If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m
Given the data in the question;
- Length of the massless beam;[tex]L = 4.00m[/tex]
- Distance of support from the left end; [tex]x = 3.00m[/tex]
- First mass; [tex]m1 = 31.3 kg[/tex]
- Distance of beam from the left end( m₁ is attached to ); [tex]x_1 = ?[/tex]
- Second mass; [tex]m_2 = 61.7 kg[/tex]
- Distance of beam from the right of the support( m₂ is attached to ); [tex]x_1 = 0.273m[/tex]
Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.
Hence, [tex]m_1g( x-x_1) = m_2gx_2[/tex]
we divide both sides by [tex]g[/tex]
[tex]m_1( x-x_1) = m_2x_2[/tex]
Next, we make [tex]x_1[/tex], the subject of the formula
[tex]x_1 = x - [ \frac{m_2x_2}{m_1} ][/tex]
We substitute in our given values
[tex]x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ][/tex]
[tex]x_1 = 3.00m - 0.538m[/tex]
[tex]x_1 = 2.46m[/tex]
Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m
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