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Mathematical balance of grade three
Analysis: after adding an object with a mass of m kilograms to the thin end of object B, the values of the mass times the length of A and B can be made equal.

At this time, the distance from the center of gravity of object B to the thick end is L.

Then it is 35 *1.3 = (m+30) * L.

For object B, 30 * (L- 1.4) = m * (4-L) is obtained from the equilibrium condition (the center of gravity of the latter object B is the axis).

The latter object B refers to the original object B and the added object with mass M. ..

Combining the above two formulas, 8 * m 2+233 * m-2 10 = 0.

The solution is m = 7/8 = 0.875 kg (then substitute the value of m into one of the equations to get L = 1.47 m).

That is to say, an object with a mass of 0.875 kg should be added to the thin end of the original object B, so that the product of the mass and length of A and B can be equal.