Chapter 2 Force Systems-Worked Examples 1. A man pull with force of 300 N on a rope attached to a building as shown in fig, what are the horizontal and vertical components of the force exerted by the rope at point A.
6m a 300N
Solution: tanb =6/8=0.75, b=tan-1(0.75)=36.870
a=900-36.870=53.130 possible to use both bor a
Fx=Fcos36.870=300N cos36.870= 300sin 53.130 Fy= -Fsin36.870 =-300N sin36.870=-300N cos53.130 2.The cable AB prevents bar OA from rotating clockwise about the pivot O. If the cable tension 750N,determine the n- and t-components of this force acting on point A of the bar.
AB2 = 1.22 + 1.52 -2*1.2*1.5 cos1200 =2.34m sina/1.2 = sin 1200/2.34 , a=26.30 Tn=Tsin a=750sin 26.30 =333 N
Tt = -Tcosa =-750cos26.30 = -672 N
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Chapter 2 Force Systems-Worked Examples 3.In the design of a control mechanism, it is determined that rod AB transmits a 260-N force P to the crank BC.Determine the x and y scalar components of P P=260 N
Solution: hypotenuse=
52 +122
=13
Px=- Pcos 210=-260(12/13)= -240 N
Py=-Psin210 -260(5/13)= - 100 N
4.Determine the resultant R of the two forces shown by a) applying the parallelogram rule for vector addition b) summing scalar components.
Solution: Laws of cosines: R2=6002+4002 -2*600*400cos 600= 529N Laws of sines: 529/sin600 =600/sin a ° , a=79.10
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Chapter 2 Force Systems-Worked Examples
a
b) Rx=SFx= 600cos 600 -400= -100 N
Y
Ry=SFy = 600sin 600+0= 520 N SO , R= -100i +520 j N
X
5.
6.If the equal tension T in the pulley cable are 400 N, express in vector notation the force R exerted on the pulley by the two tensions. Determine the magnitude of R.
Solution: RX= SFX=400+400cos600=600 N 3
Chapter 2 Force Systems-Worked Examples RY=SFY=400sin600=346 N R = 600 i +346 j R=
6002 +3462
=693 N
7.Determine the resultant of the three forces below.
Solution: F x = 350 cos 25 o + 800 cos 70o - 600 cos 60o = 317.2 + 273.6 - 300 = 290.8 N F y = 350 sin 25 o + 800 sin 70o + 600 sin 60o = 147.9 + 751 + 519.6 = 1419.3 N i.e. F = 290.8 N i + 1419.3 N j Resultant, F
F
290.82 1419.32 1449 N 1419.3 tan 1 78.4 0 290.8 F = 1449 N
78.4 o
8.The two structural members, one of which is in tension and the other in compression, exert the indicated forces on joint O. Determine the magnitude of the resultant R of the two forces and the angle which R makes with the positive x-axis.(exercise)
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Chapter 2 Force Systems-Worked Examples
9.A force F of magnitude 40N is applied to the gear. Determine the moment of F about point O.
Solution:
10.A 200 N force acts on the bracket as shown determine the moment of force about “A” (principle of moment)
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Chapter 2 Force Systems-Worked Examples
Given F=200N θ = 45o Required MA =? Solution Resolve the force into components F1 am F2 F1= F cosθ ,F1=200 cosine 45o =141.42N. F2= F sin θ, F2 = 200 sin 45o= 2.468N. We know that MA = 0 MA = moment produce due to component F1+ moment produce due to component F2. =F1 x r1+ F2 x r2. Let us consider that clock wise moment is + ve. MA = F1 x r1+ F2 x r2 = - 141.42 x 0.1 + 2.468 x (0.1 +0.1) = - 13.648 N = 13 .648 N anti clock wise. 11.Determine the moment of couple acting on the moment shown(COUPLE)
Given F1=200 N, L1=4m F2=200 N ,L2 = 2m. Required Moment of couple = M =? Working Formula M = F x r. Solution Put the values in working formula M= 200(4+2) =1200 N. m 12.Determine the moment of couple acting on the moment shown.
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Chapter 2 Force Systems-Worked Examples
Given F1=F2 =90lb F3 = F4 = 120lb. Required Moment of couple = M=? Solution The moment of couple can be determined at any point for example at A, B or D. Let us take the moment about point B MB = Σ F R. = -F1 x r1 – F2 x r2 . = - 90(3) – 120 (1) = - 390 lb ft Result MB = MA=MD =390 lb .ft counter clock wise. Moment of couple = 390 lb.ft count clockwise 13. For each case illustrated in Fig. below , determine the moment of the force about point O . solution (scalar analysis) The line of action of each force is extended as a dashed line in order to establish the moment arm d . Also illustrated is the tendency of rotation of the member as caused by the force. Furthermore, the orbit of the force about O is shown as a colored curl.
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Chapter 2 Force Systems-Worked Examples
Thus, Fig. a Fig b Fig c Fig. d Fig e
MO = (100 N)(2 m) = 200 N. m CW MO = (50 N)(0.75 m) = 37.5 N.m.CW MO = (40 lb)(4 ft + 2 cos 30_ ft) = 229 lb.ftb CW MO = (60 lb)(1 sin 45_ ft) = 42.4 lb.ft CCW MO = (7 kN)(4 m - 1 m) = 21.0 kN.m. CCW
14.Determine the resultant moment of the four forces acting on the rod shown in Fig. below about point O .
