✔ 最佳答案
1. Let E1 and E2 be the Young's Modulus of the two wires
L1 and L2 be their natural lengths.
A1 and A2 be their cross-sectional areas.
k1 and k2 be their force constant.
Hence, k1 = (E1).(A1)/L1, and k2 = (E2).(A2)/L2
Let k be the equivalent force constant of the joined wire, the
1/k = 1/k1 + 1/k2
i.e. 1/k = L1/[(E1).(A1)] + L2/[(E2).(A2)]
But k = E.A/L, where E, A and L are the Young's Mouduls, effective area and new length of the joined wire, and L = L1 + L2
A = (L1.A1 + L2.A2)/(L1 + L2).
Thus, L/E.A = L1/[(E1).(A1)] + L2/[(E2).(A2)]
i.e. (L1+L2)/E.A = L1/[(E1).(A1)] + L2/[(E2).(A2)]
from which the value of E can be calculated.
For the case when A1 = A2 = A
then the equation is simplified to:
(L1+L2)/E = L1/E1 + L2/E2
2. The tension along the wire should be uniform (assume that the mass of the wire is negligible compared with that of the hanging mass).
Since tension T is proportional to the stress, which is equal to force/area, a smaller cross0sectional area will give rise to a larger stree. This is why a wire generally breaks at the point where the cross-sectional area is the minimum.
