Resultant of Parallel Force SystemSPONSORED LINKSCoplanar Parallel Force SystemParallel forces can be in the same or in opposite directions. The sign of the direction can be chosen arbitrarily, meaning, taking one direction as positive makes the opposite direction negative. The complete definition of the resultant is according to its magnitude, direction, and line of action.
Resultant of Distributed LoadsThe resultant of a distributed load is equal to the area of the load diagram. It is acting at the centroid of that area as indicated. The figure below shows the three common distributed loads namely; rectangular load, triangular load, and trapezoidal load.
Rectangular Load
Triangular Load
Trapezoidal Load
Spatial Parallel Force SystemThe resultant of parallel forces in space will act at the point where it will create equivalent translational and rotational (moment) effects in the system.
In vector notation, the resultant of forces are as follows...
Note:Two parallel forces that are equal in magnitude, opposite in direction, and not colinear will create a rotation effect. This type of pair is called aCouple. The placement of a couple in the plane is immaterial, meaning, its rotational effect to the body is not a function of its placement. The magnitude of the couple is given by
Where F = the magnitude of the two equal opposing forces and d is the perpendicular distance between these forces.- See more at: http://www.mathalino.com/reviewer/engineering-mechanics/resultant-of-parallel-force-system#sthash.grtqP3rV.dpuf
Problem 236A parallel force system acts on the lever shown in Fig. P-236. Determine the magnitude and position of the resultant.
Solution 236HideClick here to show or hide the solution
downward
clockwise
to the right of AThus, R = 110 lb downward at 6 ft to the right of A. answer- See more at: http://www.mathalino.com/reviewer/engineering-mechanics/236-computation-resultant-parallel-forces-acting-lever#sthash.YeeCbyMT.dpuf
Problem 237Determine the resultant of the four parallel forces acting on the rocker arm of Fig. P-237.
Solution 237HideClick here to show or hide the solution
downward
clockwise
to the right of OThus, R = 50 lb downward at 4 ft to the right of point O. answer- See more at: http://www.mathalino.com/reviewer/engineering-mechanics/237-finding-resultant-parallel-forces-acting-both-sides-rocker-arm#sthash.fmVyFjgk.dpufProblem 238The beam AB in Fig. P-238 supports a load which varies an intensity of 220 N/m to 890 N/m. Calculate the magnitude and position of the resultant load.
Solution 238HideClick here to show or hide the solution
Thus, R = 3330 N downward at 3.6 m to the left of A. answer- See more at: http://www.mathalino.com/reviewer/engineering-mechanics/238-finding-resultant-trapezoidal-loading#sthash.S1MfF2sj.dpufProblem 239The 16-ft wing of an airplane is subjected to a lift which varies from zero at the tip to 360 lb per ft at the fuselage according to w = 90x1/2lb per ft where x is measured from the tip. Compute the resultant and its location from the wing tip.Solution 239HideClick here to show or hide the solution
upward
Thus, R = 3840 lb upward at 9.6 ft from the tip of the wing. answer- See more at: http://www.mathalino.com/reviewer/engineering-mechanics/239-resultant-lift-wing-airplane#sthash.KXyUte8r.dpuf
Problem 240The shaded area in Fig P-240 represents a steel plate of uniform thickness. A hole of 4-in. diameter has been cut in the plate. Locate the center of gravity the plate.Hint:The weight of the plate is equivalent to the weight of the original plate minus the weight of material cut away. Represent the original plate weight of plate by a downward force acting at the center of the 10 14 in. rectangle. Represent the weight of the material cut away by an upward force acting at the center of the circle. Locate the position of the resultant of these two forces with respect to the left edge and bottom of the plate.
Solution 240HideClick here to show or hide the solution
Thus, the centroid is located at 6.8 in. to the right of left edge and 4.9 in. above the bottom edge. answer- See more at: http://www.mathalino.com/reviewer/engineering-mechanics/240-how-locate-centroid-metal-plate-circular-hole#sthash.xTJrwNUf.dpufProblem 241Locate the amount and position of the resultant of the loads acting on the Fink truss in Fig. P-241.
Solution 241HideClick here to show or hide the solution
Magnitude of resultant
downwardLocation of resultant
to the right of AThus, R = 15 130 N downward at 3.62 m to the right of left support. answer- See more at: http://www.mathalino.com/reviewer/engineering-mechanics/241-finding-resulatnt-vertical-forces-acting-fink-truss#sthash.Y4sU3ktx.dpuf
Problem 242Find the value of P and F so that the four forces shown in Fig. P-242 produce an upward resultant of 300 lb acting at 4 ft from the left end of the bar.
Solution 242HideClick here to show or hide the solutionSum of vertical forces
Moment about point A
answer
answer- See more at: http://www.mathalino.com/reviewer/engineering-mechanics/242-finding-unknown-two-forces-given-resultant#sthash.pxkhj8eL.dpuf
Problem 243The resultant of three parallel loads (one is missing in Fig. P-243) is 13.6 kg acting up at 3 m to the right of A. Compute the magnitude and position of the missing load.
Solution 243HideClick here to show or hide the solutionSum of vertical forces
downward
Moment about point A
Thus, F = 31.4 kg downward at 2.48 m to the right of left support. answer- See more at: http://www.mathalino.com/reviewer/engineering-mechanics/243-finding-magnitude-and-position-missing-force#sthash.n8UhvewY.dpuf
