Question

A man carries a weight tied at the end of a stick \(\frac{3}{2} m\) long, which is placed on his shoulder. He holds the other end of the stick with his hand. In order that the force on his shoulder may be three times the weight hanging at the end, the distance between his hand and the shoulder should be (neglect weight of the stick) (1) \(0.25 \mathrm{~m}\) (2) \(0.3 \mathrm{~m}\) (3) \(0.4 \mathrm{~m}\) (4) \(0.5 \mathrm{~m}\)

Short answer

The distance is 0.5 m.

Step-by-step solution

Understand the Forces Involved

Identify the forces acting on the stick. There is the weight at the end of the stick denoted as \( W \). The man's shoulder exerts a force \( F_s \) and his hand exerts a force \( F_h \). Given: \( F_s = 3W \).

Set up the Equilibrium Condition

Since the system is in equilibrium, sum of the forces acting on the stick should be zero. Thus: \( W + F_s + F_h = 0 \). Also, sum of the moments about any point should be zero.

Choose the Point of Moments Calculation

Calculate the moments about the man's hand for simplicity. The total length of the stick is \( 1.5 \, \text{m} \). Denote the distance between the man's shoulder and hand as \( d \).

Write the Moment Equation

Using the fact that sum of the clockwise moments equals the sum of counter-clockwise moments: \( W \times 1.5 \, \text{m} = F_s \times d \).

Substitute the Given Force Ratio

Since \( F_s = 3W \), substitute this into the moment equation: \( W \times 1.5 \, \text{m} = 3W \times d \).

Solve for Distance \( d \)

Simplify the equation to solve for \( d \): \( 1.5 \, \text{m} = 3d \Rightarrow d = \frac{1.5 \, \text{m}}{3} = 0.5 \, \text{m} \).

Key concepts

Equilibrium of forces

In physics, equilibrium of forces occurs when all the forces acting on a system are balanced. This means the net force, or the sum of all individual forces, is zero. In our problem, the stick is in equilibrium because it's not accelerating or rotating.
There are three key forces to consider:

• The weight W bonded to the stick's end.
  
• The force F_h exerted by the man’s hand.
  
• The force F_s exerted by the man’s shoulder.
  

The given condition tells us that the shoulder force F_s is three times the weight (F_s = 3W). To maintain equilibrium, the sum of forces upward must equal the sum of forces downward. Hence, we write: W + F_s + F_h = 0. This ensures the stick remains stationary and highlights how forces balance during equilibrium.

Moments of forces

The concept of moments (or torque) refers to the rotational effect produced by a force acting at a distance from a pivot point. In simple terms, a moment is the force multiplied by the distance from the point of rotation.
L= F×d.
Here, our pivot point is the man's hand holding the stick. You calculate the moments around this point to simplify the problem.
Knowing the stick's total length (1.5 m), and denoting the distance from his shoulder to hand as d, we set up the equation:
W × 1.5 m = F_s × d.
Replacing F_s with 3W (since F_s = 3W):
W × 1.5 m = 3W × d.
By solving this simple linear equation, you find d = 1.5 m / 3 = 0.5 m.
This indicates that to achieve the specified force condition, the man's hand should be 0.5 meters from his shoulder.

Physics problem-solving

Approaching physics problems systematically helps to break down complex situations into manageable parts. Here’s a simple checklist you can use:

• Identify all forces and their directions.
  
• Write down the equilibrium conditions (sum of forces and moments equal zero).
  
• Select a suitable pivot or reference point to calculate moments.
  
• Substitute known values and solve the equations.
  

In our example, you start by determining all forces acting on the stick: the weight, the shoulder force, and the hand force. By setting up equilibrium conditions and choosing a pivot point wisely, you ensure calculations are straightforward. This methodical approach can be applied universally to various physics problems, making complex problems seem less daunting. By following these steps, you develop a deeper understanding and solve problems more efficiently.