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Snowcat [4.5K]
3 years ago
7

A 15kg beam that is 10m long is placed on a fulcrum that is 3m from the end an 80kg person sits at the end closer to the fulcrum

what does the mass of an object plaved at the opposite end need to be in order to balance the beam?
Physics
1 answer:
Andrei [34K]3 years ago
7 0

Answer:

 m₃ = 30 kg

Explanation:

This is a problem of rotational equilibrium, let's write Newton's law for rotational equilibrium.

Let's fix our reference system in the support, the positive torques are those that create an anti-clockwise turn

Let's look for the distances to the point of support

The distance of man

      x₁ = -3 m

The distance of the bar is

      x₂ = L / 2 -3

     x₂ = 10/2 -3

     x₂ = 2 m

Remote object at the end

      x₃ = L-3

     x₃ = 10-3

     x₃ = 7 m

They give us the mass of man (m1) and the mass of the bar (m2)

Let's write the torques

      W₁ x₁ - W₂ x₂ - w₃ x₃ = 0

       m₁ g 3 - m₂ g 2 - m₃ g 7 = 0

       m₃ = (3m₁ -2m₂) / 7

Let's calculate

      m₃ = (3 80 -2 15) / 7

     m₃ = 30 kg

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Answer:

the canoe moved 1.2234 m in the water

Explanation:

Given that;

A man whose mass = 69 kg

A woman whose mass = 52 kg

at opposite ends of a canoe 5 m long, whose mass is 20 kg

now let;

x1 = position of the man

x2 = position of canoe

x3 = position of the woman

Now,

Centre of mass = [m1x1 + m2x2 + m3x3] / m1 + m2 + m3

= ( 69×0 ) + ( 52×5) + ( 20× 5/2) / 69 + 52 + 20

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= 0.9219858 m  { away from the man }

To get how far, the canoe moved;

⇒ 2.5 + 0.9219858 - 2.19858

= 1.2234 m

Therefore, the canoe moved 1.2234 m in the water

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