Example 12 Three forces \( \mathrm{F}_{1}\left(2 \mathrm{~N}, 060^{\circ}\right), \mathrm{F}_{2}(4.5 \mathrm{~N} \), \( \left.180^{\circ}\right) \) and \( \mathrm{F}_{3}\left(5 \mathrm{~N}, 300^{\circ}\right) \) act on a body of mass 2 kg , initially at rest. Find i) the magnitude and direction of the resultant force. ii) the acceleration with which the body begins to move.

Certainly! Let's solve the problem step by step. ### **Given:** - **Forces Acting on the Body:** 1. \( \mathbf{F}_1 = 2\, \text{N} \) at \( 60^\circ \) 2. \( \mathbf{F}_2 = 4.5\, \text{N} \) at \( 180^\circ \) 3. \( \mathbf{F}_3 = 5\, \text{N} \) at \( 300^\circ \) - **Mass of the Body:** \( m = 2\, \text{kg} \) - **Initial State:** The body is at rest. ### **Objective:** 1. **(i)** Determine the **magnitude** and **direction** of the resultant force \( \mathbf{R} \). 2. **(ii)** Calculate the **acceleration** \( a \) of the body when it begins to move. --- ### **Part (i): Finding the Resultant Force \( \mathbf{R} \)** To find the resultant force, we'll break each force into its horizontal (x-axis) and vertical (y-axis) components, sum these components separately, and then determine the magnitude and direction of the resultant vector. #### **1. Breaking Down Each Force into Components:** - **Force \( \mathbf{F}_1 = 2\, \text{N} \) at \( 60^\circ \):** \[ \begin{align*} F_{1x} &= F_1 \cdot \cos(60^\circ) = 2 \cdot 0.5 = 1\, \text{N} \\ F_{1y} &= F_1 \cdot \sin(60^\circ) = 2 \cdot \frac{\sqrt{3}}{2} \approx 1.732\, \text{N} \end{align*} \] - **Force \( \mathbf{F}_2 = 4.5\, \text{N} \) at \( 180^\circ \):** \[ \begin{align*} F_{2x} &= F_2 \cdot \cos(180^\circ) = 4.5 \cdot (-1) = -4.5\, \text{N} \\ F_{2y} &= F_2 \cdot \sin(180^\circ) = 4.5 \cdot 0 = 0\, \text{N} \end{align*} \] - **Force \( \mathbf{F}_3 = 5\, \text{N} \) at \( 300^\circ \):** \[ \begin{align*} F_{3x} &= F_3 \cdot \cos(300^\circ) = 5 \cdot 0.5 = 2.5\, \text{N} \\ F_{3y} &= F_3 \cdot \sin(300^\circ) = 5 \cdot (-\frac{\sqrt{3}}{2}) \approx -4.330\, \text{N} \end{align*} \] #### **2. Summing Up the Components:** - **Total Horizontal Component (\( R_x \)):** \[ R_x = F_{1x} + F_{2x} + F_{3x} = 1\, \text{N} - 4.5\, \text{N} + 2.5\, \text{N} = -1\, \text{N} \] - **Total Vertical Component (\( R_y \)):** \[ R_y = F_{1y} + F_{2y} + F_{3y} \approx 1.732\, \text{N} + 0\, \text{N} - 4.330\, \text{N} \approx -2.598\, \text{N} \] #### **3. Calculating the Magnitude of the Resultant Force (\( |\mathbf{R}| \)):** \[ |\mathbf{R}| = \sqrt{R_x^2 + R_y^2} = \sqrt{(-1)^2 + (-2.598)^2} \approx \sqrt{1 + 6.75} \approx \sqrt{7.75} \approx 2.78\, \text{N} \] #### **4. Determining the Direction of the Resultant Force (\( \theta_R \)):** The direction is given by the angle the resultant makes with the positive x-axis. Since both \( R_x \) and \( R_y \) are negative, the resultant lies in the **third quadrant**. \[ \theta_{\text{reference}} = \arctan\left(\frac{|R_y|}{|R_x|}\right) = \arctan\left(\frac{2.598}{1}\right) \approx 69^\circ \] Therefore, the angle relative to the positive x-axis is: \[ \theta_R = 180^\circ + 69^\circ = 249^\circ \] **\(\boxed{\begin{aligned} \text{i) Resultant force magnitude} &= 2.78\, \text{N} \\ \text{Direction} &= 249^\circ \text{ from the positive x-axis} \end{aligned}}\)** --- ### **Part (ii): Calculating the Acceleration \( a \) of the Body** Once the resultant force is known, we can use Newton's Second Law to find the acceleration. \[ \mathbf{F}_{\text{net}} = m \cdot \mathbf{a} \quad \Rightarrow \quad \mathbf{a} = \frac{\mathbf{F}_{\text{net}}}{m} \] Given: - \( |\mathbf{F}_{\text{net}}| = 2.78\, \text{N} \) - \( m = 2\, \text{kg} \) \[ a = \frac{2.78\, \text{N}}{2\, \text{kg}} = 1.39\, \text{m/s}^2 \] **\(\boxed{\text{ii) Acceleration} = 1.39\, \text{m/s}^2}\)** --- ### **Summary:** 1. **Resultant Force:** - **Magnitude:** \( 2.78\, \text{N} \) - **Direction:** \( 249^\circ \) from the positive x-axis 2. **Acceleration of the Body:** \( 1.39\, \text{m/s}^2 \)