Advertisements
Advertisements
Question
A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10–2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.
Advertisements
Solution 1
Let AB be a mild steel wire of length 2L = lm and its cross-section area A = 0.50 x 10-2 cm2. A mass m = 100 g = 0.1 kg is suspended at mid-point C of wire as shown in figure. Let x be the depression at mid-point i.e., CD = x
`:. AD = DB = sqrt((AC^2 + CD^2)) = sqrt(L^2+x^2)`
:. Increase in length `triangleL = (AD+DB) - AB = 2sqrt(L^2+x^2) - 2L`
`= 2L[(1+x^2/L^2)^(1/2) - 1] = 2L.x^2/(2L^2) = x^2/L`
:. Longitudinal strain = `(triangleL)/(2L) = x^2/(2L^2)`
If T be the tension in the wire as shown figure then in equilibrium `2T cos theta = mg`
or `T = "mg"/(2costheta)`
`= "mg"/(2 x/(sqrt(x^2+L^2))) = (mgsqrt(x^2+L^2))/(2x) = "mgL"/(2x)`
:. Stress = `T/A = "mgL"/(2 x A)`
As Young's modulus Y = `"stress"/"strain"`
`=((mgL)/(2 x A))/((x^2)/(2L^2)) = (mgL)/(2 x A) xx (2L^2)/x^2 = (mgL^3)/(Ax^3)`
`=> x = [(mgL^3)/(YA)]^(1/3) = L[(mg)/(YA)]^(1/3)`
`= 1/2[(0.1xx9.8)/(2xx10^11xx0.50xx10^(-2)xx10^(-4))]^(1/3) = 1.074 xx 10^(-2) m`
= 1.074 cm ≈ 1.07 cm or 0.01 m
Solution 2
Length of the steel wire = 1.0 m
Area of cross-section, A = 0.50 × 10–2 cm2 = 0.50 × 10–6 m2
A mass 100 g is suspended from its midpoint.
m = 100 g = 0.1 kg
Hence, the wire dips, as shown in the given figure.
Original length = XZ
Depression = l
The length after mass m is attached to the wire = XO + OZ
Increase in the length of the wire:
Δl = (XO + OZ) – XZ
Where,
XO = OZ = `[(0.5)^2 + l^2]^(1/2)`
`:.triangle l = 2[(0.5)^2 + (l)^2]^(1/2) - 1.0`
`= 2 xx 0.5 [1+(l/0.5)^2]^(1/2) - 1.0`
Expanding and neglecting higher terms, we get
`triangle l = l^2/0.5`
`"Strain" = "Increase in length"/"Original length"`
Let T be the tension in the wire.
∴mg = 2T cosθ
Using the figure, it can be written as:
`cos theta = l/((0.5)^2 + l^2)^(1/2)`
`= l/ ((0.5)(1+(l/0.5)^2)^(1/2))`
Expanding the expression and eliminating the higher terms:
`cos theta = l/((0.5)(1+l^2/(2(0.5)^2)))`
`(1+l^2/(0.5))=~1` for small l
`:. cos theta = l/(0.5)`
`:. T = (mg)/(2(l/(0.5))) = (mgxx 0.5)/(2l) = (mg)/(4l)`
`"Stress" = "Tension"/"Area" = (mg)/(4lxxA)`
`Y = (mg xx 0.5)/(4lxxAxxl^2)`
`l = sqrt((mgxx0.5)/(4YA)) `
Young’s modulus of steel, Y = 2 x 1011 Pa
`:.l = sqrt((0.1xx9.8xx0.5)/(4xx2xx10^11xx0.50xx10^(-6)))`
= 0.0106 m
Hence, the depression at the midpoint is 0.0106 m.
RELATED QUESTIONS
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.
A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Figure. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.
Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 107 Pa? Assume that each rivet is to carry one-quarter of the load.
When the skeleton of an elephant and the skeleton of a mouse are prepared in the same size, the bones of the elephant are shown thicker than those of the mouse. Explain why the bones of an elephant are thicker than proportionate. The bones are expected to withstand the stress due to the weight of the animal.
A steel blade placed gently on the surface of water floats on it. If the same blade is kept well inside the water, it sinks. Explain.
A rope 1 cm in diameter breaks if the tension in it exceeds 500 N. The maximum tension that may be given to a similar rope of diameter 2 cm is
The breaking stress of a wire depends on
A heave uniform rod is hanging vertically form a fixed support. It is stretched by its won weight. The diameter of the rod is
Answer in one sentence.
Define strain.
A charged particle is moving in a uniform magnetic field in a circular path of radius R. When the energy of the particle becomes three times the original, the new radius will be ______.
A rod has a radius of 100 mm and a length of 10 cm. A 100 N force compress along its length. Calculate the longitudinal stress developed in the rod.
A wire is suspended from the ceiling and stretched under the action of a weight F suspended from its other end. The force exerted by the ceiling on it is equal and opposite to the weight.
- Tensile stress at any cross section A of the wire is F/A.
- Tensile stress at any cross section is zero.
- Tensile stress at any cross section A of the wire is 2F/A.
- Tension at any cross section A of the wire is F.
Is stress a vector quantity?
The value of tension in a long thin metal wire has been changed from T1 to T2. The lengths of the metal wire at two different values of tension T1 and T2 are l1 and l2 respectively. The actual length of the metal wire is ______.
A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that g = 3.1πms-2, what will be the tensile stress that would be developed in the wire?
A body of mass m = 10 kg is attached to one end of a wire of length 0.3 m. The maximum angular speed (in rad s-1) with which it can be rotated about its other end in the space station is (Breaking stress of wire = 4.8 × 107 Nm-2 and the area of cross-section of the wire = 10-2 cm2) is ______.
The area of the cross-section of the rope used to lift a load by a crane is 2.5 × 10-4m2. The maximum lifting capacity of the crane is 10 metric tons. To increase the lifting capacity of the crane to 25 metric tons, the required area of cross-section of the rope should be ______.
(take g = 10 ms-2)
What is an elastomer?
The SI unit of stress is ______.
