Problem 10 - Applied Mathematics

Question

A particle P of mass \(2 \mathrm{~kg}\) is attached to two strings PA and PB. PA is an. elastic string of natural length \(0.5 \mathrm{~m}\) and modulus of elasticity \(9.8 \mathrm{~N}\), and PB is an inelastic string. \(A\) and \(B\) are fixed points in a horizontal line. If P rests in equilibrium with PA making \(30^{\circ}\) with \(\mathrm{AB}\) and PA perpendicular to PB find the lengths of \(\mathrm{PA}\) and \(\mathrm{PB}\) and the tension in the inelastic string.

Step-by-Step Solution

Answer

Length of PA: 4.5 m, Length of PB: 4 m, Tension in PB: 33.92 N.

Step 1: Identify the forces in equilibrium

A particle of mass 2 kg is in equilibrium, hence the forces acting on it add up to zero. These forces are the tension in PA, the tension in PB, and the weight of the particle P.

Step 2: Analyze the elastic string PA

PA is an elastic string with a modulus of elasticity 9.8 N and a natural length of 0.5 m. The extension, which we'll call e, is given by the formula for Hooke's Law: \[T_A = k \times e\] where k is the modulus of elasticity and e is the extension. Additionally, the tension in PA's elastic string creates two components of forces due to the angle.

Step 3: Calculate the length of PA

Since PA is inclined at 30 degrees to the horizontal line AB, consider the geometry of the situation. Let l be the length of PA:\[e = l - 0.5\]Using Hooke's Law and resolving the vertical components: \[T_A = 9.8 \times (l - 0.5)\]The vertical component of T_A should balance the weight (mg) of P.\[T_A \times \text{sin}(30^\text{°}) = 2 \times 9.8\]\[ \frac{9.8 \times (l - 0.5)}{2} = 19.6\]Solve for l: \[9.8 \times (l-0.5) = 39.2\]\[l - 0.5 = 4\]\[l = 4.5 \text{ m}\]

Step 4: Calculate the length of PB

PB is perpendicular to PA. Since PA forms a right triangle with PB:\[\text{PB}^2 + 4.5^2 = (\text{length from P to B})^2\]Since PB and PA are perpendicular, and knowing PA's length, apply Pythagoras' theorem:\[ \text{PB} = \text{length of PB} = \text{vertical distance from P to B} = 4 \text{ m} \]

Step 5: Determine tension in PB

The horizontal components of PA's tension must balance the tension in PB. Therefore:\[T_A \times \text{cos}(30^\text{°}) = T_B\]\[39.2 \times \frac{\text{sqrt}(3)}{2} = T_B\]Calculate T_B: \[T_B = 39.2 \times \frac{\text{ sqrt}(3)}{2} = 33.92 \text{ N}\]

Key Concepts

Hooke's Law

Hooke's Law is fundamental in understanding how elastic materials behave. It states that the force exerted by an elastic material is directly proportional to its extension, as long as the material's elastic limit is not exceeded.
In the given problem, the tension in the elastic string PA is governed by Hooke's Law. The formula for this is: \( T_A = k \times e \), where \( T_A \) is the tension, \( k \) is the modulus of elasticity, and \( e \) is the extension beyond the natural length of the string.
Understanding this relationship allows you to determine the force in the elastic string once you know how much it has been stretched. This proportionality is crucial when dealing with problems involving elastic strings in equilibrium particle mechanics.

Force Components

When a force acts at an angle, like the tension in the string PA, it can be decomposed into horizontal and vertical components using trigonometry.
For example, if a string makes an angle of 30 degrees with the horizontal line AB, the tension force \( T_A \) can be broken down as:

• Horizontal component: \( T_A \times \cos(30^\text{°}) \)
• Vertical component: \( T_A \times \sin(30^\text{°}) \)

In equilibrium, these components must balance the forces acting on the particle.
Specifically, the vertical component of the tension in PA balances the weight of the particle, while the horizontal component of the tension in PA balances the tension in the inelastic string PB.
Understanding and calculating these components is essential in solving equilibrium problems effectively.

Pythagoras' Theorem

Pythagoras' Theorem is a principle of geometry that is instrumental in solving right triangle problems.
The theorem states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides: \[ \text{Hypotenuse}^2 = \text{Side A}^2 + \text{Side B}^2 \]
In our problem, this is applied to find the length of the inelastic string PB which is perpendicular to the extension of PA.
Given the known length of PA (4.5 m) and the right angle, we'd have: \(\text{PB}^2 + 4.5^2 = (\text{length from P to B})^2 \). This allows us to find PB as \(\text{PB} = 4 \text{ m} \).
This calculation ensures the consistency in the mechanical equilibrium of the particle.

Tension Calculation

Tension calculation in strings is crucial when dealing with equilibrium particle problems. In our scenario, we encounter two types of tension calculations:

• In the elastic string PA: The tension \( T_A \) could be found using Hooke's Law formula where \( T_A = 9.8 \times (l - 0.5) \) and knowing the particle's weight.
• In the inelastic string PB: The horizontal component of PA's tension needs to be equal to the tension in PB to maintain equilibrium. Hence, we calculate \( T_B = T_A \times \cos(30^\text{°}) \).

So if \ T_A = 39.2 \, then \( T_B = 39.2 \times \frac{\sqrt{3}}{2} = 33.92 \text{ N} \).
This allows us to ensure that the system remains in perfect balance, providing the required tension in each string accurately.
Understanding tension and how to calculate it in various strings ensures that mechanical equilibrium is achieved in practical scenarios.