As shown, a mass is being lifted by a strut that is supported by two cables AC a

Question

As shown, a mass is being lifted by a strut that is supported by two cables AC and CD. The dimensions given are a=7.70 ft,b=4.90 ft, c=8.30 ft, d=3.70 ft, and e=5.50 ft.

Part A - Finding the tension in cable AC

A weight of 190 lb acts at C on the strut. Find the magnitude of the tension in cable AC.

Part B - Finding the unknown weight

What is the weight of an unknown hanging mass when the compressive force in strut BC is  555 lb?

Part C - Finding the maximum weight

Find the maximum weight that the cable and strut system can support if the magnitude of the compressive force in the strut cannot exceed 1100 lb and the magnitude of tension in the cables cannot exceed 300 lb.

Any help?

Force Systems A weight of 190 lb acts at C on the strut Find the magnitude of the tension in cable AC To apply the forces. by two w- Value Units

Explanation / Answer

>> Writing all the co-ordinates;

A = (d,0,0) = (3.70,0,0) = 3.70 i

B = (0,a,0) = (0,7.7,0) = 7.70 j

C = (-5.5,12.6,8.3) = - 5.5 i + 12.6 j + 8.3 k

D = (-5.5,0,0) = - 5.5 i

>> Now, let tensions are:

Tac, Tcd and Tbc

>> As, Tac is acting in CA direction

As, CA = (3.70 i) - (- 5.5 i + 12.6 j + 8.3 k) = 9.20 i - 12.6 j - 8.3 k

Magnitude = [ 9.22 + 12.62 + 8.32 ]1/2 = 17.67

=> Unit Vector along CA = ( 9.20 i - 12.6 j - 8.3 k)/17.67 = 0.521 i - 0.713 j - 0.47 k

=> Tca = Tca(0.521 i - 0.713 j - 0.47 k)   ...(1)....

>> As, Tcd is acting in CD direction

As, CD = (- 5.5 i) - (- 5.5 i + 12.6 j + 8.3 k) = - 12.6 j - 8.3 k

Magnitude = [ 02 + 12.62 + 8.32 ]1/2 = 15.09

=> Unit Vector along CD = (0i - 12.6 j - 8.3 k)/15.09 = - 0.835 j - 0.55 k

=> Tcd = Tcd(- 0.835 j - 0.55 k)   ...(2)....

>> As, Tcb is acting in CB direction

As, CB = (7.70 j) - (- 5.5 i + 12.6 j + 8.3 k) = 5.5 i - 4.9 j - 8.3 k

Magnitude = [ 5.52 + 4.92 + 8.32 ]1/2 = 11.097

=> Unit Vector along CB = (5.5 i - 4.9 j - 8.3 k)/11.097 = 0.496 i - 0.441 j - 0.748 k

=> Tcb = Tcb(0.496 i - 0.441 j - 0.748 k)   ...(3)....

>> As, Weight, W = mg k

>> Now atC, as under euilibrium,

Tca + Tcb + Tcd + W = 0

=> Tca(0.521 i - 0.713 j - 0.47 k) + Tcb(0.496 i - 0.441 j - 0.748 k) + Tcd(- 0.835 j - 0.55 k) - mg k = 0

>> Comparing Coefficients both sides;

=> 0.521 Tca + 0.496 Tcb = 0

=> -0.713 Tca - 0.441 Tcb - 0.835 Tcd = 0

=> - 0.47 Tca - 0.748 Tcb - 0.55 Tcd - mg = 0

Part (a).

m = 190 lb, ;

As g = 32.2 m/s2

>> So, solving above three equations;

=> Tca = 12739.9 lbf   ...TENSION in Cable AC ...ANSWER..

Part (b).

Now, Tbc = 555 lb

So, simpifying above three equations

=> m = 8 lb ..ANSWER..

Part (c).Tension = 300 lb

and, Compression = 300 lb

>> As, In Ac, there will be compression force

and in BC and CD , tension will act

=> Tca = 1100 lb

Tcb = Tcd = 300 lb

>> So, simplifying last equation, m = 3.96 lb ..ANSWER...

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