Question 4.6: A man raises a 10-kg joist, of length 4 m, by pulling on a r......
A man raises a 10-kg joist, of length 4 m, by pulling on a rope. Find the tension T in the rope and the reaction at A.
Free-Body Diagram. The joist is a three-force body, since it is acted upon by three forces: its weight W, the force T exerted by the rope, and the reaction R of the ground at A. We note that
W=mg=(10\mathrm{~kg})(9.81\mathrm{~m/s^{2}})\,=\,98.1\,\,\,\mathrm{N}Three-Force Body. Since the joist is a three-force body, the forces acting on it must be concurrent. The reaction R, therefore, will pass through the point of intersection C of the lines of action of the weight W and the tension force T. This fact will be used to determine the angle α that R forms with the horizontal.
Drawing the vertical BF through B and the horizontal CD through C, we note that
A{ F}\,=\,B{F}\,=\,(A B)\,\cos\,45^{\circ}\,=\,(4\,\operatorname{m})\,\cos\,45^{\circ}\,=\,2.828\,\operatorname{m} \\ \\ C D\,=\,E F\,=\,A E\,=\,{\frac{1}{2}}(A F)\,=\,1.414\,\mathrm{m} \\ \\ B D\,=\,(C D)\;\mathrm{cot}\;(45^{\circ}\,+\,25^{\circ})\,=\,(1.414\;\mathrm{m})\;\mathrm{tan}\;20^{\circ}\,=\,0.515\;\mathrm{m} \\ \\ C E\,=\,D F\,=\,B F\,-\,B D\,=\,2.828\,\operatorname{m}\,-\,0.515\,\operatorname{m}\,=\,2.313\,\operatorname{m}We write
\tan~a={\frac{C E}{A E}}={\frac{2.313\mathrm{{\ m}}}{1.414\mathrm{{\ m}}}}=1.636α = 58.6°
We now know the direction of all the forces acting on the joist.
Force Triangle. A force triangle is drawn as shown, and its interior angles are computed from the known directions of the forces. Using the law of sines, we write
{\frac{T}{\sin31.4^{\circ}}}={\frac{R}{\sin110^{\circ}}}={\frac{98.1\,\mathrm{N}}{\sin38.6^{\circ}}}T = 81.9 N
R = 147.8 N a58.6°
