Q.A metal bar 70 cm long and 4.00 kg in mass supported on two knife-edges placed 10 cm from each end. A 6.00 kg load is suspended at 30 cm from one end. Find the reactions at the knife-edges. (Assume the bar to be of uniform cross section and homogeneous.)

The problem describes a rigid body (the metal bar) that is at rest and not accelerating, which means it is in static equilibrium. For static equilibrium, two conditions must be satisfied: the net force acting on the body must be zero, and the net torque about any point must be zero.

Let's define the forces and their positions from one end of the bar (say, the left end):

• Length of bar $L = 70 \text{ cm} = 0.70 \text{ m}$.
• Mass of bar $M_b = 4.00 \text{ kg}$, so its weight $W_b = M_b g = 4.00g \text{ N}$. This acts at the center of the bar, $0.35 \text{ m}$ from the left end.
• Mass of load $M_L = 6.00 \text{ kg}$, so its weight $W_L = M_L g = 6.00g \text{ N}$. This acts at $0.30 \text{ m}$ from the left end.
• Let $R_1$ and $R_2$ be the upward reaction forces at the knife-edges.
  • $R_1$ is at $0.10 \text{ m}$ from the left end.
  • $R_2$ is at $0.70 \text{ m} - 0.10 \text{ m} = 0.60 \text{ m}$ from the left end. We use $g = 9.8 \text{ m/s}^2$.

1. Sum of vertical forces is zero:

   The upward forces must balance the downward forces.

   $R_1 + R_2 - W_b - W_L = 0$

   $R_1 + R_2 = (4.00g) + (6.00g) = 10.00g = 10.00 \times 9.8 = 98.0 \text{ N}$ (Equation 1)

2. Sum of torques about any point is zero:

   To find $R_1$ and $R_2$, we choose a pivot point. Selecting the position of $R_1$ as the pivot ($0.10 \text{ m}$ from the left end) eliminates $R_1$ from the torque equation.

   • Distance of $W_b$ from $R_1$: $0.35 \text{ m} - 0.10 \text{ m} = 0.25 \text{ m}$ (clockwise torque).
   • Distance of $W_L$ from $R_1$: $0.30 \text{ m} - 0.10 \text{ m} = 0.20 \text{ m}$ (clockwise torque). …

Figure 6.26 is a free-body diagram of a rigid rod in static equilibrium. The rod AB rests horizontally on two knife-edges, labelled K₁ and K₂. These supports are narrow, so they exert only vertical forces on the rod — there is no horizontal friction or binding. The upward reaction forces at K₁ and K₂ are labelled R₁ and R₂ respectively.

The rod’s own weight, W, acts vertically downward at its centre of gravity G. In addition, a load of weight W₁ is suspended from the rod at some point between the supports. The figure shows all these forces as arrows: R₁ and R₂ pointing up, W and W₁ pointing down. The distances between the points of application — from K₁ to G, from G to K₂, and from K₁ to the point where W₁ hangs — are marked along the rod.

The physical idea is that for a rigid body to be in complete equilibrium, two conditions must hold simultaneously. First, the net force on the body must be zero — the sum of all upward forces must equal the sum of all downward forces. Second, the net torque about any point must be zero — the clockwise and anticlockwise moments must balance. This figure is the textbook’s vehicle for teaching exactly that.

From the figure, the force-balance equation is straightforward:

The torque-balance equation is written by choosing a convenient axis — usually one of the knife-edges, because the reaction at that point then produces zero torque about itself. If we take torques about K₁, the clockwise moments come from W and W₁ (each acting at their respective distances from K₁), and the anticlockwise moment comes from R₂ (acting at the full distance from K₁ to K₂). Setting the sum equal to zero gives:

where $d_{12}$ is the distance between K₁ and K₂, $d_{1G}$ is the distance from K₁ to G, and $d_{1L}$ is the distance from K₁ to the point where W₁ is suspended. A similar equation can be written about K₂.

