A loading car is at rest on a track forming an angle of 25° with the vertical. The gross weight of the car and its load is 5500 lb, and it is applied at a point 30 in. from the track, halfway between the two axles. The car is held by a cable attached 24 in. from the track. Determine the tension in the cable and the reaction at each pair of wheels.
Free-Body Diagram. A free-body diagram of the car is drawn. The reaction at each wheel is perpendicular to the track, and the tension force T is parallel to the track. For convenience, we choose the x axis parallel to the track and the y axis perpendicular to the track. The 5500-lb weight is then resolved into x and y components.
W_{x}\,=\,+(5500\;\mathrm{lb})\;\mathrm{cos}\;25^{\circ}\,=\,+4980\;\mathrm{lb}\\ \\ W_{y}\,=\,-(5500\;\mathrm{lb})\;\mathrm{sin}\;25^{\circ}\,=\,-2320\;\mathrm{lb}Equilibrium Equations. We take moments about A to eliminate T and R_1 from the computation.
+\mathrm l\Sigma M_{A}=0:\qquad-(2320\ \mathrm{lb})(25\ \mathrm{in.})\ -\ (4980\ \mathrm{lb})(6\ \mathrm{in.})\ +\ R_{2}(50\ \mathrm{in.})\,=\,0 \\ \\ \qquad\qquad\qquad R_{2}\,=\,+1758\;\mathrm{lb}\qquad\qquad\qquad\qquad\qquad\qquad R_{2}\,=\,1758\;\mathrm{lb}\nearrowNow, taking moments about B to eliminate T and R_2 from the computation, we write
+\mathrm l\Sigma M_{B}=0:\qquad (2320\ \mathrm{lb})(25\ \mathrm{in.})\ -\ (4980\ \mathrm{lb})(6\ \mathrm{in.})\ -\ R_{1}(50\ \mathrm{in.})\,=\,0 \\ \\ \qquad\qquad\qquad R_{1}\,=\,+562\;\mathrm{lb}\qquad\qquad\qquad\qquad\qquad\qquad R_{1}\,=\,+562\;\mathrm{lb}\nearrowThe value of T is found by writing
\searrow +\Sigma F_{x}\,=\,0\colon\qquad+4980\;\mathrm{lb}\,-\,T=\,0 \\ \\\qquad\qquad\qquad\qquad T=\ +4980\,\mathrm{lb} \qquad\qquad\qquad\qquad T=\ 4980\,\mathrm{lb}\nwarrowThe computed values of the reactions are shown in the adjacent sketch.
Check. The computations are verified by writing
The solution could also have been checked by computing moments about any point other than A or B.