Question

What will be the minimum diameter of a steel wire, which is used to raise a load of 4000N if the stress in the rod is not to exceed 95 MN/m2?

Answer 1

AnswEr :

From the Question,

• Force (F) = 4 × 10³ N

• Stress = 95 MN/m² = 95 × 10^-3 N/m²

To finD

Diameter of the steel wire

Stress : Force acting per unit area of the object

$\sf Stress = \dfrac{F}{A} \\ \\ \dashrightarrow \sf Stress = \dfrac{F}{\pi r^2}$

Putting the values, we get :

$\dashrightarrow \: \sf \: 95 \times {10}^{ - 2} = \dfrac{4 \times {10}^{3} }{3.14 \: \times {r}^{2} } \\ \\ \dashrightarrow \: \sf \: {r}^{2} = \dfrac{4 \times {10}^{5} }{3.14 \times 95} \\ \\ \dashrightarrow \: \sf \: r {}^{2} = \dfrac{4 \times {10}^{7} }{298.3} \\ \\ \dashrightarrow \: \sf \: {r}^{2} = 13.41 \\ \\ \dashrightarrow \: \boxed{ \boxed{ \sf r =3.66 \: m }}$

WKT,

D = 2r = 2(3.66) = 7.32 m

Thus,the diameter of the steel wire should be atleast 7.32 m

Answer 2

Given :

• Force = 4000 N
• Stress = 95 MN/m² = $\sf{95 \: Nmm^{-2}}$

To Find :

• Minimum Diameter of Steel Wire

Solution :

As we know that :

$\large{\boxed{\sf{Stress \: = \: \dfrac{Force}{Area}}}} \\ \\ \implies {\tt{Stress \: = \: \dfrac{Force}{\pi r^2}}} \\ \\ \implies \tt{95 \: = \: \dfrac{4000}{\pi r^2}} \\ \\ \implies \tt{r^2 \: = \: \dfrac{4000}{95 \: \times \: 3.14}} \\ \\ \implies \tt{r^2 \: = \: \dfrac{4000}{298.3}} \\ \\ \implies \tt{r^2 \: = \: 13.4} \\ \\ \implies \tt{r \: = \: \sqrt{13.4}} \\ \\ \implies \tt{r \: = \: 3.66} \\ \\ \underline{\sf{\therefore \: Radius \: of \: steel \: wire \: is \: 3.66 \: mm}}$

So,

$\implies \tt{Diameter \: = \: 2 \: \times \: Radius} \\ \\ \implies \tt{Diameter \: = \: 3.66 \: \times \: 2} \\ \\ \implies \tt{Diameter \: = \: 7.32} \\ \\ \underline{\sf{\therefore \: Diameter \: of \: Steel \: wire \: required \: is \: 7.32 \: mm}}$