A rock is attached to the left end of a uniform meter stick that has the same mass as the rock. How far from the left end of the stick should the triangular object be placed so that the combination of meter stick and rock is in balance

This is a static equilibrium exercise, to solve it we must fix a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive

∑ τ = 0

The mass of the rock is M and placed at a distance, L the mass of the rod M₁, is considered to be placed in its center of mass, which by uniform e is in its geometric center (x = 0) and the triangular mass M₂, with a distance L₂

The triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.

M L + M₁ 0 - m₂ L₂ = 0

M L - m₂ L₂ = 0

L₂ = [tex]\frac{M}{M_{2}}[/tex] L

From this answer we have several possibilities

* if the two masses are equal then L₂ = L

* If the masses are different, with M₂> M then L₂ = \frac{M}{M_{2}} L

snehashish65 snehashish65 · Answer 2 · 2021-11-17T10:50:44+01:00

The triangular object should be placed at the distance of [tex]\dfrac{M}{M_{2}} \times L[/tex] from the left end of the stick.

The given problem can be resolved using the concept of static equilibrium. We must consider a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive. So, static equilibrium says,

[tex]\sum \tau = 0[/tex]

Here, [tex]\tau[/tex] is the torque.

Let the mass of the rock is M and it is placed at a distance L. Such that the mass of the rod is [tex]M_{1}[/tex] which is considered to be placed in its center of mass. By uniformity in its geometric center (x = 0) and the triangular mass [tex]M_{2}[/tex] lies at the distance [tex]L_{2}[/tex].

Also, triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.

Then,

[tex]ML+M_{1} \times 0 -M_{2} \times L_{2}=0\\\\ML=M_{2}L_{2}\\\\L_{2}=\dfrac{M}{M_{2}} \times L[/tex]

Thus, we can conclude that triangular object should be placed at the distance of [tex]\dfrac{M}{M_{2}} \times L[/tex] from the left end of the stick.

Learn more about the equilibrium of forces here:

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