Physics
A uniform metre rule is pivoted at its mid-point. A weight of 50 gf is suspended at one end of it. Where should a weight of 100gf be suspended to keep the rule horizontal?
Force
164 Likes
Answer
Let us assume that a 50 gf weight produces an anticlockwise moment about the middle point ( 50 cm ).
Now, if a weight of 100 gf produces a clockwise moment about the middle point and d cm be the distance from the middle.
As we know, the principle of moments states that
Anticlockwise moment = Clockwise moment.
Therefore, a weight of 100gf will be suspended at a distance of 25cm to keep the ruler balanced.
Answered By
107 Likes
Related Questions
Figure shows two forces each of magnitude 10N acting at the points A and B at a separation of 50 cm, in opposite directions. Calculate the resultant moment of the two forces about the point (i) A, (ii) B and (iii) O, situated exactly at the middle of the two forces.
The diagram shows a uniform bar supported at the middle point O. A weight of 40 gf is placed at a distance 40cm to the left of the point O. How can you balance the bar with a weight of 80 gf?
A uniform metre rule balances horizontally on a knife edge placed at the 58 cm mark when a weight of 20gf is suspended from one end. (i) Draw a diagram of the arrangement. (ii) What is the weight of the rule?
A steering wheel of diameter 0.5m is rotated anti-clockwise by applying two forces each of magnitude 6N. Draw a diagram to show the application of forces and calculate the moment of forces applied.