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Answer: A square metal plate 0.180 m on each side is
Chapter 10, Problem 3E(choose chapter or problem)
A square metal plate \(0.180 \mathrm{~m}\) on each side is pivoted about an axis through point O at its center and perpendicular to the plate (Fig. E10.3). Calculate the net torque about this axis due to the three forces shown in the figure if the magnitudes of the forces Are \(F_{1}=18.0 \mathrm{~N}, F_{2}=26.0 \mathrm{~N}\) and \(F_{3}=14.0 \mathrm{~N}\). The plate and all forces are in the plane of the page.
Equation Transcription:
°
Text Transcription:
0.180 m
F_1=18.0 N, F_2=26.0 N,
F_3=14.0 N
45degree
Questions & Answers
QUESTION:
A square metal plate \(0.180 \mathrm{~m}\) on each side is pivoted about an axis through point O at its center and perpendicular to the plate (Fig. E10.3). Calculate the net torque about this axis due to the three forces shown in the figure if the magnitudes of the forces Are \(F_{1}=18.0 \mathrm{~N}, F_{2}=26.0 \mathrm{~N}\) and \(F_{3}=14.0 \mathrm{~N}\). The plate and all forces are in the plane of the page.
Equation Transcription:
°
Text Transcription:
0.180 m
F_1=18.0 N, F_2=26.0 N,
F_3=14.0 N
45degree
ANSWER:
Step 1 of 3
Before solving this question, let us recall that torque is taken as positive if the force acts counterclockwise and negative if the force acts clockwise.
Given, \(F_{1}=18.0 \mathrm{~N}\)
Its perpendicular distance from the point \(\mathrm{O}\) is \(=0.180 \mathrm{~m} / 2=0.09 \mathrm{~m}\)
But this force acts clockwise, so the torque will be negative.
Therefore, torque due to this force \({ }^{\tau_{1}}=-18.0 \mathrm{~N} \times 0.09 \mathrm{~m}=-1.62 \mathrm{~N} . \mathrm{m}\) Given, \(F_{2}=26.0 \mathrm{~N}\)