Math  /  Geometry

QuestionGiven L=2000 mm,S=4400 mmL=\mathbf{2 0 0 0 ~ m m}, S=\mathbf{4 4 0 0} \mathbf{~ m m}, and ψ=5\psi=5^{\circ}, determine the moment of the force F=2.7NF=\mathbf{2 . 7} \mathbf{N} applied at an angle of θ=42\theta=\mathbf{4 2}{ }^{\circ} about point OO. Neglect the thickness of the member.
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Studdy Solution
Convert the distance from millimeters to meters for consistency in units:
d=2647.92mm=2.64792m d = 2647.92 \, \text{mm} = 2.64792 \, \text{m}
Now calculate the moment:
M=2.7N2.64792m7.149384Nm M = 2.7 \, \text{N} \cdot 2.64792 \, \text{m} \approx 7.149384 \, \text{Nm}
The moment of the force F F applied at an angle θ=42 \theta = 42^\circ about point O O is approximately:
7.15Nm \boxed{7.15 \, \text{Nm}}

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