ext{ N}$
3. Tension (to the right): $T$
4. Friction (to the left): $1.8 ext{ N}$

2.3.1 The tension in the string:

For crate P, applying Newton's Second Law in the horizontal direction:

$F_{net} = T - F_f$ $T - 1.8 = (1.25)(0.1)$ $T - 1.8 = 0.125$ $T = 0.125 + 1.8$ $T = 1.925 ext{ N}$

2.3.2 Angle θ:

For crate Q, the forces acting are:

• The applied force $F = 7.5 ext{ N}$ at angle θ
• The friction force of 2.2 N

Using Newton's Second Law:

$F_{net} = F imes ext{cos}( heta) - F_f$ Substituting known values:

$7.5 imes ext{cos}( heta) - 2.2 = (2)(0.1)$ $7.5 imes ext{cos}( heta) - 2.2 = 0.2$ $7.5 imes ext{cos}( heta) = 0.2 + 2.2$ $7.5 imes ext{cos}( heta) = 2.4$ ext{cos}( heta) = rac{2.4}{7.5} $ext{cos}( heta) = 0.32$

Calculating θ:

$heta = ext{cos}^{-1}(0.32) ext{ which gives } heta ext{ approximately } 70.53^ ext{o}.$