Blocks A and B, of masses m and m , are at rest on a frictionless surface, as shown above, with block A...

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Blocks A and B, of masses m and m , are at rest on a frictionless surface, as shown above, with block A fixed to the table. Block C of mass m is suspended by a string that is tied to block B over an ideal pulley. Which of the following gives the magnitude of the force exerted by block A on block B ?

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To determine the magnitude of the force exerted by block A on block B, we can analyze the forces acting on block C and how they relate to blocks A and B. 1. **Identify the forces**: - Block C has a gravitational force acting on it, which is its weight, given by: $$ F_g = m \cdot g $$ - This force acts downward and is what causes tension in the string tied to block B. 2. **Tension in the string**: - Assuming the pulley is ideal, the tension $ T $ in the string is equal to the weight of block C when it is at rest: $$ T = m \cdot g $$ 3. **Force on Block B**: - Block B is experiencing tension $ T $ upward due to the string and the force exerted by block A downward. According to Newton's second law, if Block B is stationary, the net force acting on it must equal zero. Therefore, the force exerted by block A on block B (let's denote it as $ F_{AB} $) must balance the tension in the string: $$ F_{AB} = T $$ 4. **Calculating $ F_{AB} $**: - From our earlier conclusion, we can substitute $ T $ with the expression for the weight of Block C: $$ F_{AB} = m \cdot g $$ 5. **Final result**: - Hence, the magnitude of the force exerted by block A on block B is: The magnitude of the force exerted by block A on block B is: mg

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Blocks A and B, of masses m and m , are at rest on a frictionless surface, as shown above, with block A...

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