non-concurrent forces - learneasy · non - concurrent forces (components) monday, 16 may 2011 6:55...

Problem 7.1a (p112)Mp = 0+(15 * 3) - (Fr * 5) = 0Fr = 45/5 = 9 kNNow take Fy=0;Fp + Fr -15 = 0Fp = 15 - Fr = 15 - 9 = 6 kN

Problem 7.1h (p113)

(2 * 1.5) + (3 * 1) - (Fr * 3) + (5 * 4.5) = 0-Mp=0

-3 + 3 - 3Fr +22.5 = 0Fr = 22.5/3 = 7.5 kN

Fy=0-2 + Fp -3 + 7.5 - 5 = 0Fp = 5 - 7.5 + 3 + 2 = 2.5 kN

Non-Concurrent ForcesThursday, 12 May 20112:40 PM

Non-Concurrent Forces

Span = 20" (508mm), 2kg mass, 4" (101.6mm) from right.Find Ra & Rb.Force = 2 * 9.81 = 19.62 NMoment equation:

Ma=0+ (19.62 * 406.4) - (Fb * 508) = 07973.568 = Fb * 508 Fb = 7973.568 / 508 = 15.696 N (15.696/9.81 = 1.6 kg)3.527392 lb

Fy=0Fa - 19.62 + 15.696 = 0Fa = 19.62 - 15.696 = 3.924 N 3.924/9.81 = 0.4 kg0.88185 lb

Beam DemoMonday, 16 May 20117:53 PM


Take components;

Ax = Fcos = 33 * cos(303) = 17.973 kN

Ay = Fsin = 33 * sin(303) = -27.676 kNDo Moment eqn:

ML=0+ (27.676 * 3.22) - (Fr * 5.25) = 0

Fr = 89.1167 / 5.25 = 16.9746 kN89.1167 = Fr * 5.25

Fy=0FLy -27.676 +16.9746 = 0FLy = 27.676 -16.9746 = 10.7014 kN

Fx=0FLx + 17.973 = 0FLx = - 17.973 kN

Resolve components at pin joint back to a single force.Pythag:10.7014^2 + 17.973^2 = 437.5487 Hyp = 437.5487 ^0.5 = 20.9177 kN

ComponentsMonday, 16 May 20118:19 PM


Hyp = 437.5487 ^0.5 = 20.9177 kN

Tan B = FLx / Fly = -17.973 / 10.7014 = -1.6795 B = atan (-1.6795) = 59.229 degsIn 360 notation: Angle = 90 + 59.229 = 149.229 degsOR....Straight to 360 notation? (Sorry atan can't determine the correct quadrant)atan(Fly/Flx) = atan(10.7014/-17.973) = -30.77021774213884 OR...CAD.


+ (Fa * 0.25) - (8400 * 0.3415) + (2900 * 0) + (119 * 9.81 * 0.125) = 0+ (Fa * 0.25) - 2868.6 + 0 + 145.9238 = 0

= 2722.77 NmFa = 2722.6762 / 0.25 = 10890.7 N = 10.89 kN

+ (Fa * 0.25) = + 2868.6 - 145.9238

Find force A

Non - Concurrent ForcesMonday, 16 May 20116:55 PM


360o notation: Force B is at 150o

Fbx = Fcos= 8400*cos(150) = -7274.6 N

Fby = Fsin8400*sin(150) = 4200 NThere are 5 forces; So there are 5 sets of moments (but Fc is zero)...( ) ( ) ( ) ( ) ( ) =0So moment eqn at pivot point is;

+ (Fa*0.25) + (Fw*0.125) - (Fbx*0.25) - (Fby*0.25) + (2900 *0) = 0

+ (Fa*0.25) + (Fw*0.125) - (Fcos*0.25) - (Fsin*0.25) + (2900 *0) = 0You must ignore minus sign inside the brackets!+ (Fa * 0.25) + (119*9.81*0.125) - (7274.6*0.25) - (4200*0.25) = 0

= 2722.7262 NmFa = 2722.7262/0.25 = 10890.9048 N

(Fa * 0.25) = - 145.9238 + 1818.65 + 1050

Check it…+ ( 10890.9048* 0.25) + (119*9.81*0.125) - (7274.6*0.25) - (4200*0.25) = -0.0001 (Close enough)

Find force A. This time use components of force B!

