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Resultant of Concurrent Force System
Resultant of a force system is a force or a couple that will have the same effect to the body, both in translation and rotation, if all the forces are removed and replaced by the resultant. The equation involving the resultant of force system are the following 1.
2.
3.
The x-component of the resultant is equal to the summation of forces in the xdirection.
The y-component of the resultant is equal to the summation of forces in the ydirection.
The z-component of the resultant is equal to the summation of forces in the zdirection.
Note that according to the type of force system, one or two or three of the equations above will be used in finding the resultant. Resultant of Coplanar Concurrent Force System
The line of action of each forces in coplanar concurrent force system are on the same plane. All of these forces meet at a common point, thus concurrent. In x-y plane, the resultant can be found by the following formulas:
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Resultant of Spatial Concurrent Force System
Spatial concurrent forces (forces in 3-dimensional space) meet at a common point but do not lie in a single plane. The resultant can be found as follows:
Direction Cosines
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Vector Notation of the Resultant
Where
Sample Problems Three vectors A, B, and C are shown in the figure below. Find one vector (magnitude and direction) that will have the same effect as the three vectors shown in Fig. P-013 below.
Solution
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Problem 014
From Fig. P-014, P is directed at an angle α from x-axis and the 200 N force is acting at a slope of 5 vertical to 12 horizontal. a. Find P and α if the resultant is 500 N to the right along the x-axis. b. Find P and α if the resultant is 500 N upward to the right with a slope of 3 horizontal to 4 vertical. c. Find P and α if the resultant is zero.
Solution 014 Part a: The resultant is 500N to the right along the x-axis
By Cosine law of the shaded triangle
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By Sine law
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Part b: The resultant is 500 N upward to the right with a slope of 3 horizontal to 4 vertical
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answer
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Part c: The resultant is zero
The resultant is zero if P and the 200 N force are equal in magnitude, oppositely directed, and collinear.
Thus, P = 200 N at α = 157.38°
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Problem 015
Forces F, P, and T are concurrent and acting in the direction as shown in Fig. P-015. a. Find the value of F and α if T = 450 N, P = 250 N, β = 30°, and the resultant is 300 N acting up along the y-axis. b. Find the value of F and α if T = 450 N, P = 250 N, β = 30° and the resultant is zero. c. Find the value of α and β if T = 450 N, P = 250 N, F = 350 N, and the resultant is zero.
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Solution 015 Part a: Unknown force and direction with non-zero resultant
and
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Part b: Unknown force and direction with zero resultant
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and
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Part c: Unknown direction of two forces with zero resultant
and
→ Equation (1)
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→ Equation (2)
Equation (1) + Equation (2)
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From Equation (1)
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Coplanar Parallel Force System
Parallel 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.
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Resultant of Distributed Loads
The 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 System
The resultant of parallel forces in space will act at the point where it will create equivalent translational and rotational (moment) effects in the system.
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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 a Couple Couple.. 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.
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236 Computation of the resultant of parallel forces acting on the lever Problem 236
A parallel force system acts on the lever shown in Fig. P-236. Determine the magnitude and position of the resultant.
Solution 236
downward
counterclockwise
to the right of A Thus, R = 110 lb downward at 6 ft to the right of A.
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237 Finding the resultant of parallel forces acting on both sides of the rocker arm Problem 237
Determine the resultant of the four parallel forces acting on the rocker arm of Fig. P237.
Solution 237
downward
clockwise
to the right of O Thus, R = 50 lb downward at 4 ft to the right of point O.
answer
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238 Finding the resultant of trapezoidal loading
Problem 238
The 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 238
Thus, R = 3330 N downward at 3.6 m to the left of A.
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239 Resultant of lift on the wing of an airplane Problem 239
The 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 = 90x 1/2 lb per ft where x is measured from the tip. Compute the resultant and its location from the wing tip. Solution 239
upward
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Thus, R = 3840 lb upward at 9.6 ft from the tip of the wing. Another Solution
okay!
okay!
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240 How to locate the centroid of metal plate with circular hole
Problem 240
The shaded area in Fig P-240 represents a steel plate of uniform thickness. A hole of 4in. 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 thecenter 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 240
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Thus, the centroid is located at 6.8 in. to the right of left edge and 4.9 in. above the bottom edge. answer
241 Finding the resultant of vertical forces acting on the Fink truss Problem 241
Locate the amount and position of the resultant of the loads acting on the Fink truss in Fig. P-241.
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Solution 241
Magnitude of resultant
downward Location of resultant
to the right of A Thus, R = 15 130 N downward at 3.62 m to the right of left support.
Answer
242 Finding the unknown two forces with given resultant Problem 242
Find 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.
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Solution 242
Sum of vertical forces
Moment about point A
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243 Finding the magnitude and position of the missing force Problem 243
The 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.
