Find the tension in the two cords shown in the ... - JAMB Physics 2023 Question

Given the forces acting on the mass (\(m\)):

1. The weight of the mass (\(mg\)) acts vertically downward.
2. The tension in \(T_1\) acts at an angle of 38° to the horizontal.
3. The tension in \(T_2\) acts horizontally.

The vertical component of \(T_1\) (\(T_{1v}\)) is given by \(T_1 \cdot \sin(38^\circ)\), and the horizontal component of \(T_1\) (\(T_{1h}\)) is given by \(T_1 \cdot \cos(38^\circ)\).

The net force in the vertical direction is equal to the weight of the mass (\(mg\)):

\[ T_{1v} = mg \]

The net force in the horizontal direction is zero (assuming no acceleration in the horizontal direction):

\[ T_{1h} = T_2 \]

Now, set up the equations:

1. In the vertical direction:
\[ T_1 \cdot \sin(38^\circ) = mg \]

2. In the horizontal direction:
\[ T_1 \cdot \cos(38^\circ) = T_2 \]

Given that \(m = 220 \, \text{kg}\) and \(g = 9.8 \, \text{m/s}^2\), you can solve these equations to find \(T_1\) and \(T_2\).

Let's solve these equations:

1. \[ T_1 = \frac{mg}{\sin(38^\circ)} \]
2. \[ T_2 = T_1 \cdot \cos(38^\circ) \]

Substitute the values and calculate:

\[ T_1 \approx \frac{(220 \, \text{kg} \cdot 9.8 \, \text{m/s}^2)}{\sin(38^\circ)} \]

\[ T_2 = T_1 \cdot \cos(38^\circ) \]

Now, let's calculate these values:

\[ T_1 \approx 3502 \, \text{N} \]
\[ T_2 \approx 2760 \, \text{N} \]

So, the correct option is: \(T_1 \approx 3502 \, \text{N}, \quad T_2 \approx 2760 \, \text{N}\)