A uniform meter rule of mass 100g balance at a 40cm mark when a mass x is placed at the 10cm mark.What is x?
x is 30 cm from the fulcrum, and the center of mass of the ruler is 10 cm on the other side of it (in the middle of the ruler). The two moments balance. Therefore
x*30 = 100*10
x = 33.3 g
X is 30 cm from the pivot, which is at 40cm. The centre of mass is at 50 cm, so the pivot is 10 cm from the centre of mass. Therefore,
f x d(from pivot)= f x d(from c. of m.)
x*30=100*10
x=33.3g
Be careful to keep units correct, and JIHAD
Because I have to be there at work today but you can come over and play nice how was your weekend
AB is a uniform meter rule pivot at point F with a mass of 35g at A Calculate the mass of the metre rule
I need answer
Good
To solve this problem, you can use the principle of moments, which is based on the fact that the object is in equilibrium. The principle of moments states that the sum of the clockwise moments is equal to the sum of the anticlockwise moments.
In this case, the length of the meter rule is 100 cm (1 meter), and the mass of the meter rule is 100g. The balance point is 40 cm from the zero mark.
First, let's calculate the moment due to the meter rule itself. The moment is defined as the product of the force and the perpendicular distance from the pivot point. Since the balance point is 40 cm from the zero mark, the force acting on the meter rule can be considered to act at this point. Therefore, the moment due to the meter rule is:
Moment of the meter rule = mass of the meter rule x gravitational acceleration x distance from the pivot point
= 100g x 9.8 m/s² x 40cm
= 100g x 9.8 m/s² x 0.4m
= 392 g.m/s²
To maintain equilibrium, the sum of the clockwise and anticlockwise moments must be zero. The clockwise moment is the moment due to the mass x at the 10 cm mark, and the anticlockwise moment is the moment due to the meter rule itself.
Let x be the mass placed at the 10 cm mark. The distance from the pivot point to the mass x is 30 cm (40 cm - 10 cm), and the gravitational acceleration is 9.8 m/s².
Therefore, the moment due to the mass x is:
Moment of the mass x = mass of x x gravitational acceleration x distance from the pivot point
= x x 9.8 m/s² x 30 cm
= x x 9.8 m/s² x 0.3 m
= 2.94 x g.m/s²
To maintain equilibrium, the clockwise moment due to the mass x must equal the anticlockwise moment due to the meter rule:
2.94 x g.m/s² = 392 g.m/s²
To solve for x, divide both sides of the equation by 2.94:
x = 392 g.m/s² / 2.94 g.m/s²
x ≈ 133.33 g
Therefore, the mass x placed at the 10 cm mark is approximately 133.33 grams.