Question

A uniform metre rule of weight 10gf is pivoted at its 0 mark.

(i) What moment of force depresses the rule?

(ii) How can it be made horizontal by applying least force?

Answer: (i) 500gf cm (ii) By applying a force 5gf upward at the 100 cm mark.

Please solve this. Very very urgent. ​

Answer 1

Explanation:

(i) Anticlockwise moment= 10gf × 50 cm = 500gf

(ii) From the principle of moments,

Anticlockwise moment = Clockwise moment

10gf × 50 cm = W × 100cm

So, W = 10gf x 50cm/100cm = 5gf

By applying a force 5 gf upwards at the 100 cm mark, rule can be made horizontal

Hope it helps you.!!:)

BORAHAE!

Answer 2

Magnitude of force = 5 gf .

Given:

The given data suggests that the uniform meter rule of weight 10gf is pivoted at zero mark.

Solution:

So the moment of force is given by the product of magnitude of force and perpendicular distance corresponding to the force action line in relation to rotational axis.

Thereby the moment of force =$50 \mathrm{cm} \times 10 \mathrm{gf}$

And therefore the least force applied will have the perpendicular distance at the highest, therefore we have the distance of 100 cm mark highest where the force application be least.

$\begin{gathered}\begin{array}{l}{\text {Moment of force = Magnitude of force} \times \text { Mark distance }} \\ {500=\text { Force magnitude } \times 100}\end{array}\end{gathered}$

Magnitude of force = 5 gf which is to be applied upwards.