### Step 1: Express the Balancing Condition Mathematically For static balance, vector sum of mass-radius products (moments) must be zero. Let the fourth mass be \( m_4 \) at radius \( R_4 = 178\,\text{mm} \) and angle \( \theta_4 \). First, write the mass-radius product in vector form for each mass: - \( m_1 R_1 \) at \( ^\circ \) (along +y) - \( m_2 R_2 \) at \( 225^\circ \) - \( m_3 R_3 \) at \( 330^\circ \) - \( m_4 R_4 \) at \( \theta_4 \) (unknown) Express each as a complex vector: \[ \begin{align*} m_1 R_1 e^{j^\circ} &= 1.81 \times 381 \times e^{j^\circ} \\ m_2 R_2 e^{j225^\circ} &= 2.27 \times 254 \times e^{j225^\circ} \\ m_3 R_3 e^{j330^\circ} &= .907 \times 305 \times e^{j330^\circ} \\ m_4 R_4 e^{j\theta_4} &= m_4 \times 178 \times e^{j\theta_4} \end{align*} \] Sum in x and y directions, set total to zero for balance: \[ \sum m_i R_i \cos\theta_i + m_4 R_4 \cos\theta_4 = \] \[ \sum m_i R_i \sin\theta_i + m_4 R_4 \sin\theta_4 = \] **Explanation:** This step sets up the vector equations for balancing, expressing each mass-radius as a 2D vector (complex number) and forming the required equations for the system to be in static balance. --- ### Step 2: Substitute Values and Solve for \( m_4 \) and \( \theta_4 \) Calculate each term: - \( m_1 R_1 = 1.81 \times 381 = 690.61 \) (at \( ^\circ \): \( x = , y = 690.61 \)) - \( m_2 R_2 = 2.27 \times 254 = 576.58 \) (at \( 225^\circ \): \( x = 576.58 \cos 225^\circ, y = 576.58 \sin 225^\circ \)) - \( m_3 R_3 = .907 \times 305 = 276.64 \) (at \( 330^\circ \): \( x = 276.64 \cos 330^\circ, y = 276.64 \sin 330^\circ \)) Calculate components: - \( \cos 225^\circ = -.7071, \sin 225^\circ = -.7071 \) - \( \cos 330^\circ = .866, \sin 330^\circ = -.5 \) So, - \( m_2 R_2: x = 576.58 \times -.7071 = -407.89, y = 576.58 \times -.7071 = -407.89 \) - \( m_3 R_3: x = 276.64 \times .866 = 239.61, y = 276.64 \times -.5 = -138.32 \) Sum x and y: \[ \text{Total } x = - 407.89 + 239.61 = -168.28 \] \[ \text{Total } y = 690.61 - 407.89 - 138.32 = 144.40 \] Set up equations: \[ -168.28 + m_4 \times 178 \cos\theta_4 = \] \[ 144.40 + m_4 \times 178 \sin\theta_4 = \] Solve for \( m_4 \) and \( \theta_4 \): \[ \frac{144.40}{-168.28} = \frac{m_4 \times 178 \sin\theta_4}{m_4 \times 178 \cos\theta_4} = \tan\theta_4 \] \[ \tan\theta_4 = -.8584 \implies \theta_4 = \arctan(-.8584) = -40.92^\circ \text{ (or 319.08°)} \] Now, find \( m_4 \): \[ m_4 \times 178 \cos(-40.92^\circ) = 168.28 \] \[ \cos(-40.92^\circ) = .757 \implies m_4 = \frac{168.28}{178 \times .757} = 1.26 \text{ kg} \] --- ## **Final Answer** - **Balancing mass \( m_4 = 1.26 \) kg** - **Angular position \( \theta_4 = 319.1^\circ \) (or equivalently, \( -40.9^\circ \))** --- **Explanation:** We first wrote the vector sum equations for static balance, then substituted values and solved the simultaneous equations to find the required mass and angular position for the balancing mass.