(i) Since the bar remains horizontal when the mass is suspended, the tension in each wire must be equal. Let the tension in each wire be T. The total downward force acting on the bar is the weight of the suspended mass, given by mg, where m = 10.0 kg and g is the acceleration due to gravity (approximately 9.81 m/s²). The total upward force acting on the bar is the sum of the tensions in the two wires, which is 2T. Since the bar is in equilibrium, these forces must balance each other:
2T = mg
T = mg/2
T = (10.0 kg)(9.81 m/s²)/2
T ≈ 49.05 N
So the tension in each wire is approximately 49.05 N.
(ii) To calculate the extension of the steel wire, we can use the formula for the extension of a wire under tension:
ΔL = (FL)/(AE)
Where ΔL is the extension, F is the force acting on the wire (which is the tension T), L is the length of the wire, A is the cross-sectional area, and E is the Young's modulus. For the steel wire, we have:
L = 2.00 m
Diameter (d) = 0.60 mm = 0.60 × 10⁻³ m
Thus, the radius (r) = d/2 = 0.30 × 10⁻³ m
A = πr² = π(0.30 × 10⁻³ m)² ≈ 2.827 × 10⁻⁷ m²
E (steel) = 2.0 × 10¹¹ Pa
ΔL (steel) = (49.05 N)(2.00 m)/[(2.827 × 10⁻⁷ m²)(2.0 × 10¹¹ Pa)] ≈ 1.735 × 10⁻³ m
ΔL (steel) ≈ 1.735 mm
The extension of the steel wire is approximately 1.735 mm.
To calculate the energy stored in the steel wire, we can use the formula for the energy stored in a wire under tension:
E = (1/2)FΔL
E (steel) = (1/2)(49.05 N)(1.735 × 10⁻³ m) ≈ 0.0425 J
The energy stored in the steel wire is approximately 0.0425 J.
(iii) Since the tension in the brass wire is also 49.05 N, we can use the formula for the extension of a wire under tension to relate its diameter to the given values:
ΔL (brass) = (FL)/(AE)
We know the tension in the brass wire (T) is equal to the tension in the steel wire, so we have:
F = 49.05 N
E (brass) = 1.0 × 10¹¹ Pa
Since the ratio of the Young's moduli and the extensions of the wires are equal (i.e., E_brass/E_steel = ΔL_brass/ΔL_steel), we can write:
ΔL (brass) = (ΔL (steel))(E (brass)/E (steel))
ΔL (brass) ≈ (1.735 × 10⁻³ m)(1.0 × 10¹¹ Pa)/(2.0 × 10¹¹ Pa) ≈ 0.8675 × 10⁻³ m
ΔL (brass) ≈ 0.8675 mm
Now, we can use the formula for the extension of a wire under tension and solve for the diameter (d) of the brass wire:
A (brass) = (FL)/(EΔL)
π(d/2)² = (49.05 N)(2.00 m)/(1.0 × 10¹¹ Pa)(0.8675 × 10⁻³ m)
d² = (4(49.05 N)(2.00 m))/(π(1.0 × 10¹¹ Pa)(0.8675 × 10⁻³ m)) ≈ 1.104 × 10⁻⁶ m²
d ≈ 1.05 × 10⁻³ m
d ≈ 1.05 mm
The diameter of the brass wire is approximately 1.05 mm.
(iv) If the brass wire is replaced with another brass wire of diameter 1.00 mm, we need to calculate where the mass should be suspended along the bar so that the bar remains horizontal.
To do this, we first calculate the value of ΔL (brass) with the new diameter, using the formula for the extension of a wire under tension:
A (brass) = π(1.0 × 10⁻³ m/2)² ≈ 7.854 × 10⁻⁷ m²
ΔL (new brass) = (49.05 N)(2.00 m)/[(7.854 × 10⁻⁷ m²)(1.0 × 10¹¹ Pa)] ≈ 0.9975 × 10⁻³ m
ΔL (new brass) ≈ 0.9975 mm
Since the ratio of the extensions is equal to the ratio of the distances of the mass from the brass wire (x) and the distance between the two wires (0.20 m), i.e., ΔL(new brass)/ΔL(steel) = x/(0.20 m - x), we can calculate the value of x:
x/(0.20 m - x) = (0.9975 × 10⁻³ m)/(1.735 × 10⁻³ m)
x ≈ 0.1104 m
x ≈ 110.4 mm
So the mass should be suspended approximately 110.4 mm from the brass wire to maintain equilibrium with the new brass wire of diameter 1.00 mm.
A light rigid bar is suspended horizontally from two vertical wires, one of steel and one of brass. Each wire is 2.00m long. The diameter of the steel wire is 0.60mm and the length of the bar AB is 0.20m. When a mass of 10.0kg is suspended from the centre of AB the bar remains horizontal. (i)what is the tension in each wire? (ii)calculate the extension of the steel wire and the energy stored in it. (iii)calculate the diameter of the brass wire. (iv)if the brass wire were replaced by another brass wire of diameter 1.00mm, where should the mass be suspended so that AB would remain horizontal? The young modulus for steel= 2.0 x 10^11 Pa, the young modulus for brass=1.0 x 10^11 Pa.
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