A uniform meter scale is suspended from 40 cm mark when weight of 25 gf and 10 gf are suspended at 5 cm mark and 75cm mark respectively. Calculate the weight of the meter scale.
52.5gf
W*50 + 25*5 + 10*75 = (W+25+10)40
10 W = 525
W = 52.5 gf
whatever a gf is :)
To calculate the weight of the meter scale, we can use the principle of moments. The principle states that the sum of the moments acting clockwise must be equal to the sum of the moments acting counterclockwise.
Let's assume the weight of the meter scale is W gf.
Clockwise moments:
Moment due to the weight of 25 gf at 5 cm mark: 25 gf × 5 cm = 125 gf·cm.
Moment due to the weight of 10 gf at 75 cm mark: 10 gf × 75 cm = 750 gf·cm.
Counterclockwise moments:
Moment due to the weight of the meter scale at 40 cm mark: W gf × 40 cm = 40W gf·cm.
According to the principle of moments, the sum of clockwise moments is equal to the sum of counterclockwise moments:
125 gf·cm + 750 gf·cm = 40W gf·cm.
Simplifying the equation:
875 gf·cm = 40W gf·cm.
Divide both sides of the equation by 40 cm:
875 gf·cm ÷ 40 cm = W gf.
Calculating W:
21.875 gf = W.
Therefore, the weight of the meter scale is approximately 21.875 gf.
To calculate the weight of the meter scale, we need to analyze the equilibrium of the system.
Let's assume the weight of the meter scale is "W" gf.
The weight of the 25 gf object acts downward at the 5 cm mark.
The weight of the 10 gf object acts downward at the 75 cm mark.
To maintain equilibrium, the sum of the clockwise moments (M) must be equal to the sum of the anti-clockwise moments. The formula for calculating the moment is given as:
Moment (M) = Force (F) x Perpendicular distance from the pivot (d)
The pivot in this case is the 40 cm mark.
Now, we can find the moments generated by the 25 gf and 10 gf weights:
Moment of the 25 gf weight (M1) = 25 gf x (40 cm - 5 cm) = 25 gf x 35 cm
Moment of the 10 gf weight (M2) = 10 gf x (75 cm - 40 cm) = 10 gf x 35 cm
Since the moments are proportional to the weights, we can write the equation:
M1 = M2
25 gf x 35 cm = 10 gf x 35 cm
And solving this equation, we get:
875 gf cm = 350 gf cm
Now, we add the moment generated by the weight of the meter scale (M3) at the 40 cm mark.
M3 = W gf x (40 cm - 40 cm) = 0
Therefore, the equation becomes:
875 gf cm + 0 = 350 gf cm
This is not possible. The equation is not balanced, which means there is an error in the given data or the question itself. Please double-check the question or provide additional information if available.