The figure thus makes concrete the abstract idea that equilibrium is a double condition — forces and torques — and that the choice of axis for torque is arbitrary but strategically chosen to simplify algebra. The rod, the knife-edges, and the suspended load are not just a picture; they are a worked example of how to apply $\sum F = 0$ and $\sum \tau = 0$ to a real rigid body.

When an object is at rest and remains at rest, it is said to be in static equilibrium. For an object to be in static equilibrium, two fundamental conditions must be met:

1. Translational Equilibrium: The net force acting on the object must be zero. This means the sum of all forces in any direction (e.g., vertical, horizontal) must be zero. If the net force were non-zero, the object would accelerate.
2. Rotational Equilibrium: The net torque acting on the object about any point must be zero. Torque is the rotational equivalent of force; if the net torque were non-zero, the object would undergo angular acceleration (start rotating).

In this problem, the metal bar is supported and remains stationary, so we can apply these two conditions to find the unknown reaction forces.

Let's break down the solution step-by-step. We will use $g = 9.8 \text{ m/s}^2$ for the acceleration due to gravity.

1. Visualize and Draw a Free-Body Diagram (FBD)

   First, it's essential to visualize the setup and draw a clear diagram showing all forces acting on the bar and their respective positions. Let's define the left end of the bar as the origin ($x=0$).

   • Bar length: $L = 70 \text{ cm} = 0.70 \text{ m}$.
   • Bar mass: $M = 4.00 \text{ kg}$.
   • Weight of the bar ($W_B$): Since the bar is uniform and homogeneous, its center of mass is at its geometric center. $W_B = Mg = 4.00 \text{ kg} \times 9.8 \text{ m/s}^2 = 39.2 \text{ N}$. This force acts downwards at $x_{CM} = L/2 = 70/2 = 35 \text{ cm} = 0.35 \text{ m}$.
   • Load mass: $m = 6.00 \text{ kg}$.
   • Weight of the load ($W_L$): $W_L = mg = 6.00 \text{ kg} \times 9.8 \text{ m/s}^2 = 58.8 \text{ N}$. This force acts downwards at $x_L = 30 \text{ cm} = 0.30 \text{ m}$ from one end (let's assume the left end).
   • Knife-edges: They are placed 10 cm from each end.
     • Knife-edge 1 (left): Position $x_1 = 10 \text{ cm} = 0.10 \text{ m}$. It exerts an upward reaction force $R_1$.
     • Knife-edge 2 (right): Position $x_2 = 70 - 10 = 60 \text{ cm} = 0.60 \text{ m}$. It exerts an upward reaction force $R_2$.

   Here's a summary of forces and their positions from the left end:

   • $R_1$ (upwards) at $x = 0.10 \text{ m}$
   • $W_L = 58.8 \text{ N}$ (downwards) at $x = 0.30 \text{ m}$
   • $W_B = 39.2 \text{ N}$ (downwards) at $x = 0.35 \text{ m}$
   • $R_2$ (upwards) at $x = 0.60 \text{ m}$

2. Apply the Condition for Translational Equilibrium

   For the bar to be in vertical equilibrium, the sum of all upward forces must equal the sum of all downward forces.

   Let's consider upward forces as positive and downward forces as negative.

This equation has two unknowns, $R_1$ and $R_2$, so we need another equation. …

Concept: Static Equilibrium of a Rigid Body (Translational + Rotational)

Step 1: Set up the free-body diagram

Take the left end as $x=0$. With $g=9.8\text{ m/s}^2$:

• Bar weight $W_B = 4.00\times9.8=39.2$ N, acting at the centre, $x=0.35$ m.
• Load weight $W_L = 6.00\times9.8=58.8$ N, acting at $x=0.30$ m.
• Knife-edge reactions $R_1$ (up) at $x=0.10$ m and $R_2$ (up) at $x=0.60$ m.

Step 2: Apply translational equilibrium ($\sum F_y=0$)

Step 3: Apply rotational equilibrium about $R_1$ (eliminates $R_1$ from the torque equation)

$R_2(0.50) = W_L(0.20) + W_B(0.25)$ …