Physics
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?
Force
107 Likes
Answer
From the given figure we can see that,
Anticlockwise moment = 40 gf × 40 cm
Clockwise moment = 80 gf × d cm
As we know, the principle of moments states that
Anticlockwise moment = Clockwise moment.
Therefore the bar will be in balanced position if weight of 80gf is placed at a point of 20cm to the right of O.
Answered By
61 Likes
Related Questions
When a boy weighing 20 kgf sits at one end of a 4m long see-saw, it gets depressed at its end. How can it be brought to the horizontal position by a man weighing 40 kgf.
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?
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?
Figure shows a uniform metre rule placed on a fulcrum at its mid-point O and having a weight 40gf at the 10 cm mark and a weight of 20 gf at the 90 cm mark.
(i) Is the metre rule in equilibrium? If not, how will the rule turn?
(ii) How can the rule be brought in equilibrium by using an additional weight of 40gf?