Problem 42 - Master Resource Book in JEE Main Physics

Question

A mass of \(10 \mathrm{~kg}\) is suspended from a spring balance. It is pulled by a horizontal string so that it makes an angle of \(60^{\circ}\) with the vertical. The new reading of the balance is \([\) Karnataka CET 2008] (a) \(10 \sqrt{3} \mathrm{~kg}-\mathrm{wt}\) (b) \(20 \sqrt{3} \mathrm{~kg}-\mathrm{wt}\) (c) \(20 \mathrm{~kg}-\mathrm{wt}\) (d) \(10 \mathrm{~kg}-\mathrm{wt}\)

Step-by-Step Solution

Answer

The new reading of the balance is (c) \(20 \mathrm{~kg}-\mathrm{wt}\).

Step 1: Identify the Forces Acting on the Mass

When a mass is suspended by both a vertical spring balance and a horizontal string, the system is in equilibrium. In this case, the vertical force of tension in the spring balance, the horizontal tension in the string, and the weight of the mass must all be considered.

Step 2: Draw the Free Body Diagram

Visualize the forces acting on the mass. The vertical component of the tension is due to the spring balance, the horizontal tension is from the string, and the downward force is from gravity acting on the mass.

Step 3: Analyze Forces in the Vertical Direction

The vertical component of the tension balances the gravitational force. This can be represented as: \( T_v = T \cos \theta = mg \), where \( T \) is the tension in the spring balance, \( \theta \) is the angle, and \( m \) is the mass in kg. With \( \theta = 60^{\circ} \), the equation becomes: \[ T \cos 60^{\circ} = 10 \times 9.8 \]

Step 4: Calculate the Component Forces

Solving for the vertical component of the tension gives us: \( T \times \frac{1}{2} = 98 \) (since \( \cos 60^{\circ} = \frac{1}{2} \)). Thus, \( T = 196 \text{ N} \).

Step 5: Convert Tension to Mass-Wt Reading

The spring balance is a measure of mass, so we convert the tension back to a mass equivalent by dividing the tension by the gravitational acceleration: \( \frac{T}{g} = 196 / 9.8 = 20 \text{ kg} \).

Step 6: Determine the Correct Answer

From the options given, the tension in terms of kg-wt is \( 20 \mathrm{~kg}-\mathrm{wt} \). This matches option (c).

Key Concepts

Free Body Diagram

When we encounter problems in physics involving forces acting on an object, a Free Body Diagram (FBD) becomes a vital tool. It is essentially a visual representation that captures all the forces acting upon the object. In this exercise, we have a mass that is affected by a vertical spring balance and a horizontal string. To represent these, the FBD will show:

• A downward force due to gravity, the weight of the mass, which is given by: \( W = mg = 10 \times 9.8 = 98 \text{ N} \).
• A vertical tension force from the spring balance that counteracts the weight of the object.
• A horizontal tension force exerted by the string.

Depicting this scenario through a Free Body Diagram helps in visualizing how the forces interact and balance each other. This is crucial for understanding how the forces resolve and maintain equilibrium.

Tension in a Spring

The tension in a spring, or more broadly in any supporting mechanism like a balance, plays a vital role in problems involving suspended masses. In equilibrium, this tension must counterbalance the weight of the mass.
In the provided exercise, the mass is inclined, creating an angle of \(60^{\circ}\) with the vertical. Normally, the tension in the spring keeping this mass suspended vertically would need to equal the gravitational force (or weight). However, with the string pulling horizontally, the tension in the spring also has a horizontal component to consider.
The vertical component of the tension is crucial for maintaining equilibrium. We use:\[ T_v = T \cos \theta \]Given \( \theta = 60^{\circ} \), and knowing that the weight \( mg = 98 \text{ N} \), we can calculate the total tension before balancing it out:\[ T \cos 60^{\circ} = 98 \text{ N} \quad \text{thus} \quad T \times \frac{1}{2} = 98 \text{ N} \]Solving gives us \( T = 196 \text{ N} \). This tension needs to be converted back to a mass reading (kg-wt) for the spring scale, giving us the equivalent mass-wt as \(20 \text{ kg-wt}\).

Vertical and Horizontal Force Components

Analyzing the forces acting on an object often involves breaking these forces into their vertical and horizontal components. Understanding these components is essential in establishing how the forces balance each other to maintain equilibrium.
For the given mass scenario, we start with the gravitational force acting vertically downwards. The mass is affected by:

• Vertical force component of tension from the spring, balancing the weight of the mass.
• Horizontal tension due to the string, which affects the total tension the spring must support.

The concept of breaking forces into components simplifies the analysis. For the vertical component, \[ T_v = T \cos \theta \]where \( \theta = 60^{\circ} \), which helps to establish that the vertical force balances the weight.
Similarly, the horizontal component \[ T_h = T \sin \theta \]affects the overall balance of forces.
By analyzing both components, you can ensure that the forces are balanced, confirming the system's equilibrium as the spring indicates the correct mass equivalent.