Non - Concurrent Forces (Components)Monday, 16 May 20116:55 PM


ML=0+ (13) - (Fr * 10) = 0Fr = 13/10 = 1.3 kN

Fy=0FL + Fr = 0FL = -Fr = -1.3kN (Reaction is downwards, beam wants to go upwards)

Applied MomentsMonday, 16 May 20119:04 PM


Do moment equation at B because we

want force at A. Mb = 0

+(173) - (154*9.81*0.23) + (Fa*0.46) = 0

= 174.4702 NmFa*0.46 = -173 + 347.4702

Fa = 174.4702/0.46 = 379.283 N

Fy=0Fa - Fw + Fb = 0379.283 - 154*9.81 = -FbFb = 1510.74 - 379.283 = 1131.457 N

Eg. Applied MomentThursday, 19 May 201112:10 PM


Take total load: 1.8 * 3 = 5.4 kNApply the load at the centre of the distributed region:

Ma = 0+ (5.4*1.5) + (13*6) - (Fb * 10) = 010Fb = 8.1 + 78 = 86.1 Fb = 86.1 /10 = 8.61 kN (ooops, should have taken moment at B)

(13*4) - (5.4*8.5) + (Fa * 10) = 0-10Fa = 52+45.9 = 97.9 Fa = 9.79 kN

Check it...

Fy=0 ?-5.4 - 13 + 8.61 + 9.79 = 0

Mb = 0

A distributed load is spread over a region.It has the same effect (on the reactions) as putting all the weight in the centre of that region.

Distributed LoadsMonday, 23 May 20117:41 PM


HINT:Take moment on top of point that is…(a) Not your input info (Force A)(b) Not your output info (Support cable)(c) Maybe a pin or fixed joint

kN * m- (4.7 * 1.15) - (125 * 9.81*0.322) + (Fc * 0.5) = 0- (4.7 * 1.15) - (125 * 9.81*0.322/1000) = -5.7999 kNm

Fc = -5.7999 / -0.5 = 11.5998 kN

Planar Bodies...Monday, 23 April 20127:09 PM


Usually - Find reaction at R first. (KN * m)

ML = +(26.4*0.8) + (6.5*1.6) -(19.3*1.385) -(Fr * 3.2) = 0+(26.4*0.8) + (6.5*1.6) -(19.3*1.385) = 4.7895 (Fr * 3.2) =4.7895 Fr = 4.7895 /3.2 = 1.4967 kN

ML = +(2.2*0.8) + (6*1.6) -(1.3*1.385) -(Fr * 3.2) = 0 = + 1.76 + 9.6 - 1.8005 - (Fr * 3.2) = 0(Fr * 3.2) =9.5595 kNmFr = 9.5595 / 3.2 = 2.9873 kN

Find Reactions at L and R where Forces A= 2.2kN, B=1.3kN, C=6kN

More Planar Bodies...Monday, 23 April 20127:57 PM


Find distance from rear wheel to mass...x = (2900+2240)*cos(23) = 4731.4 mmNow Moment at front wheel: Mf = +((4.7314-3.35)*1900*9.81) -(5000*9.81*2.2) +(Fr*3.35) =0+((4.7314-3.35)*1900*9.81) -(5000*9.81*2.2) = -82162.0854 (Fr*3.35) = 82162.0854 Fr = 82162.0854 /3.35 = 24525.9956 N(OK it's safe)

Front wheels.... Use... Sum Fy = 0Fy = -(1900*9.81) -(5000*9.81) +(24525.9956) +(Ff) = 0-(1900*9.81) -(5000*9.81) +(24525.9956) = -43163.0044

Ff = 43163.0044 N


(4400*9.81*0.45) + (Ft*3.7587) + (Ft * 3.5619) = 0-Ft (3.7587 + 3.5619) = (4400*9.81*0.45) = 19423.8 NmFt = 19423.8 / (3.7587 + 3.5619) = 2653.3071 N

(This is the same tension in each rope)

Use CAD to find perpendicular distances from pivot to ropes: 3.7587 and 3.5619 m

Q18: Angle A=30 and B=34 °. Obelisk is 4.4 tonnes. Both rope tensions are equal. What rope tension will topple the obelisk?

ObeliskMonday, 18 March 20138:45 PM


non-concurrent forces - learneasy · non - concurrent forces (components) monday, 16 may 2011 6:55...

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