Weight On Hanging Supports Question

  • Thread starter Thread starter thelovemonkey
  • Start date Start date
  • Tags Tags
    Weight
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 3K views
thelovemonkey
Messages
12
Reaction score
0

Homework Statement



The figure below shows a weight, W, supported by two wires, the stresses n the wires AB and AC are not to exceed 100M N/m^2 and 150 N/m^2 resp. The X-sectional Area of AB is 400 mm^2 and AC is 200 mm^2. you are required to calculate the largest load which can be supposted by the two wires.


http://img130.imageshack.us/img130/9804/26382989.jpg


in summary

P(ab)= 100 x 10^6 N/m^2
P(ac)= 150 x 10^6 N/m^2
A(ab)= 400mm^2 => 0.0004 m^2
A(ac)= 200mm^2 => 0.0002 m^2

Homework Equations



P = T/A
=> T = PA

The Attempt at a Solution



since system is in equillibrium summation of horizontal forces:

Fh =0
T(ac) Cos45 = T(ab) Cos30
T(ac)= 1.225 T(ab)

since system is in equillibrium summation of vertical forces:

Fv = 0

T(ab) Sin30 + T(ac) Sin45 - W = 0

but
T(ac) = 1.225 T(ab)

so
T(ab) Sin30 + 1.225 T(ab) Sin45 = W
1.336 T(ab) = W

____________________

T = PA

T(ab) = P(ab) x A(ab)

T(ab) = 100 x 10^6 N/m^2 X 0.0004 m^2
T(ab) = 40000 N

but

1.336 T (ab) = W
W = 53.440 KN


Out of curiosity i checked to see if i got the same answer if i used an equation in terms of W and T(ac)...

T(ab) = 0.816 T(ac)

1.115 T(ac) = W

T(ac) = 150 x 10^6 N/m^2 X 0.0002 m^2
T(ac) = 30000 N

1.115 T(ac) = W
W = 1.115 (30000)
W = 33.450 KN


i don't see why i am getting 2 different values for the Weights... the math appears flawless... or is it o.O
 
Last edited by a moderator:
Physics news on Phys.org
thelovemonkey said:

Homework Statement



The figure below shows a weight, W, supported by two wires, the stresses n the wires AB and AC are not to exceed 100M N/m^2 and 150 N/m^2 resp. The X-sectional Area of AB is 400 mm^2 and AC is 200 mm^2. you are required to calculate the largest load which can be supposted by the two wires.


http://img130.imageshack.us/img130/9804/26382989.jpg


in summary

P(ab)= 100 x 10^6 N/m^2
P(ac)= 150 x 10^6 N/m^2
A(ab)= 400mm^2 => 0.0004 m^2
A(ac)= 200mm^2 => 0.0002 m^2

Homework Equations



P = T/A
=> T = PA

The Attempt at a Solution



since system is in equillibrium summation of horizontal forces:

Fh =0
T(ac) Cos45 = T(ab) Cos30
T(ac)= 1.225 T(ab)

since system is in equillibrium summation of vertical forces:

Fv = 0

T(ab) Sin30 + T(ac) Sin45 - W = 0

but
T(ac) = 1.225 T(ab)

so
T(ab) Sin30 + 1.225 T(ab) Sin45 = W
1.336 T(ab) = W

____________________

T = PA

T(ab) = P(ab) x A(ab)

T(ab) = 100 x 10^6 N/m^2 X 0.0004 m^2
T(ab) = 40000 N

but

1.336 T (ab) = W
W = 53.440 KN


Out of curiosity i checked to see if i got the same answer if i used an equation in terms of W and T(ac)...

T(ab) = 0.816 T(ac)

1.115 T(ac) = W

T(ac) = 150 x 10^6 N/m^2 X 0.0002 m^2
T(ac) = 30000 N

1.115 T(ac) = W
W = 1.115 (30000)
W = 33.450 KN


i don't see why i am getting 2 different values for the Weights... the math appears flawless... or is it o.O

Your math looks fine...however, you are missing the point of the problem.

Hint: What is the stress in each wire if you use a W of 53.440 kN? Are they acceptable? What is the stress in each wire if you use a W of 33.450 kN? Are they acceptable?

CS
 
Last edited by a moderator:
thelovemonkey: You are doing well; however, your solution contains a mistake: 1.336 is incorrect. Try that again.
 
I think he just made a typo... (1.336 versus 1.366)...his equations look right...

CS
 
omg i got it... the 54 blah blah blah Newtons is too much for the wire AC to handle as in it produced a pressure over 150 X 10^6 N/m^2... i did it on my calc and am too tired to do it again and post the answer.

thanks guys... you're the best (Y)
extra kudos to stewartcs