Question

Determine the mass of a metre rule using a spring balance or by balancing it on a knife edge

at some point away from the middle and a 50g weight on the other side. Next pivot (F) the

metre rule at the 40cm, 50cm and 60cm mark, 104 each time suspending a load L or the left

end and effort E near the right end. Adjust E and or its position so that the rule is balanced.

Tabulate the position of L, F and E and the magnitudes of L and E and the distances of load

arm and effort arm. Calculate MA=L/E and VR = effort arm/load arm. It will be found that MA

VR in the third case. Try to explain why this is so. Also try to calculate the real load and real

effort in these cases​

Answer 1

Answer:

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Explanation:

Let W and W  

′

 be the respective weights of the metre stick and the coin.

The mass of the metre stick is concentrated at its mid-point, i.e., at the 50 cm mark.

Mass of the metre stick=m  

′

 

Mass of each coin=5g

When the coins are placed 12cm away from the end P, the centre of mass gets shifted by 5cm from point R toward the end P. The centre of mass is located at a distance of 45cm from point P.

The net torque will be conserved for rotational equilibrium about point R.

10g(45−12)−m  

′

g(50−45)=0

Thus m  

′

=66g