A laboratory balance essentially uses the method of moments to compare an unknow

Question

A laboratory balance essentially uses the method of moments to compare an unknown mass with a known mass. Some balances have constant and equal lever arms, and other do not. Figure 4(a) shown the balance used in our lab. Its sketch diagram is shown in Fig. 4(b), where m1 is the unknown mass, m2 is the known mass and mc is the mass of the metal scale bar. Explain how the condition Frightarrow = 0 is satisfied for the balance in Fig. 4 (b). Explain how the condition tau rightarrow = 0 is satisfied for the balance in Fig. 4 (b). The torques are calculated about the rotating axis through the pivot point O. Docs the individual torque depend on the position of the rotating axis? Does the second equilibrium condition ( tau rightarrow = 0) of a rigid body depend on the position of the rotating axis? Why?

Explanation / Answer

(A) There will be a normal force on the lever at the pivot O. in upward direction

N = m1g + m2g + mcg

=> m1g + m2g +m3g - N = 0

(B)

Since Moment is acting on Point O so r = 0 for normal

So   

r1*m1g = rc*mcg + r2*m2g

=>   r1*m1g - rc*mcg - r2*m2g = 0 ----------- eqn 2

2. No it does not.

Right as +ve and left as -ve

and upward as -ve & downward as +ve)

Taking torque about Any point A of Force Acting on Point X

then

AX = AO + OX

Torque about A

AM1*m1*g + AM2*m2g + AMC*(mcg) + AO*(-N)

=> (AO+r1)*m1g + (AO-r2)m2g + (AO-rc)*mcg - AO*N

= AO*(m1g+m2g+m3g-N) + (r1m1g-r2m2g -rcmcg)

Since from eqn 1 & 2

= AO*0 + 0 = 0

Proved

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