A uniform metre rule of mass 120g is pivoted at 60cm Mark.At what point should a mass of 50g be suspended for it to balance horizontally
To find the point at which a mass of 50g should be suspended on the uniform metre rule for it to balance horizontally, we can use the principle of moments.
The principle of moments states that for an object to be in equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point.
In this case, the metre rule is balanced horizontally, so the sum of the clockwise moments must be equal to the sum of the anticlockwise moments.
First, let's label the distances on the metre rule. The pivot point is at the 60cm mark, which means the distance from the pivot to the 60cm mark is 60cm. We also know that the mass of the metre rule is 120g and the mass we need to suspend is 50g.
Let's assume that the distance from the pivot to the point where the 50g mass is suspended is x cm.
The clockwise moment is given by:
Clockwise moment = Mass x Distance
For the 120g metre rule:
Clockwise moment of the metre rule = 120g x 60cm
For the 50g mass:
Clockwise moment of the 50g mass = 50g x x cm
The anticlockwise moment is given by:
Anticlockwise moment = Mass x Distance
For the 120g metre rule:
Anticlockwise moment of the metre rule = 120g x (100cm - 60cm)
Now we can set up the equation using the principle of moments:
Clockwise moments = Anticlockwise moments
120g x 60cm = 50g x x cm + 120g x (100cm - 60cm)
We can simplify the equation:
7200 = 50x + 4800
Rearranging the equation:
50x = 2400
Now we can solve for x:
x = 2400 / 50
x = 48 cm
Therefore, the mass of 50g should be suspended at the 48cm mark on the metre rule for it to balance horizontally.
balance the torques. Each part of the rule may be considered a point mass halfway between the fulcrum and the end of the rule.
60*30 = 40*20 + 50x