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A rod of negligible mass is pivoted at a point that is off-center, so that length ?1 is different from length ?2. The figures show two cases in which masses are suspended from the ends of the rod. In each case the unknown mass m is balanced by a known mass M1 or M2 so that the rod remains horizontal. TTL Mi e1 Tn

What is the value of m in terms of the known masses?

1. m = p M1 M2

2. m = M1 M2

3. m = M1 + M2 2

4. m = M1 M2 2

5. m = M1 + M2

Respuesta :

Answer:

[tex]m = \sqrt{M_1 M_2}[/tex]

Explanation:

Two casess are mentioned in the given figure

first case

two mass m and M_1

m is l_1 distance from center and M_1 is l_2 distance from center

second case

two mass m and M_2

m is l_2 distance from center and M_2 is l_1 distance from center

The balance in the  1st case requires [tex]m l_1 = M_1 l_2.[/tex]

And the balance in second case  

[tex]M_2 l_1 = m l_2.[/tex]

Cancel[tex] l_1 and l_2 [/tex]

from  the above equations.

So[tex] m^2 = M_1 M_2[/tex], i.e.

[tex]m = \sqrt{M_1 M_2}[/tex]

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