A metre rule of mass 15g pivoted at 30cm mark is balanced horizontally when a load of 25g was hung at the 18cm mark
I)sketch the set-up
II) if the position of the pivot is moved to the 33cm mark, by how much must the load be reduced to maintain the horizontal balance of the rule
I) Sketching the setup:
To visualize the setup, draw a horizontal line representing the meter rule. Mark the 0 cm position on the left end, and the 100 cm position on the right end. Place a dot at the 30 cm mark, representing the pivot point. Also, mark another dot at the 18 cm mark to indicate the position where the load of 25 g is hanging.
--0-----------------------[.]----------[.]
II) Calculating the needed reduction in load:
To maintain the horizontal balance of the rule when the position of the pivot is moved to the 33 cm mark, we need to determine the new load needed.
Starting with the original setup:
- The mass of the meter rule is 15 g.
- The load is 25 g, and it is located at the 18 cm mark.
- The pivot is at the 30 cm mark.
Now, consider the clockwise and counterclockwise torques acting on the system:
- The counterclockwise torque is produced by the load and is equal to (25 g) * (18 cm).
- The clockwise torque is produced by the meter rule and is equal to (15 g) * (distance from pivot to the center of mass).
Since the system is balanced, the sum of the counterclockwise and clockwise torques is zero.
- For the original setup, the distance from the pivot to the center of mass can be calculated using the concept of the center of mass. Assuming the meter rule has uniform density, the center of mass is at the midpoint, which is 50 cm. Therefore, the distance from the pivot to the center of mass is 20 cm (30 cm - 50 cm).
By setting up the equation for the balance of torques, we have:
(25 g) * (18 cm) = (15 g) * (20 cm)
Now, let's solve for the new load required when the pivot is moved to the 33 cm mark.
- The distance from the pivot to the center of mass changes. For the new setup, the distance is 17 cm (33 cm - 50 cm).
Setting up the equation for the balance of torques, we get:
(New Load) * (17 cm) = (15 g) * (20 cm)
To find the new load, divide both sides of the equation by 17 cm:
New Load = (15 g) * (20 cm) / (17 cm)
Calculating this, we get:
New Load = 17.65 g
So, to maintain the horizontal balance of the rule when the position of the pivot is moved to the 33 cm mark, the load needs to be reduced by approximately 17.65 grams.