Final answer:
The support force exerted by the clamp on a meterstick in rotational equilibrium with three equal masses located at specific points is the combined weight of the masses and the meterstick itself. However, based on the provided positions, the masses cannot be in balance, indicating an issue with the statement or a misunderstanding of the equilibrium condition.
Step-by-step explanation:
To determine the support force exerted by the clamp, which is acting as a fulcrum, on a meterstick with three equal masses located at specific points, we must consider the condition for rotational equilibrium. This condition states that the sum of all torques around the pivot point must be zero. Since the meterstick is in equilibrium, the torques produced by the weights of the masses and the weight of the meterstick itself are balanced.
Let's denote the mass of the meterstick as M and its weight as w = M × g (where g is the acceleration due to gravity). The weight of the meterstick acts at its center, which is at the fulcrum, so it exerts no torque. The three masses, each being m, are located at distances of 35 cm, 5 cm, and 45 cm from the fulcrum, producing torques of m × g × 35 cm, m × g × 5 cm, and m × g × 45 cm, respectively.
For equilibrium, the sum of torques clockwise equals the sum of torques counterclockwise:
- Torque from mass at 15 cm (counterclockwise): m × g × 35 cm
- Torque from mass at 45 cm (counterclockwise): m × g × 5 cm
- Torque from mass at 95 cm (clockwise): m × g × 45 cm
Setting up the equation for equilibrium:
m × g × 35 + m × g × 5 = m × g × 45
This simplifies to:
40 × m × g = 45 × m × g
We can cancel out m × g since they appear on both sides:
40 = 45
This result indicates there is an error in the statement, as the masses on one side cannot balance the mass on the other side if they are all equal and at the stated positions. Assuming we were trying to find the net force on the fulcrum rather than balance conditions, the support force Fs exerted by the clamp (fulcrum) would be the weight of all three masses plus the weight of the meterstick, as the meterstick is in equilibrium and the fulcrum must support the entire system. Therefore:
Fs = 3 × m × g + M × g
Conclusively, given the above constraints, if we calculate the support force, we must consider the correct positioning of masses or adjust the masses such that equilibrium is actually achieved.