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Question

0.5 mg

2 mg

0.75 mg

1 mg

Solution

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Step-by-Step Solution Step 1: Identify the forces on the beam Let the tension (or force) at point A be $T_A$, the weight acting at its center be $mg$ (acting at 50 cm from the end), and another weight $2 mg$ (acting at 75 cm from the end, i.e., 25 cm from the midpoint). These distances are typically measured from the pivot or reference point chosen for torque calculation. Step 2: Write the torque equilibrium condition about point B Since the beam is in rotational equilibrium about point B, the net torque about B is zero. Assuming all forces act perpendicular to the beam, the torque equation can be written as: $T_A \times 100 - mg \times 50 - (2mg) \times 25 = 0$ Here 100, 50, and 25 (all in cm) represent the perpendicular distances from point B to the points of application of $T_A$, $mg$, and $2mg$ respectively. Step 3: Simplify and solve for $T_A$ Rearranging the above equation, we get: $100\,T_A = 100\,mg$ Therefore, $T_A = mg$ Step 4: Conclude the required value of $T_A$ Since $T_A$ equals $mg$, and comparing to the multiple-choice options, this corresponds to “1 mg.” Answer The correct answer is 1 mg.

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Parent Quiz 748262

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