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The upward force exerted by the left sawhorse is approximately [tex]\(51.9 \, \text{N}\).[/tex]
To find the upward force exerted by the left sawhorse, we need to consider the torques and forces acting on the board. The torque due to the board's weight is countered by the torque from the saw and the left sawhorse.
The torque ([tex]\( \tau \)[/tex]) is calculated as [tex]\( \tau = r \times F \)[/tex], where r is the lever arm and F is the force.
The weight of the board ([tex]\( W_{\text{board}} \)[/tex]) acts at its center and produces a torque. The saw and the left sawhorse also contribute torques. The sum of these torques must be zero for equilibrium.
[tex]\[ \text{Torque}_{\text{board}} = \text{Torque}_{\text{saw}} + \text{Torque}_{\text{sawhorse}} \][/tex]
[tex]\[ (m_{\text{board}} \times g \times \frac{L_{\text{board}}}{2}) = (m_{\text{saw}} \times g \times d_{\text{saw}}) + (F_{\text{sawhorse}} \times L_{\text{board}}) \][/tex]
Solving for [tex]\( F_{\text{sawhorse}} \):[/tex]
[tex]\[ F_{\text{sawhorse}} = \frac{m_{\text{board}} \times g \times \frac{L_{\text{board}}}{2} - m_{\text{saw}} \times g \times d_{\text{saw}}}{L_{\text{board}}} \][/tex]
Substitute the given values:
[tex]\[ F_{\text{sawhorse}} = \frac{(10.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \frac{6.00 \, \text{m}}{2}) - (4.45 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 1.80 \, \text{m})}{6.00 \, \text{m}} \][/tex]
[tex]\[ F_{\text{sawhorse}} \approx 51.9 \, \text{N} \][/tex]
The upward force exerted by the left sawhorse is approximately [tex]\(51.9 \, \text{N}\).[/tex]
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Rewritten by : Barada
