A uniform metre rule of mass 120g is pivoted at the 60cm mark .at what pont on the meter rule should a mass of 50g be suspended for it to balance horizontal
l dont know
the long side of the rule has 72g, the short side 48g
So, you need
72*30 = 48*20 + 50x
x = 24
so, the mass must be at the 84cm mark
84
To find the point on the meter rule where a mass of 50g should be suspended for it to balance horizontally, we need to consider 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 equal the sum of the anticlockwise moments about the same point.
In this case, let's take the pivot point at the 60cm mark as our reference point.
The moment (M) of a force about a point is given by the equation M = F × d, where F is the force and d is the perpendicular distance from the force to the reference point.
The 120g meter rule is already balanced at the 60cm mark, so the anticlockwise moment produced by the 120g mass is equal to the clockwise moment produced by the 120g mass:
(120g) × (60cm) = (50g) × (x cm)
Simplifying the equation:
(120g) × (60cm) = (50g) × (x cm)
7200 = 50x
Now we can solve for x:
x = 7200 / 50
x = 144 cm
Therefore, a mass of 50g should be suspended at the 144cm mark on the meter rule to balance it horizontally.