A uniform half metre rule is freely pivoted at 15cm mark and it balances horizontally when a body of mass 40g is hung from the 2cm mark. Draw a clear force diagram of the arrangement and calculate the mass of the rule
To draw a force diagram of the arrangement, we need to consider all the forces acting on the half metre rule. We can assume that the gravity acts vertically downwards.
Force diagram:
```
|\ |
| \ |
| \ 40g |
| \ |
--------------- |
| | |
| | |
| | |
| 15cm mark |
| |
--------------------- Rule
2cm mark
```
In this diagram, the 40g mass is represented as a force arrow pointing downwards from the 2cm mark on the ruler. The pivot point is represented as a dot at the 15cm mark. The rule itself is represented by the horizontal line.
To calculate the mass of the rule, we can use the principle of torque. Since the half metre rule balances horizontally, the sum of the clockwise torques must be equal to the sum of the counterclockwise torques.
The clockwise torque is given by the formula: Clockwise Torque = (Force of gravity on 40g mass) * (Distance from pivot)
The counterclockwise torque is given by the formula: Counterclockwise Torque = (Force of rule) * (Distance from pivot to the center of mass of the rule)
The torque due to the force of the rule is zero since it is balanced horizontally.
Let's assume the center of mass of the rule is at the 25cm mark.
Using the given information:
Clockwise Torque = (0.04 kg) * (9.8 m/s^2) * (0.02 m - 0.15 m)
Counterclockwise Torque = Mass of the rule * (9.8 m/s^2) * (0.25 m - 0.15 m)
Setting the torques equal to each other:
(0.04 kg) * (9.8 m/s^2) * (0.02 m - 0.15 m) = Mass of the rule * (9.8 m/s^2) * (0.25 m - 0.15 m)
Simplifying the equation:
(-0.13 kg*m/s^2) = Mass of the rule * (9.8 m/s^2) * (0.1 m)
Solving for the mass of the rule:
Mass of the rule = (-0.13 kg*m/s^2) / ((9.8 m/s^2) * (0.1 m))
Mass of the rule = -0.13 kg / 0.98 kg
Mass of the rule ≈ -0.1327 kg (assuming a positive mass)
Since mass cannot be negative, we can conclude that the mass of the rule is approximately 0.1327 kg.
To draw the force diagram, we need to consider the forces acting on the uniform half meter rule when it is balanced horizontally.
1. The weight of the half-meter rule acts downward at its center of gravity, which is at the 15 cm mark.
2. There is an upward force exerted at the pivot point of the rule, which allows it to balance horizontally.
3. The weight of the 40 g mass hanging from the 2 cm mark acts downward at that point.
Here's a simplified representation of the force diagram:
|
|
| F <-- Pivot force
|
<-------------------------|----------> 15 cm
|
|
|
| W <------- Weight of rule
|
|
| F <-- Weight of 40 g mass
|
<-------------------------|----------> 2 cm
|
To calculate the mass of the rule, we can use the principle of moments. The equation for the principle of moments is:
Sum of clockwise moments = Sum of anticlockwise moments
Let's consider the pivot point as the stationary point for calculating moments. The distance of the rule's center of gravity from the pivot point is 15 cm.
Anti-clockwise moments:
Weight of the rule (W) * distance from the pivot point (15 cm)
Clockwise moments:
Weight of the 40g mass * distance from the pivot point (13 cm)
Since the rule is balanced, the sum of the clockwise moments will be equal to the sum of the anti-clockwise moments:
(W * 15 cm) = (40g * 13 cm)
To calculate the mass of the rule (W) in grams, we can multiply both sides by 1g (since 1g = 1g mass):
(W * 15 cm) = (40g * 13 cm)
W = (40g * 13 cm) / 15 cm
W = 34.66g
Therefore, the mass of the rule is approximately 34.66 grams.