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.
(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.
(i) To ensure that the bar remains horizontal, the tension in each wire must balance each other. Let's denote the tension in the steel wire as T_s, and the tension in the brass wire as T_b.
Since the bar is in equilibrium, the sum of the vertical forces on the bar must be zero. Considering the forces acting on the bar, we have:
T_s*sin(theta) = T_b*sin(theta)
Since the angles involved are very small, we can approximate sin(theta) as theta:
T_s*theta = T_b*theta
Therefore, the tension in each wire must be equal: T_s = T_b.
(ii) To calculate the extension of the steel wire and the energy stored in it, we can use Hooke's Law. The extension of an object is given by:
Extension = (Force * Length) / (Young's modulus * Area)
First, we need to calculate the force on the steel wire. The force is equal to the weight of the mass:
Force = mass * gravity
where mass = 10.0 kg and gravity = 9.8 m/s^2.
Next, we need to find the cross-sectional area of the steel wire. The area of a cylindrical shape is given by:
Area = (pi * diameter^2) / 4
Given that the diameter of the steel wire is 0.60 mm, we can convert it to meters by dividing by 1000:
Area = (pi * (0.60/1000)^2) / 4
Now, we can use Hooke's Law to calculate the extension:
Extension = (Force * Length) / (Young's modulus * Area)
Finally, the energy stored in the steel wire can be calculated using the formula:
Energy = (1/2) * (Force * Extension)
(iii) The diameter of the brass wire can be calculated using the same formula as in part (ii), but by substituting the known values for length, mass, force, and Young's modulus for brass. We can then solve for the diameter of the brass wire.
(iv) To ensure that the bar remains horizontal with the new brass wire of diameter 1.00 mm, the tension in both wires must balance each other. Thus, we need to shift the position of the mass on the bar. We can use the concept of torques to find the new position.
The torque on the bar (relative to point A) is given by:
Torque = (Force * distance_from_A_to_mass)
To maintain horizontal equilibrium, the torques produced by each wire must cancel each other out. Therefore, we need to determine the distance from point A to the new position of the mass.
Let's denote this distance as d. Using the above equation, we can find the value of d that satisfies Torque_steel = Torque_brass, given the known tension in the steel wire and the new tension in the brass wire (which can be calculated by using the same formula as in part (ii) with the new diameter of the brass wire).
Once we find the value of d, we can determine the location on the bar where the mass should be suspended to maintain horizontal equilibrium.
To solve this problem, we can use the concepts of equilibrium and Hooke's Law for the extension of a wire.
(i) To determine the tension in each wire, we need to consider the forces acting on the bar at equilibrium. Since the bar remains horizontal, the tension in each wire must balance the weight of the bar and the suspended mass. Let's denote the tension in the steel wire as Ts and the tension in the brass wire as Tb.
Since the bar remains horizontal, the vertical components of the tensions in both wires must balance the weight of the bar and the mass:
2Ts cos(θ) + 2Tb cos(θ) = mg ----(Equation 1)
Where m is the mass (10.0 kg), g is the acceleration due to gravity (9.8 m/s²), and θ is the angle between the wires and the horizontal position of the bar (which is 90 degrees).
Since the bar remains horizontal, the horizontal components of the tensions in the wires must balance each other:
2Ts cos(θ) = 2Tb cos(θ) ----(Equation 2)
Simplifying Equation 2, we can cancel out the common terms:
Ts = Tb
Now substituting Ts = Tb into Equation 1, we get:
4Ts cos(θ) = mg
Simplifying further:
Ts = mg / (4cos(θ))
Substituting the given values:
T = (10.0 kg * 9.8 m/s²) / (4cos(90°))
T ≈ 24.5 N
Therefore, the tension in each wire is approximately 24.5 N.
(ii) To calculate the extension of the steel wire and the energy stored in it, we'll use Hooke's Law, which states that the extension of a wire is directly proportional to the force applied and inversely proportional to its cross-sectional area.
The extension of the steel wire (ΔLs) can be calculated using the formula:
ΔLs = (F * Ls) / (A * Ys) ----(Equation 3)
Where F is the tension in the steel wire (24.5 N), Ls is the length of the steel wire (2.00 m), A is the cross-sectional area of the steel wire, and Ys is the Young's modulus of steel (2.0 x 10^11 Pa).
To calculate the cross-sectional area of the steel wire, we need to first determine its radius. The diameter is given as 0.60 mm, so the radius (rs) is half of that, i.e., 0.30 mm = 0.00030 m.
The cross-sectional area (As) of the steel wire is then given by:
As = π * rs²
Substituting the known values:
As = π * (0.00030 m)²
Now we can substitute the values of F, Ls, As, and Ys into Equation 3 to calculate ΔLs.
The energy stored in the steel wire (Us) can be calculated using the formula:
Us = (1/2) * F * ΔLs
Substituting the calculated ΔLs and the known value of F:
Us = (1/2) * (24.5 N) * ΔLs
(iii) To calculate the diameter of the brass wire, we can use the same approach as for the steel wire.
The diameter of the brass wire (db) is given as "d" (in the question).
The radius (rb) is half of the diameter, so rb = db/2.
The cross-sectional area (Ab) of the brass wire is given by:
Ab = π * rb²
Now we can substitute the known value of Ab and solve for db.
(iv) To determine where the mass should be suspended when the brass wire is replaced by another with a diameter of 1.00 mm, we need to ensure that the tension in both wires remains balanced and the bar remains horizontal.
Using the same approach as in part (i), we set up the equations to balance the vertical and horizontal components of the tensions:
2Ts cos(θ) + 2Tb cos(θ) = mg ----(Equation 4)
2Ts cos(θ) = 2Tb cos(θ) ----(Equation 5)
Since Ts and Tb are equal (as shown in part (i)), we can rewrite Equation 4 as:
4Ts cos(θ) = mg
Substituting the known values, we can calculate Ts (which is also equal to Tb) and then determine where the mass should be suspended to keep the bar horizontal.