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3 years ago

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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?

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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• $\mathbf s_3(y,z)=y\,\mathbf j+z\,\mathbf k$ with $0\le y\le 6$ and $0\le z\le2$ ;

• $\mathbf s_4(u,v)=u\cos v\,\mathbf i+(6-u\cos v)\,\mathbf j+u\sin v\,\mathbf k$ with $0\le u\le2$ and $0\le v\le\frac\pi2$ ; and

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$\mathbf n_5=\dfrac{\partial\mathbf s_5}{\partial y}\times\dfrac{\partial\mathbf s_5}{\partial u}=2\cos u\,\mathbf i+2\sin u\,\mathbf k$

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