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aalyn [17]
3 years ago
14

A cat walks along a plank with mass M= 6.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a di

stance d1= 0.850 m to the left of sawhorse B. When the cat is a distance d2= 1.11 m to the right of sawhorse B, the plank just begins to tip.
If the cat has a mass of 2.9 kg, how far to the right of sawhorse B can it walk before the plank begins to tip?
Physics
1 answer:
dimulka [17.4K]3 years ago
6 0

Answer:

d₂ = 1.466 m

Explanation:

In this case we must use the rotational equilibrium equations

        Στ = 0

        τ = F r

we must set a reference system, we use with origin at the easel B and an axis parallel to the plank , we will use that the counterclockwise ratio is positive

      + W d₁ - w_cat d₂ = 0

      d₂ = W / w d₁

      d₂ = M /m d₁

      d₂ = 5.00 /2.9    0.850

      d₂ = 1.466 m

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b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

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Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

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the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

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