A uniform metre rule of 80kg is pivoted at the 30cm mark.find the mass that would be placed at the 5cm mark for the metre rule to be in equilibrium
Let's first draw a diagram to help us visualize the problem:
```
<----30 cm----><----50 cm----><----20 cm---->
| |
0 100 cm
|_____________________________________________|
```
The pivot is at the 30 cm mark, which means that the weight of the metre rule (80kg) is acting downwards at that point. We need to find the mass that would be placed at the 5 cm mark such that the metre rule is in equilibrium (i.e. not tipping over in either direction).
To do this, we can use the principle of moments. The moment of a force about a point is equal to the force multiplied by the perpendicular distance from the point to the line of action of the force. In this case, the weight of the metre rule (W = 80kg x 9.81 m/s^2 = 784.8 N) is acting downwards at a distance of 30 cm from the pivot. Therefore, the moment of the weight about the pivot is:
M(W) = W x d = 784.8 N x 0.3 m = 235.44 Nm
To counteract this moment and keep the metre rule in equilibrium, we need to place a mass (m) at the 5 cm mark that exerts a moment in the opposite direction. The distance from the pivot to this mass is 25 cm (30 cm - 5 cm = 25 cm).
The moment due to this mass is:
M(m) = m x g x d' = m x 9.81 m/s^2 x 0.25 m
Setting M(W) = -M(m) (since the moments must balance each other out), we get:
235.44 Nm = -m x 9.81 m/s^2 x 0.25 m
m = -235.44 Nm / (-9.81 m/s^2 x 0.25 m) ≈ 96.1 kg
Therefore, a mass of approximately 96.1 kg placed at the 5 cm mark would keep the metre rule in equilibrium. Note that the negative sign in the equation just indicates that the moment due to the mass is in the opposite direction to the moment due to the weight of the metre rule.
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