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Answer :
Final answer:
The upward force that the left sawhorse exerts on the board is 725.4 N.
Explanation:
The upward force that the left sawhorse exerts on the board is 725.4 N. To find the upward force that the left sawhorse exerts on the board, we need to analyze the forces acting on the system. The board weights 10.5 kg * 9.8 m/s2 = 102.9 N, which acts vertically downward. The weight of the saw is 4.45 kg * 9.8 m/s2 = 43.5 N, and it acts vertically downward at a distance of 1.80 m from the left end. The left sawhorse exerts an upward force to balance the weight of the board and the saw.
Using the principle of moments, we can set up the following equation:
(10.5 kg * 9.8 m/s2 + 4.45 kg * 9.8 m/s2) * (6.00 m - x) = (x) * (10.5 kg * 9.8 m/s2)
Simplifying the equation and solving for x, we get:
x = 70.245 m
Therefore, the upward force that the left sawhorse exerts on the board is:
F = (x) * (10.5 kg * 9.8 m/s2) = (70.245 m) * (10.5 kg * 9.8 m/s2) = 725.4 N
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Rewritten by : Barada
Final Answer:
b) 98.1 N
The left sawhorse exerts an upward force of 98.1 N, determined by balancing torques due to board and saw weights. Therefore the right answer is option is b).
Explanation:
The upward force exerted by the left sawhorse can be determined by considering the equilibrium of forces acting on the board. The sum of the torques about any point must be zero for rotational equilibrium. In this case, we can choose the left end of the board as the pivot point. The weight of the board, the weight of the saw, and the force from the left sawhorse contribute to the torques.
First, find the torque due to the weight of the board. The weight acts at the center, 3.00 m from the left end, giving a torque of [tex]\(τ_{\text{board}} = (10.5 \, \text{kg} \cdot 9.8 \, \text{m/s}^2) \cdot 3.00 \, \text{m}\).[/tex]
Next, calculate the torque due to the weight of the saw. The saw is 1.80 m from the left end, giving a torque of [tex]\(τ_{\text{saw}} = (4.45 \, \text{kg} \cdot 9.8 \, \text{m/s}^2) \cdot 1.80 \, \text{m}\).[/tex]
To find the upward force from the left sawhorse, set the sum of the torques equal to zero: [tex]\(τ_{\text{board}} + τ_{\text{saw}} - F_{\text{left}} \cdot 6.00 \, \text{m} = 0\).[/tex]
Solving for [tex]\(F_{\text{left}}\), we get \(F_{\text{left}} = (τ_{\text{board}} + τ_{\text{saw}}) / 6.00 \, \text{m}\).[/tex]
Substitute the known values and solve to find[tex]\(F_{\text{left}} \approx 98.1 \, \text{N}\).[/tex] Therefore, the correct answer is option b) 98.1 N.
