A uniform metre rule is ab is pivoted 30cm from end a. It balances horizontally when a weight of 5n is hug at a.calculate the weight of the metre rule.
center of meter rule weight w at 50 cm
5 N at 0
(5+w) up at 30
moments about a (zero end)
30 (5+w) = 50 w
150 + 30 w = 50 w
20 w = 150
w = 7.5 Newtons
Well, well, well. Looks like we've got a balancing act going on here! So, let me get this straight. We've got a uniform meter rule, and it's all fancy with its pivot point at 30cm from end A. And somehow, magically, it manages to balance horizontally when a weight of 5N is hanging from point A. Impressive!
Now, you want to know the weight of the meter rule itself, right? This might sound like a riddle, but fear not, my friend. I've got an answer for you!
Since we're dealing with equilibrium here, we can use a little trick called the principle of moments. We can say that the clockwise moment created by the weight hanging at point A is equal to the anticlockwise moment created by the weight of the meter rule itself at point B.
So, let's do some math: The distance from point B to the pivot is (100cm - 30cm) = 70cm. Now, let's denote the weight of the meter rule as "W". Since the meter rule is balanced, the clockwise moment created by the weight at point A (5N) is equal to the anticlockwise moment created by the weight of the meter rule (W) at point B.
So, 5N x 30cm = W x 70cm
Now, let's do a little cross-multiplication and solve for W:
5N x 30cm = W x 70cm
150Ncm = 70W
W = 150/70 N
And there you have it! The weight of the meter rule is approximately 2.14 N. So, we've got a balancing act with a 5N weight on one side and a 2.14N weight on the other. It's all about equilibrium, my friend!
To solve this problem, we can use the concept of moments. Moment is a measure of the turning effect of a force.
Given:
Distance from pivot ab = 30 cm = 0.3 m
Weight hung at a = 5 N
Let's assume the weight of the meter rule is W. Since the rule balances horizontally, the clockwise moments are equal to the anticlockwise moments.
Clockwise moment = Weight x Distance
Anticlockwise moment = Weight of the meter rule x Distance
5 N x 0.3 m = W x (1 m - 0.3 m)
1.5 N·m = 0.7W N·m
Now, we can solve for the weight of the meter rule (W):
W = 1.5 N·m / 0.7 N·m
W = 2.14 N (rounded to two decimal places)
Therefore, the weight of the meter rule is approximately 2.14 N.
To calculate the weight of the meter rule, we need to consider the principle of moments. The principle of moments states that for an object to be in rotational equilibrium, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments.
In this case, the meter rule is balanced horizontally, meaning there is no net clockwise or anticlockwise moment. The weight of the meter rule and the weight hanging at the pivot point contribute to these moments.
Let's break down the problem step by step:
1. The pivot point is located 30 cm away from end A.
2. A weight of 5 N is hung at point A.
Now, let's calculate the weight of the meter rule using the principle of moments:
1. Consider the clockwise moments: The weight at point A (5 N) creates a clockwise moment. Since it is acting at a distance of 30 cm from the pivot point, the moment is given by: Moment_A = 5 N * 30 cm = 150 Ncm.
2. Consider the anticlockwise moments: The weight of the meter rule creates an anticlockwise moment. Let's call the weight of the meter rule "W." Since it is acting at a distance of 0 cm (at the pivot point), the moment is given by: Moment_W = W * 0 cm = 0 Ncm.
According to the principle of moments, the clockwise moments and anticlockwise moments must balance each other:
Moment_A = Moment_W
150 Ncm = 0 Ncm + W * 0 cm
Therefore, we can conclude that the weight of the meter rule is 0 N.
The weight of the meter rule is 0 N because it is balanced horizontally by the 5 N weight hung at point A.