Question

A mass of \(\mathrm{M} \mathrm{kg}\) is suspended by a weight-less string, the horizontal force that is required to displace it until the string makes an angle of \(60^{\circ}\) with the initial vertical direction is (A) \(\mathrm{Mg} / \sqrt{3}\) (B) \(\mathrm{Mg} \cdot \sqrt{2}\) (C) \(\mathrm{Mg} / \sqrt{2}\) (D) \(\mathrm{Mg} \cdot \sqrt{3}\)

Short answer

The horizontal force required to displace the mass until the string makes an angle of 60 degrees with the initial vertical direction is \(F = \frac{Mg}{\sqrt{3}}\).

Step-by-step solution

Draw the forces diagram

First, draw a diagram with the mass M, the string, and the angles. The forces acting on the mass are tension (T) in the string (that has a vertical and a horizontal component), gravitational force (Mg), and the horizontal force F.

Apply equilibrium of forces

We will analyze the vertical and horizontal forces when the mass is in equilibrium. In the vertical direction, the vertical component of tension balances the gravitational force: \(T_v = Mg\) In the horizontal direction, the horizontal component of tension balances the horizontal force: \(T_h = F\) Since the string makes an angle of 60 degrees with the initial vertical direction, the angle between the string and the horizontal direction is 30 degrees (90 - 60 degrees). Thus, we can find the relation between the vertical and horizontal components of tension using trigonometry.

Relate tension components

We can find the relationship between the vertical and horizontal components of tension using the angle between the string and the horizontal direction: Using sine function, we have: \(T_h = T \cdot \sin(30^{\circ})\) Using cosine function, we have: \(T_v = T \cdot \cos(30^{\circ})\) Now, we know that \(T_v = Mg\), and \(T_h = F\). Therefore, \(Mg = T \cdot \cos(30^{\circ})\) and \(F = T \cdot \sin(30^{\circ})\)

Solve for F

Now, we need to find the value of F in terms of M and g. Divide the equations: \(F = T \cdot \sin(30^{\circ})\) and \(Mg = T \cdot \cos(30^{\circ})\) We obtain: \(F / Mg = \sin(30^{\circ}) / \cos(30^{\circ})\) We know that \(\sin(30^{\circ}) = 1/2\) and \(\cos(30^{\circ}) = \sqrt{3}/2\), so: \(F / Mg = (1/2) / (\sqrt{3}/2)\)

Simplify and find the answer

Finally, simplify the expression for F: \(F = Mg \cdot ((1/2) / (\sqrt{3}/2))\) Multiply both sides by 2: \(F = Mg / \sqrt{3}\) This corresponds to the choice (A) in the problem. Therefore, the horizontal force required to displace the mass until the string makes an angle of 60 degrees with the initial vertical direction is: \(F = \frac{Mg}{\sqrt{3}}\)

Key concepts

Trigonometry in Physics

In physics, trigonometry is often used to break forces into components that make calculations easier. When dealing with angles, such as in this exercise where a string makes an angle with the vertical, trigonometric functions like sine and cosine become very handy.
They help us understand how the tension force in the string has both horizontal and vertical parts. These are called components of the force. By knowing the angle, we can use these functions to find the size of each part:

• Sine function: This gives us the horizontal component of the force. It is written as \(T_h = T \cdot \sin(\theta)\). For our problem, with \(\theta = 30^{\circ}\), we use \(\sin(30^{\circ}) = \frac{1}{2}\).
• Cosine function: This gives us the vertical component of the force. It is written as \(T_v = T \cdot \cos(\theta)\). In this case, \(\cos(30^{\circ}) = \frac{\sqrt{3}}{2}\).

Using these functions allows us to simplify complex physics problems, making it easier to find solutions like the required horizontal force to hold the mass in place.

Tension in Strings

Tension in a string acts along the length of the string and can be broken down into components that align with the directions of the actual forces being applied. When a string is under tension, it has a certain amount of force pulling at each end, keeping the object attached suspended.

In the problem given, the tension (T) in the string has both a vertical component, which balances the gravitational force (Mg), and a horizontal component, which is balanced by the applied force (F). Here’s how it works:

• The vertical component of tension (\(T_v\)) must equal the gravitational force to keep the mass from falling. So, \(T_v = Mg\).
• The horizontal component of tension (\(T_h\)) is what we equate to the applied force, so \(T_h = F\).

Understanding the tension and its components is important for solving problems involving objects held by strings. This helps us see how the sum of forces in both directions leads to equilibrium and how each part works together to keep everything balanced.

Newton's Laws of Motion

Newton's Laws of Motion are a cornerstone of classical physics. They tell us how things move and how forces interact. In this exercise, we focus on Newton's first and second laws, which are key to understanding equilibrium of forces.

• Newton's First Law: An object in equilibrium doesn’t change its motion. If it's at rest, it stays that way unless a net external force acts on it. Here, the mass is at rest because the forces (tension, gravity, and the applied horizontal force) balance each other.
• Newton's Second Law: This law tells us that the acceleration of an object depends on the net force acting on it and its mass. In formula terms, that's \(F = ma\). But in equilibrium (no acceleration), the net force is zero. Hence, the forces in any direction—like horizontal or vertical—must sum to zero.

These laws help us set up the equations that describe how the tension, gravity, and horizontal force come together to keep the mass in equilibrium. Knowing they balance explains why we find the horizontal force \(F = \frac{Mg}{\sqrt{3}}\) for the problem.