SOLUTION Assuming that positive moments act in the +k direction, i.e., counterclockwise, we have + (MR)O = _Fd; (MR)O = -50 N(2 m) + 60 N(0) + 20 N(3 sin 300 m) -40 N(4 m + 3 cos 300m) (MR)O = -334 N. m = 334 N.m CW 8
Chapter 2 Force Systems-Worked Examples For this calculation, note how the moment-arm distances for the 20-N and 40-N forces are established from the extended (dashed) lines of action of each of these forces. 15.Determine the moment produced by the force F in Fig. tree shown below a about point O . Express the result as a Cartesian vector. solution As shown in Fig. b , either rA or rB can be used to determine the moment about point O . These position vectors are rA = {12k} m and rB = {4i + 12j} m Force F expressed as a Cartesian vector is
Thus
OR
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Chapter 2 Force Systems-Worked Examples NOTE: As shown in Fig. b , MO acts perpendicular to the plane that contains F, rA, and rB . Had this problem been worked using MO = Fd, notice the difficulty that would arise in obtaining the moment arm 16.Two forces act on the rod shown in Fig. a . Determine the resultant moment they create about the flange at O . Express the result as a Cartesian vector.
solution Position vectors are directed from point O to each force as shown in Fig. b . These vectors are
17. Determine the moment of the force in Fig. below a about point O .(PINCIPLE OF MOMENT)
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Chapter 2 Force Systems-Worked Examples
solution i The moment arm d in Fig. a can be found from trigonometry. d = (3 m) sin 750 = 2.898 m Thus, MO = Fd = (5 kN)(2.898 m) = 14.5 kN. m Since the force tends to rotate or orbit clockwise about point O , the moment is directed into the page. solution ii The x and y components of the force are indicated in Fig. b . Considering counterclockwise moments as positive, and applying the principle of moments, we have + MO = -Fxdy - Fydx = -(5 cos 450KN)(3 sin 300m) - (5 sin 450 KN)(3 cos 300m)= -14.5 KN. m = 14.5 KN.m solution iii The x and y axes can be set parallel and perpendicular to the rod’s axis as shown in Fig. c . Here Fx produces no moment about point O since its line of action passes through this point. Therefore, + MO = -Fy dx = -(5 sin 750kN)(3 m) = -14.5 kN.m = 14.5 kN.m CW 18.Force F acts at the end of the angle bracket in Fig a . Determine the moment of the force about point(p .m)
solution i (scalar analysis) The force is resolved into its x and y components, Fig. b , then + MO = 400 sin 300N(0.2 m) - 400 cos 300 N(0.4 m) = -98.6 N.m = 98.6 N.m or MO = {-98.6k} N.m solution ii (vector analysis) Using a Cartesian vector approach, the force and position vectors, Fig. c , are r = {0.4i - 0.2j} m 11
Chapter 2 Force Systems-Worked Examples F = {400 sin 300i - 400 cos 300 j} N = {200.0i - 346.4j} N The moment is therefore
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Chapter 2 Force Systems-Worked Examples
6.Replace the two forces and couple by a wrench. Find the moment M of the wrench and the coordinates of point P in the y-z plane through which the force of the wrench passes.
ANS.
R=SF = 200i +150j N Assume positive wrench so the direction cosines of m are those of r or 0.8, 0.6,0 SMP =200(0.3 -Z)j -200(0.3-y)k +150 z i +150 (0.2)k -30i =(-30 +150Z)i +(60-200Z)j + (-30+200y)k N.m Equate the direction cosines of SMP and SF (-30+150z)/M=0.8 (60-200z)/M=0.6 (-30+200y)/M=0 Where M equals the magnitude of SMP Solving For y=0.15m or y=150mm z=0.264m or z=264mm //M// =(-30+150(0.264))/0.8=12 N.m
M =12(0.8i + 0.6 j) N.m 13
Chapter 2 Force Systems-Worked Examples 7.Determine and locate the resultant R of the two force and one couple acting on the I-beam.
You saw in the preceding lesson that the resultant of two forces may be determined either graphically or from the trigonometry of an oblique triangle. A. When three or more forces are involved , the determination of their resultant R is best carried out by first resolving each force into rectangular components. Two cases may be encountered, depending upon the way in which each of the given forces is defined: Case 1. The force F is defined by its magnitude F and the angle a it forms with the x axis. The x and y components of the force can be obtained by multiplying F by cos a and sin a, respectively [Example 1]. Case 2. The force F is defined by its magnitude F and the coordinates of two points A and B on its line of action ( Fig. 2.23 ). The angle a that F forms with the x axis may first be determined by trigonometry. However, the components of F may also be obtained directly from proportions among the various dimensions involved, without actually determining a [Example 2]. B. Rectangular components of the resultant. The components R x and R y of the resultant can be obtained by adding algebraically the corresponding components of the given forces [Sample Prob. 2.3]. You can express the resultant in vectorial form using the unit vectors i and j , which are directed along the x and y axes, respectively: R 5 Rxi 1 Ryj Alternatively, you can determine the magnitude and direction of the resultant by solving the right triangle of sides R x and R y for R and for the angle that R forms with the x axis.
88888888S8O88.LVING PROBLEMS
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Chapter 2 Force Systems-Worked Examples
ON YO U89988999999999999999999999 99999R Determine the x and y components of each of the forces shown 9.
.
10.Knowing that the tension in cable BC is 725 N, determine the resultant of the three forces exerted at point B of beam AB.
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Chapter 2 Force Systems-Worked Examples
11.The forces F1, F2, and F3, all of which act on point A of the bracket, are specified in three different ways. Determine the x and y scalar components of each of the three forces.
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Chapter 2 Force Systems-Worked Examples
couple 16.Replace the 10-kN force acting on the steel column by an equivalent force–couple system at point O. This replacement is frequently done in the design of structures.
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