Question 3.3: A 30-lb force acts on the end of the 3-ft lever as shown. De......
A 30-lb force acts on the end of the 3-ft lever as shown. Determine the moment of the force about O.
The force is replaced by two components, one component P in the direction of OA and one component Q perpendicular to OA . Since O is on the line of action of P , the moment of P about O is zero and the moment of the 30-lb force reduces to the moment of Q , which is clockwise and, thus, is represented by a negative scalar.
Q\,=\,(30\,{\mathrm l b})\,\sin\,\ 20^{\circ}\,=\,10.26\,\,\mathrm{l b}\, \\ \\ \begin{array}{l}{{M_{O}=\ -Q(3\mathrm{~ft})=\ -(10.26\ \mathrm lb)(3\mathrm{~ft})=\ -30.8\ \mathrm lb\cdot\mathrm{~ft}}}\end{array}
Since the value obtained for the scalar { M}_{O} is negative, the moment { M}_{O} points into the paper. We write
{ M}_{O}\,=\,30.8\,{\ \mathrm {lb}}\,\cdot\,\mathrm{ft}\,\mathrm{i}\qquad \qquad
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