# Physics question

• Jun 18th 2006, 11:14 AM
salvavida
Urgent Math-Physics question
Could you please urgently help me to solve the following question. I have to turn it in until tomorrow (monday) night.
Thank you.

A mass of 1019Kg is supported by two wires A and B each made of a different material. Before loading the wires are of equal length. After loading they are still of equal length. Find the load carried by each wire.

Wire Diameter E

Wire A 1 mm 206,000 Nmm 2

Wire B 2 mm 116,000 Nmm 2
• Jun 18th 2006, 11:42 AM
CaptainBlack
Quote:

Originally Posted by salvavida
Could you please urgently help me to solve the following question. I have to turn it in until tomorrow (monday) night.
Thank you.

A mass of 1019Kg is supported by two wires A and B each made of a different material. Before loading the wires are of equal length. After loading they are still of equal length. Find the load carried by each wire.

Wire Diameter E

Wire A 1 mm 206,000 Nmm 2

Wire B 2 mm 116,000 Nmm 2

You know that the sum of the loads/tensions on/in the two wires sums to
1019g Newtons:

$F_A+F_B=1019 g\ \ \ \dots(1)$

and that the original lengths and the extensions of the wires are equal.

If $Y_A$ and $Y_B$ are the Youngs moduli of the two wires (Presumably your $E$'s)
and $A_A$ and $A_B$ their crossectional areas then if $F_A$ and $F_B$ are the
tensions in the two wires we have:

$
Y_A=\frac{F_A/A_A}{\Delta l_A/l_A}
$

and:

$
Y_B=\frac{F_B/A_B}{\Delta l_B/l_B}
$

but the original length and the extension of each wire are
equal so:

$\frac{Y_A}{Y_B}=\frac{F_A/A_A}{F_B/A_B}\ \ \ \dots(2)$.

Equations $(1)$ and $(2)$ should allow you to solve
for the two loads $F_A$ and $F_B$

RonL
• Jun 18th 2006, 02:03 PM
salvavida
Urgent help for Math-Physics question
Dear RonL,

Thanks,
Salvavida
• Jun 18th 2006, 02:22 PM
CaptainBlack
We found that:

$\frac{Y_A}{Y_B}=\frac{F_A/A_A}{F_B/A_B}\ \ \ \dots(2)$.

plugging the values of Youngs modulus for the two materials, and the
crossectional areas gives:

$\frac{206}{116}=\frac{F_A/(\pi\ 0.5^2)}{F_B/(\pi\ (1^2)}$,

so:

$F_A=\frac{206}{116}\frac{F_B}{4}\approx0.444\ F_B\ \ \ \dots(3)$.

$
F_A+F_B=1019 g\ \ \ \dots(1)
$
,

so substituting from $(3)$:

$1.444\ F_B=1019g=1019\times 9.81=9996.39$,

solving:

$F_B=6920.79\ N$,

and so:

$F_A=9996.39-F_B=3075.6$.

RonL