Question;weights,faulty balance

[konichiwa2x]

Hi,

A body weighs 8gm when placed in one pan and 18gm when placed on the other pan of a false balkance. If the beam is horizontal when both the pans are empty, then what is the true weight of the body?

How do I do this? Please help.

 

[Andrew Mason]

Hi,

A body weighs 8gm when placed in one pan and 18gm when placed on the other pan of a false balkance. If the beam is horizontal when both the pans are empty, then what is the true weight of the body?

How do I do this? Please help.

Let the pans have mass m1 and m2. Let the body have a mass of x.

When the body is placed on pan1 one has to add 8 g. to pan 2 to balance it. When the body is placed on pan2 one has to add 18 g. to pan 1 to balance it.

I would suggest that you write out the relationship between m1, m2, x and the added masses in each weighing and solve. You won't solve for m1 and m2 but you should be able to solve for m1-m2 and x.

AM

 

[konichiwa2x]

If the beam is horizontal when both the pans are empty, doesn't that mean both the pans have equal mass?

Anway, is this what you meant?
m1 = km2
m1 + x = m2 + 8
m2 + 18 = m2 + x

solving, x = 13? Is this correct?

My book says the correct answer is 12gm..

 

[Andrew Mason]

If the beam is horizontal when both the pans are empty, doesn't that mean both the pans have equal mass?

Not quite, because it is a false balance. But I think I may have misled you a little on my first answer.

A balance works by equalizing torque. You can say that m1r1 = m2r2 where r1 and r2 are the distances of m1 and m2 from the fulcrum. If r1 and r2 are equal, the balance balances only if m1 = m2. We are told that this is not the case here. So all we can say is that:

m1r1 = m2r2

Therefore,

(m1 + x)r1 = (m2 + 8)r2

(m1 + 18)r1 = (m2 + x)r2

Work that out for x. The answer in the book is correct.

AM

 

[konichiwa2x]

ok thanks a lot!