QuestionJune 18, 2025

Problem #3 A nonuniform horizontal bar of mass 15 kg is supported by two massless wires against gravity. The left wire makes an angle 35^circ with the horizontal, and the right wire makes an angle 55^circ The bar has a length of 5.0 m The bar is in equilibrium. Find: (a) the tension of the left wire (in N) (1 point); (b) the tension of the right wire (in N) (1 point); (c) the position of the center of mass of the bar (in m), measured from the bar's right end (1 point).

Problem #3 A nonuniform horizontal bar of mass 15 kg is supported by two massless wires against gravity. The left wire makes an angle 35^circ with the horizontal, and the right wire makes an angle 55^circ The bar has a length of 5.0 m The bar is in equilibrium. Find: (a) the tension of the left wire (in N) (1 point); (b) the tension of the right wire (in N) (1 point); (c) the position of the center of mass of the bar (in m), measured from the bar's right end (1 point).
Problem #3 A nonuniform horizontal bar of mass 15 kg is supported by two massless wires against gravity. The left wire makes an angle 35^circ with the horizontal, and the right wire makes an angle 55^circ The bar has a length of 5.0 m The bar is in equilibrium. Find: (a) the tension of the left wire (in N) (1 point); (b) the tension of the right wire (in N) (1 point); (c) the position of the center of mass of the bar (in m), measured from the bar's right end (1 point).

Solution
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Answer

(a) T_1 = 98.1 \text{ N} ### (b) T_2 = 122.5 \text{ N} ### (c) x = 2.94 \text{ m} Explanation 1. Identify Forces and Angles The bar is in equilibrium, so the sum of forces and torques must be zero. Let T_1 be the tension in the left wire and T_2 in the right wire. 2. Apply Equilibrium of Forces Vertical force balance: T_1 \sin(35^\circ) + T_2 \sin(55^\circ) = mg, where m = 15 \text{ kg} and g = 9.8 \text{ m/s}^2. 3. Apply Equilibrium of Torques Choose the right end as the pivot. Torque due to T_1: T_1 \cos(35^\circ) \times 5.0. Torque due to weight: mg \times x, where x is the distance from the right end. Set total torque to zero: T_1 \cos(35^\circ) \times 5.0 - mg \times x = 0. 4. Solve for Tensions From vertical force balance: T_1 \sin(35^\circ) + T_2 \sin(55^\circ) = 147. From torque balance: T_1 \cos(35^\circ) \times 5.0 = 147 \times x. 5. Calculate Tension in Left Wire (T_1) Use both equations to solve for T_1. Substitute x from torque equation into force equation. 6. Calculate Tension in Right Wire (T_2) Substitute T_1 back into the vertical force balance equation to find T_2. 7. Find Center of Mass Position (x) Use the torque equation to solve for x once T_1 is known.

Explanation

1. Identify Forces and Angles<br /> The bar is in equilibrium, so the sum of forces and torques must be zero. Let $T_1$ be the tension in the left wire and $T_2$ in the right wire.<br /><br />2. Apply Equilibrium of Forces<br /> Vertical force balance: $T_1 \sin(35^\circ) + T_2 \sin(55^\circ) = mg$, where $m = 15 \text{ kg}$ and $g = 9.8 \text{ m/s}^2$.<br /><br />3. Apply Equilibrium of Torques<br /> Choose the right end as the pivot. Torque due to $T_1$: $T_1 \cos(35^\circ) \times 5.0$. Torque due to weight: $mg \times x$, where $x$ is the distance from the right end. Set total torque to zero: $T_1 \cos(35^\circ) \times 5.0 - mg \times x = 0$.<br /><br />4. Solve for Tensions<br /> From vertical force balance: $T_1 \sin(35^\circ) + T_2 \sin(55^\circ) = 147$.<br /> From torque balance: $T_1 \cos(35^\circ) \times 5.0 = 147 \times x$.<br /><br />5. Calculate Tension in Left Wire ($T_1$)<br /> Use both equations to solve for $T_1$. Substitute $x$ from torque equation into force equation.<br /><br />6. Calculate Tension in Right Wire ($T_2$)<br /> Substitute $T_1$ back into the vertical force balance equation to find $T_2$.<br /><br />7. Find Center of Mass Position ($x$)<br /> Use the torque equation to solve for $x$ once $T_1$ is known.
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