A uniform meterstick balances on a fulcrum placed at the 40-cm mark when a weight W is placed at the 20-cm mark. What is the weight of the meterstick?Select one:a.W/2b.0.4Wc.2Wd.W
Question
A uniform meterstick balances on a fulcrum placed at the 40-cm mark when a weight W is placed at the 20-cm mark. What is the weight of the meterstick?Select one:a.W/2b.0.4Wc.2Wd.W
Solution
The meterstick is in equilibrium, which means the sum of the torques (or moments) about any point must be zero. The torque due to a force is given by the product of the force and the distance from the pivot point (fulcrum).
Let's denote the weight of the meterstick as M and assume that it acts at the center of the meterstick, which is at the 50-cm mark.
The torque due to the weight W is W * 20 cm (since it's 20 cm from the fulcrum), and the torque due to the weight of the meterstick is M * 10 cm (since it's 10 cm from the fulcrum).
Setting these two torques equal to each other (since the meterstick is in equilibrium), we get:
W * 20 cm = M * 10 cm
Solving for M (the weight of the meterstick), we get:
M = 2W
So, the weight of the meterstick is twice the weight W. Therefore, the correct answer is c. 2W.
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