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.
To solve this problem, we will need to apply the principles of equilibrium, Hooke's Law, and the formula for calculating stress and strain.
(i) Tension in each wire:
Since the bar remains horizontal, the tension in both wires must balance each other. Let's assume T1 is the tension in the steel wire and T2 is the tension in the brass wire.
Since the system is in equilibrium, the sum of the vertical forces acting on it must be zero. In this case, the weight of the mass (10.0 kg) is the only vertical force. So, T1 + T2 = mg ----(Equation 1)
(ii) Extension of the steel wire and energy stored:
According to Hooke's Law, the extension of a wire is directly proportional to the applied force. The equation to calculate the extension (ΔL) of a wire can be given as:
ΔL = (F * L) / (A * Y)
Where:
F is the force applied (weight of the mass)
L is the original length of the wire
A is the cross-sectional area of the wire
Y is the Young's modulus of the wire
Using this formula, we can calculate the extension of the steel wire (ΔLsteel) and the energy stored in it (Esteel).
Extension of steel wire (ΔLsteel) = (F * Lsteel) / (Asteel * Ysteel) ----(Equation 2)
Energy stored in steel wire (Esteel) = (1/2) * (F * ΔLsteel) ----(Equation 3)
(iii) Diameter of the brass wire:
Assuming the cross-sectional shape of the brass wire is circular, we can calculate its diameter (dbrass) using the formula for the cross-sectional area of a circle:
A = π * (d^2) / 4
Where:
A is the cross-sectional area
d is the diameter of the wire
(iv) Position of mass for horizontal bar with a different brass wire:
If we replace the brass wire with another brass wire of a different diameter, we need to find the new position where the mass should be suspended to maintain the horizontal position of the bar.
Let's assume x is the distance from point A to the new position where the mass should be suspended.
To maintain the horizontal position, the clockwise and anticlockwise torques acting on the bar must balance each other.
The torque due to the tension in the steel wire (T1) is T1 * Lsteel * sinθ, and the torque due to the tension in the brass wire (T2) is T2 * (Lsteel - x) * sinθ. Here, θ is the angle between the steel wire and the horizontal.
Therefore, T1 * Lsteel * sinθ = T2 * (Lsteel - x) * sinθ
From Equation 1, we know that T1 = T2, so we can simplify the equation to:
Lsteel * sinθ = (Lsteel - x) * sinθ
By simplifying further, we get:
Lsteel = Lsteel - x
This means x = 0. Therefore, the position of the mass should be at point A (the left end of the bar) in order to keep the bar horizontal after replacing the brass wire.
I hope this explanation helps in solving the problem.
Yes
Tension = force by area th
i)50N
ii)0.0018, 0.044J
iii)0.85mm
iv)0.084m from B