Question

The two wires shown in figure (14-E2) are made of the same material which has a breaking stress of 8 x 10⁸ N/m². The area of cross-section of the upper wire is 0.006 cm² and that of the lower wire is 0.003 cm². The mass m₁ = 10 kg, m₂ = 20 kg and the hanger is light. 

(a) Find the maximum load that can be put on the hanger without breaking a wire. Which wire will break first if the load is increased ? 

(b) Repeat the above part if m₁ = 10 kg and m₂ = 36 kg.

Answer 1

Breaking stress = 8 x 10⁸ N/m² 

C/S Area of the upper wire A= 0.006 cm² =0.006/10000 m² 

C/S area of the lower wire A' = 0.003 cm² =0.003/10000 m² 

(a) Let the length of each wire =L and the maximum load on the hanger W  

Total load on the lower wire =W+m₁ 

The stress on the lower wire =(W+m₁)g/A'  

= (W+10)g/(0.003/10000) N/m² 

= (W+10)g*10⁷/3 N/m² 

If we equate it to breaking stress, 

(W+10)g*10⁷/3 = 8 x10⁸ 

→W =(3*80/g)-10 kg =240/10 -10 =24-10 =14 kg  

The load on the upper wire =W+m₁+m₂ =W+10+20 =W+30 kg 

The stress on the upper wire =(W+30)g/A 

=(W+30)g/(0.006/10000) N/m² 

=(W+30)g*10⁷/6 N/m² 

Equating it to breaking stress, (W+30)g*10⁷/6 = 8 x10⁸  

→W =6*80/10-30 =48-30 =18 kg 

Since the lower wire reaches the breaking stress with a lower weight, the maximum weight W =14 kg. The lower wire will break first.  

(b) If m₂ = 36 kg 

The stress on upper wire =(W+10+36)g/A  

=(W+46)g/(0.006/10000)  

=(W+46)g*10⁷/6 

Equating with breaking stress, (W+46)g*10⁷/6 =8 x10⁸ 

→W =6*80/10-46 =48-46 =2 kg 

So now the maximum load on the hanger = 2 kg and the upper wire breaks first.

Answer 2

For same stress, the max load that can be put is 14 kg. If the load is increased the lower wire will break first.

The maximum load that can be put is 2 kg. Upper wire will break first if load is increased.