Question

A uniform metre rule placed on a fulcrum at its midpoint O &having a weight 40gf at 10 cm mark & a weight of 20 gf at the 90cm mark. How to bring a rule in equilibrium by using weight of 40gf

Answer 1

Balance the torque
40*40 = 40*20 + 40*X
X = 20
Hence keep the weight at 70cm mark to make rotational equilibrium

Answer 2

Answer:

The weight of 40gf must be placed at 70 cm mark so that the anticlockwise moments and  clockwise moments are equal and the scale is in equilibrium.

Explanation:

We first draw a diagram for the given question: (refer attachment)

We will now calculate the anticlockwise and clockwise moments:

Anti-clockwise moment= Force×Perpendicular Distance  

=40gf×(50-10)cm= 1600gfcm  

Clockwise moment= Force×Perpendicular Distance  

=20gf×(90-50)dm=800gfcm  

According to the Principle of Moments, for a balance to be in equilibrium, the anti-clockwise  

moments and the clockwise moments must be equal.  

To bring the scale to balance using a weight of 40gf, we must place the weight on the right  

side of the balance. This is because the clockwise moments are less than the anti-clockwise  

moments.  

We know that according to the Principle of Moments,  

Anti-clockwise Moments= Clock-wise Moments

Let the 40gf weight be placed at a distance x cm from the middle O.

Anticlockwise moments= 40gf×10cm

Clockwise Moments= 20gf×40cm+ 40gf×xcm  

40gf×10cm= 20gf×40cm+40gf×x cm  

1600gfcm=800gfcm+40gf×x cm  

1600gfcm-800gfcm= 40gf×x cm  

800gfcm= 40gf×x cm  

800gfcm/40gf= xcm  

x= 20cm  

So the weight of 40gf must be placed at 70 cm mark so that the anticlockwise moments and  clockwise moments are equal and the scale is in equilibrium.