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Solution 243
Sum of vertical forces
downward
Moment about point A
Thus, F = 31.4 kg downward at 2.48 m to the right of left support.
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Resultant of Non-Concurrent Force System
The resultant of non-concurrent force system is defined according to magnitude, inclination, and position.
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The magnitude of the resultant can be found as follows
The inclination from the horizontal is defined by
The position of the resultant can be determined according to the principle of moments.
Where, Fx = component of forces in the x-direction Fy = component of forces in the y-direction Rx = component of thew resultant in x-direction Ry = component of thew resultant in y-direction R = magnitude of the resultant θx = angle made by a force from the x-axis MO = moment of forces about any point O d = moment arm MR = moment at a point due to resultant force ix = x-intercept of the resultant R iy = y-intercept of the resultant R Problem 262 | Resultant of Non-Concurrent Force System Problem 262
Determine completely the resultant of the forces acting on the step pulley shown in Fig. P-262.
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Solution 262
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Thus, R = 1254.89 lb downward to the right at θx = 44.21° and passes through the axle.
Problem 263 | Resultant of Non-Concurrent Force System
Problem 263
Determine the resultant of the force system shown in Fig. P-263 and its x and y intercepts.
Solution 263
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Thus, R = 161.314 lb upward to the right at θx = 21.69° and intercepts at (1.668, 0) and (0, -0.671). Problem 264 | Resultant of Non-Concurrent Force System
Problem 264
Completely determine the resultant with respect to point O of the force system shown in Fig. P-264.
Solution 264
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Thus, R = 544.68 N upward to the right at θx = 28.25°. The intercepts of R are (-4.57, 0) and (0, 2.46). Problem 265 | Resultant of Non-Concurrent Force System Problem 265
Compute the resultant of the three forces shown in Fig. P-265. Locate its intersection with X and Y axes.
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Solution 265
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Thus, R = 957.97 lb downward to the right at θx = 32.19°. The x-intercept is at 2.90 ft to the right of O and the y-intercept is 1.83 ft above point O. Problem 266 | Resultant of Non-Concurrent Force System Problem 266
Determine the resultant of the three forces acting on the dam shown in Fig. P-266 and locate its intersection with the base AB. For good design, this intersection should occur within the middle third of the base. Does it? Solution 266
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Righting moment
Overturning moment
Moment at the toe (downstream side - point p oint B)
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Location of Ry as measured from the toe
(within the middle third) Thus, R = 27 424.02 lb downward to the right at θx = 79.91° and passes through the base at 8.44 ft to the left of B which is within the middle third. Related post: Foundation (Soil) pressure of gravity dam. Problem 267 | Resultant of Non-Concurrent Force System
Problem 267
The Howe roof truss shown in Fig. P-267 carries the given loads. The wind loads are perpendicular to the inclined members. Determine the magnitude of the resultant, its inclination with the horizontal, and where it intersects AB.
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Solution 267
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Thus, R = 10 778.16 N downward to the right at θx = 68.2° passing 4.46 m to the right of A.
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Problem 268 | Resultant of Non-Concurrent Force System Problem 268
The resultant of four forces, of which three are shown in Fig. P-268, is a couple of 480 lb·ft clockwise in sense. If each square is 1 ft on a side, determine the fourth force completely.
Solution 268
Let F4 = the fourth force and for couple resultant, R is zero.
Thus,
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Assuming F4 is above point O
d is positive, thus, the assumption is correct that F 4 is above point O. Therefore, the fourth force is 200 lb acting horizontally to the left at 5.8 ft above point O. answer
Problem 269 | Resultant of Non-Concurrent Force System Problem 269
Repeat Prob. 268 is the resultant is 390 lb directed down to the right at a slope of 5 to 12 passing through point A. Also determine the x and y intercepts of the missing force F.
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Solution 269
Let F4 = the fourth force
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Resolve F4 into components at the x-axis
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Resolve F4 into components at the y-axis
Thus, F4 = 219.32 lb downward to the right at θx = 43.15° with x-intercept i x = 3.27 to the right of O, and y-intercept i y = 3.06 ft above point O.
Problem 270 | Resultant of Non-Concurrent Force System Problem 270
The three forces shown in Fig. P-270 are required to cause a horizontal resultant acting through point A. If F = 316 lb, determine the values of P and T. Hint: Apply M R = ΣMB to determine R, then M R = ΣMC to find P, and finally M R = ΣMD or Ry = ΣY to compute T.
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Solution 270
For horizontal resultant, Ry = 0 and Rx = R
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Problem 271 | Resultant of Non-Concurrent Force System Problem 271
The three forces in Fig. P-270 create a vertical resultant acting through point A. If T is known to be 361 lb, compute the values of F and P.
Solution 271
Click here to show or hide the solution For vertical resultant, R x = 0 and Ry = R
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