A meterstick (L = 1 m) has a mass of m = 0.174 kg. Initially it hangs from two short strings: one at the 25 cm mark and one at the 75 cm mark.
answer
To solve this problem, we can use the principles of torques and balance. Torque is the rotational equivalent of force, so we can think of it as the force that causes an object to rotate.
Let's assume that the meterstick is in equilibrium when it is hanging from two short strings. This means that the sum of the torques acting on it is zero.
First, we need to calculate the torques created by each string.
Torque = Force x Distance
For the string at the 25 cm mark:
The force exerted by the string at the 25 cm mark is the weight of the meterstick, which can be calculated as follows: Weight = mass x gravity.
Given that the mass is 0.174 kg and the gravitational acceleration is approximately 9.8 m/s^2, we can calculate the weight:
Weight = 0.174 kg x 9.8 m/s^2.
Next, we need to calculate the distance from the 25 cm mark to the point of rotation (fulcrum), which is the 75 cm mark.
Distance = 75 cm - 25 cm.
Now we can calculate the torque exerted by this string:
Torque1 = Weight x Distance.
Similarly, for the string at the 75 cm mark, we need to calculate the force exerted by the string, which is the weight of the meterstick.
Then we can calculate the distance from the 75 cm mark to the fulcrum and the torque exerted by this string:
Torque2 = Weight x Distance.
Finally, since the meterstick is in equilibrium, the sum of the torques exerted by each string is zero:
Torque1 + Torque2 = 0.
Now you can plug in the values and solve for the unknowns.