Find the minimum diameter of a 51.5-m-long nylon string that will stretch no more than 1.01 cm when a load of 68.9 kg is suspended from its lower end. Assume that Ynylon = 3.51·10^8 N/m2
physics - bobpursley, Saturday, October 29, 2011 at 10:32am
Hookes Law:
elongation= force/Y * length/area
.0101=68.9*9.8/3.51*10^8 * 51.5/area
solve for area, then solve for diameter
I have solved the area and after that I have solved for diameter d=sqrt4*A/3.14 with this formula and I've founded the diameter 0.11 ! But the result is not correct! Please can you help me :(
Find the minimum diameter of a 51.5-m-long nylon string that will stretch no more than 1.01 cm when a load of 68.9 kg is suspended from its lower end. Assume that Ynylon = 3.51·10^8 N/m2
physics - bobpursley, Saturday, October 29, 2011 at 10:32am
Hookes Law:
elongation= force/Y * length/area
.00101=68.9*9.8/3.51*10^8 * 51.5/area
solve for area, then solve for diameter
I have solved the area and after that I have solved for diameter d=sqrt4*A/3.14 with this formula and I've founded the diameter 0.11 ! But the result is not correct! Please can you help me :(
physics - drwls, Sunday, October 30, 2011 at 7:34am
The allowed elongation is 0.0101 m, not 0.00101
physics - maria, Sunday, October 30, 2011 at 7:39am
yes I know. I solved with the value of elongation 0.0101 m. and I've found diameter 0.11 ! But it's wrong :( I tried many times and i always found 0.11 diameter.. I don't know where is the problem..
1 answer