Question

2) Determine the tension in each rope-cable if the weight of the crate is 1500 lb. z 2 ft 2 ft B 4 ft 5.5 ft 2.5 ft 6 ft

Answer 1

The solution for calculating tension in each rope cable is given below as image file.
28 B C 1467 TAS D X 6th ..0:30.6 -0.9231 +0.2316') 2) weight of trake (W) = 150015 TACK SSH А FDA Crano N 2.5ft het us find unit vector at points A, B, D , C. A. of + 63 +2.5k B = -ał toj + C = 22 +03+5.5k D = oitototy. Now we take A B -* = -27-6571-5# magnitude JARI (-2)²+(-6)²+(15)² = AB -27-6) tist IAR 6.5 6.5 6.5 6.5 UAB ม Ac = ? ai 65 +30 JATI N221 2+ (26)2+(372 7.

UAL Date S M T W T F S in ) Notes m2-fo=87-0-8545°40-4297 TAL AD b=0 Ditt 6;²+2.50 DAL=6:5 TUDA of +0.9233+0.3052 Sum of forces &f=0 (For equilibrin TAB = ITABI UAB FA = 1 Foal XUDA TAC = ITACUAC. TAE - TAEIX UAE TAB-0-305 ART-0923 AR110-23103 TAC 0.286 Aci 0 :857 A5 +0.429ACK Foa. of t 0.923 DAT + 0-385DAK TAE = oitos -1500k Bf=0 can i, j, k sum equal to zero ) @ 1500 0-308 AB t 0.286 AC 33 -0.923 AB - 0.857 AC +0.923 DA=0 0:231AB + 0.429 AC + 0-385 DA-1500 500 0:231AB10:29 Act 3 Divide egn) 0:923. 0-38SDA 0.928ACT DA/=0 Cance) by 0.385 DA o 6AB - 1011 AC-PA = -3,896.104 -16AB - 2.038.A.C. -3896.104. Fromegncape = 1.066ABE - AB

-16AB - 2.038 (1066) AB = -3896-104 3.772 AB -38 96 104 AB 1,032.901) > Answer AC = 1066an) - 1056 x 61032 901). AC 1,101. 072 Substitute on eq nCB) 0.92.3 (1032.901) - 0.557 (1,101-072) 7 00G230A 9 DA = 2:055.270

Answer 2

Z 2f1f6 -Y (1) 42 (k) B с 4 ft FABIO 5.5ft +xli) FAC 0 FAD 1500 lb 2.5ft 6 ft -X (i) * ZWk) Figure [(a)] +46) - •> Let Fas & FAB are the forces in cables AB & AC. Let FAD be the force in stout AD FBD as •> step 1) - To find force vectors: Here, we can express each forces in shown in figure [(a)] How cartesian rector from of forces are: Let À,Ì, be the unit vectors along X, Y&Z directions respectively. LE Here, point vectors are : Äs xit f t z = oît 6 + 2-5 ^ Similarly 7 = -29+ oſ + 4k ² = 2 i to ĝ+ 5.5 ĥ D 01 + 0 + O Ав

= 6.5 TAB! .. AB > Vector AB - Z - Ā = (-2î + 0,^+4 k) - 10+6;4+2:56) AB = -21 166+1.5 Ê Magnitude of TAB = LABI = x+y*+z2 -√2+6²+1:5² TABI Now, Unit vector AB can be calculated as : u = AB - [ie] + [A] + [6] 2) + ) + (SX ūãe = -0.308 î – 0.923ſ +0.231 û Similarly, a Āč = <- A = (21 + 054+ 5.5k) - (034 +679 +2-5€) • AC = zî -69 + 3 Å IACI= 7 :ūt Ac - (++ üc = 0.2861 -0.857ỹ + 0.429 T Also, DA = D- A = 10 + 054 + O T) - (0.9 + 69 +2,5€) D = oît 69 +2:5² -|PA| = 10*+6*+2-52 UD ()+()+(205) IDA o î + 0.923ỹ +0.325 K Ac TAR - 6.5 PÀ DA

step2 : To find cartesian rectors i Let F, Ę, ç, be the cartesian vectors. FAB F = UAB. FAB F = (-0.308 î - 0.923ỹ +0.231 k). F FAB FI = -0.308 Face 1 – 0.923 F. + 0.231 For Similarly, F = MAC FAC 5,- (+ 0.2869 -0.8575 +0.4296) Fc F = (+0.286 FASÎ – 0.857 Fecſ + 0.429 FACT Also, Fz= UDA. EDA F3 = (01 +0.923; +0.3256) FPA F3 = 0.923 FDA Í + 0.325 And, FDA . 1500 [ Due to coate loading] step 3: Equilibrium condition : EF=0 => F, + 5z + Fz + FA 70.3 -0.308 FAB 0.923 FARÀ AR+ 0.231 À ) + +1+0.286 FACÎ - 0.857 FA 5 +0.429 FACTê) + +(0.923 FPA Ĵ + 0.325 FDA E -1500 Ħ] =0 AR

