Question

A uniform metre rule of weight 10 gf is pivoted a
0 mark
(1) What moment of force depresses the rule ?​

Answer 1

Answer:

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

Step-by-step explanation:

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} = 500 gf\ cm.

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

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

Answer 2

Anticlockwise moment = 10 gf × 50 cm

= 500 gf cm

Anticlockwise moment = Clockwise moment

According to principle of moment:

10 gf × 50 cm = W × 100 cm

So, W = (10 gf × 50 cm) / 100 cm

= 5 gf

Rule can be made horizontal by applying a force 5 gf upwards at the 100 cm mark.