Question 4.SP.3: A loading car is at rest on a track forming an angle of 25° ......
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 acts 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.
STRATEGY: Draw a free-body diagram of the car to determine the unknown forces, and write equilibrium equations to find their values, summing moments at A and B and then summing forces.
MODELING:
Free-Body Diagram. The reaction at each wheel is perpendicular to the track, and the tension force T is parallel to the track. Therefore, for convenience, choose the x axis parallel to the track and the y axis perpendicular to the track (Fig. 1). Then, resolve the 5500-lb weight into x and y components.
W_{x} = +(5500 lb) cos 25° = +4980 lb
W_{y} = −(5500 lb) sin 25° = −2320 lb
ANALYSIS:
Equilibrium Equations. Take moments about A to eliminate T and R_{1} from the computation.
+↺ΣM_{A} = 0\colon −(2320 lb)(25 in.) − (4980 lb)(6 in.) + R_{2}(50 in.) = 0
R_{2}= +1758 lb R_{2}= 1758 lb ◂
Then, take moments about B to eliminate T and R_{2} from the computation.
+↺ΣM_{B} = 0\colon (2320 lb)(25 in.) − (4980 lb)(6 in.) − R_{1} (50 in.) = 0
R_{1}= +562 lb R_{1} = +562 lb ◂
Determine the value of T by summing forces in the x direction.
+ΣF_{x} = 0\colon +4980 lb − T = 0
T = +4980 lb T = 4980 lb ◂
Fig. 2 shows the computed values of the reactions.
REFLECT and THINK: You can verify the computations by summing forces in the y direction.
+ΣF_{y} = +562 lb + 1758 lb − 2320 lb = 0
You could also check the solution by computing moments about any point other than A or B.