→ TAC = 0.2857TACI – 0.8571TACİ +0.4286Tack

Answer 3

Tension vectors;

Z 2f1f6 -Y (1) 42 (k) B с 4 ft FABIO 5.5ft +xli) FAC 0 FAD 1500 lb 2.5ft 6 ft -X (i) * ZWk) Figure [(a)] +46) - •> Let Fas & FAB are the forces in cables AB & AC. Let FAD be the force in stout AD FBD as •> step 1) - To find force vectors: Here, we can express each forces in shown in figure [(a)] How cartesian rector from of forces are: Let À,Ì, be the unit vectors along X, Y&Z directions respectively. LE Here, point vectors are : Äs xit f t z = oît 6 + 2-5 ^ Similarly 7 = -29+ oſ + 4k ² = 2 i to ĝ+ 5.5 ĥ D 01 + 0 + O Ав

= 6.5 TAB! .. AB > Vector AB - Z - Ā = (-2î + 0,^+4 k) - 10+6;4+2:56) AB = -21 166+1.5 Ê Magnitude of TAB = LABI = x+y*+z2 -√2+6²+1:5² TABI Now, Unit vector AB can be calculated as : u = AB - [ie] + [A] + [6] 2) + ) + (SX ūãe = -0.308 î – 0.923ſ +0.231 û Similarly, a Āč = <- A = (21 + 054+ 5.5k) - (034 +679 +2-5€) • AC = zî -69 + 3 Å IACI= 7 :ūt Ac - (++ üc = 0.2861 -0.857ỹ + 0.429 T Also, DA = D- A = 10 + 054 + O T) - (0.9 + 69 +2,5€) D = oît 69 +2:5² -|PA| = 10*+6*+2-52 UD ()+()+(205) IDA o î + 0.923ỹ +0.325 K Ac TAR - 6.5 PÀ DA

and

step2 : To find cartesian rectors i Let F, Ę, ç, be the cartesian vectors. FAB F = UAB. FAB F = (-0.308 î - 0.923ỹ +0.231 k). F FAB FI = -0.308 Face 1 – 0.923 F. + 0.231 For Similarly, F = MAC FAC 5,- (+ 0.2869 -0.8575 +0.4296) Fc F = (+0.286 FASÎ – 0.857 Fecſ + 0.429 FACT Also, Fz= UDA. EDA F3 = (01 +0.923; +0.3256) FPA F3 = 0.923 FDA Í + 0.325 And, FDA . 1500 [ Due to coate loading] step 3: Equilibrium condition : EF=0 => F, + 5z + Fz + FA 70.3 -0.308 FAB 0.923 FARÀ AR+ 0.231 À ) + +1+0.286 FACÎ - 0.857 FA 5 +0.429 FACTê) + +(0.923 FPA Ĵ + 0.325 FDA E -1500 Ħ] =0 AR

→ TAC = 0.2857TACI – 0.8571TACİ +0.4286Tack

Force vector in the strut;

$\vec{F}_{DA} = F_{DA}\left [\frac{\vec{r}_{DA}}{|r_{DA}|} \right ] = F_{DA}\left [ \frac{6\hat{j}+2.5\hat{k}}{\sqrt{6^2+2.5^2}} \right ]$

$\Rightarrow \vec{F}_{DA} =0.9231F_{DA}\hat{j} + 0.3846F_{DA}\hat{k}$

Weight;

$\vec{W}=-1500\hat{k}$

By equilibrium of forces;

$\sum F_x = 0$

$\Rightarrow -0.3077T_{AB}+0.2857T_{AC}=0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...(1)$

and

$\sum F_y = 0$

$\\\Rightarrow -0.9231T_{AB}-0.8571T_{AC}+0.9231F_{DA}=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...(2)$

and

$\sum F_z = 0$

$\\\Rightarrow 0.2308T_{AB}+0.4286T_{AC}+0.3846F_{DA}-1500=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...(3)$

Solve (1), (2) and (3);

...(Answer)