Consider an ideal lever as shown in the figure. An ideal lever is a light rigid rod of negligible mass that is pivoted at a point along its length, known as the fulcrum. A common example of a lever system is a seesaw used by children in parks. In this system, two forces, F1 and F2 act on the lever at distances d1 and d2, respectively, from the fulcrum. These forces produce moments about the pivot, determining the rotational equilibrium of the system.
Assume that the lever system is in equilibrium. Let R be the reaction force at the pivot (fulcrum), acting opposite to the applied forces F1 and F2. Since the system is in both translational and rotational equilibrium, the following conditions must be satisfied:
For translational equilibrium:
R - F1 - F2 = 0 ... (1)
For rotational equilibrium:
d1F1 - d2F2 = 0 ... (2)
(Here, one direction of moment is taken as positive and the other as negative.)
Equation (2) can be written as:
d1F1 = d2F2
This is known as the principle of moments for the lever system, which states that the moment produced by one force is balanced by the moment produced by the other force.
The ratio of forces determines the mechanical advantage (M.A.) of the system, given by:
M.A. = F₂ / F₁
Formally, the principle of moments has been defined as,
When a number of forces act on a rigid body and the body is in equilibrium, the sum of clockwise moments is equal to the sum of anticlockwise moments about the same point. In other words, the algebraic sum of all moments acting on the body is zero.
Sample Problems
Question 1: In the figure given below, the distance and forces are given as d1 = 1 m, d2 = 2 m, and F1 = 2 N. Find the value of F₂.
Answer:
Since the body is in equilibrium, the algebraic sum of moments must be zero.
d1F1 - d1F2= 0
Given: d1 = 1m, d2 = 2m, F1 = 2N and F2 = ?
plugging these values in the equation,
d1F1 - d1F2= 0
⇒ (1)(2) - (2)(F2) = 0
⇒ F2 = 1N
Question 2: In the figure given below, the distance and forces are given as d1 = 1.5m, d2 = 5 m, and F1 = 10 N. Find the value of F₂.
Answer:
Since the body is in equilibrium, the algebraic sum of moments must be zero.
d1F1 - d1F2= 0
Given: d1 = 1.5m, d2 = 5m, F1 = 10N and F2 = ?
plugging these values in the equation,
d1F1 - d1F2= 0
⇒ (1.5)(10) - (5)(F2) = 0
⇒ F2 = 3N
Question 3: In the figure given below, the distance and forces are given as d1 = 1.5m, d2 = 2 m, d3 = 5 m, F1 = 10 N, F2 = 5 N, and F3 = 4 N. Find out whether the system will be in rotational equilibrium or not.
Answer:
To analyze the equilibrium, principle of moment is used.
Calculating the moments for the system
d1F1 + d2F2 - d3F3
⇒ (1.5)(10) + (2)(5) - (5)(4)
⇒ 15 + 10 -20
⇒ 5N
Since the total sum of moments is not zero. The system is not in rotational equilibrium.
Question 4: In the figure given below, the distance and forces are given as d1 = 5m, d2 = 4m, d3 = 10m, F1 = 10N F2 = 5N F3 = 4N. Find out whether the system will be in rotational equilibrium or not.
Answer:
To analyze the equilibrium, principle of moment is used.
Calculating the moments for the system
d1F1 + d2F2 - d3F3
⇒ (5)(10) + (4)(5) - (10)(4)
⇒ 50 + 20 -40
⇒ 30
Since the total sum of moments is not zero. The system is not in rotational equilibrium.
Question 5: A 3m ladder of weight 20 kg is leaning on a frictionless wall. Its feet rest on the floor 1m from the wall. Find the reaction forces from the wall and the floor.
Answer:
Using Pythagoras theorem,
QR = 2√2m
Ladders weight acts from the center of the ladder, force from flow is F2 and from the wall is F1.
F2. is decomposed into normal force N and friction F.
Translational equilibrium,
In the horizontal direction, F- F1. = 0
In the vertical direction, N - W = 0
N = W
⇒ N = (20)(9.8)
⇒ N = 196 N
For rotational equilibrium around point A,
2√2F - (W/2) = 0
⇒ F = W/2√2
F = 34.6N = F1
F2 =
\sqrt{F^2 + N^2} F2 = 199 N
Unsolved Problems
Question 1: A uniform beam of length 6 m is supported at its ends. A load of 30 N is placed 2 m from one end. Find the reactions at the supports.
Question 2: A lever is in equilibrium with forces acting at distances 0.5 m and 1.5 m from the fulcrum. If one force is 20 N, find the other force.
Question 3: A ladder of length 4 m rests against a smooth wall, making an angle of 45° with the ground. If its weight is 100 N, find the reaction at the wall.
Question 4: Three forces of 10 N, 15 N, and 25 N act on a rigid body at distances 2 m, 3 m, and 4 m, respectively, from a point. Check whether the body is in rotational equilibrium.
Question 5: A uniform rod of length 2 m and weight 50 N is supported at one end. A weight is hung at the other end to keep it horizontal. Find the value of the weight required.