Parallel Forces in Equilibrium

PARALLEL FORCES

Statics refers to the bodies in equilibrium. Equilibrium deals with the absence of a net force. When the net equals zero, the forces are in equilibrium provided they are concurrent (they intersect). If they are non-concurrent, the body may rotate even if the vector sum of the forces equals zero. Hence, there must be another condition to set forces in equilibrium – that under the influence of forces, the body must have no tendency toward translational or rotary motion.

An example of non-concurrent forces where the vector sum may be equal to zero but it still causes the body to move is parallel forces. They act in the same or opposite directions. Their lines of action are parallel.

Forces acting in the same or opposite directions are parallel.

TORQUE (MOMENT OF FORCE)

Torque or moment of force refers to the turning effect of the force upon a body about a point (fulcrum). It is the product of the magnitude of the force and perpendicular distance from the line of action of the force to the fulcrum. This perpendicular distance is called moment arm or lever arm.

• The greater the distance from the axis to the point where we apply the force, the greater the torque.
• Maximum torque occurs when the direction of the applied force is perpendicular to a line drawn between the axis and the point where the force is applied.
• When the line and the force are in the same direction, so that the force acts directly toward or away from the axis of rotation, there is no torque.

Mathematical Equation:

Torque = Force X Moment arm

M = F X d

CONDITIONS FOR A BODY TO BE IN EQUILIBRIUM
UNDER THE INFLUENCE OF FORCES

1. The sum of the forces pulling the body in one direction must equal the sum of the forces pulling the body in the opposite direction: ΣF = 0
2. The sum of moments tending to rotate the body clockwise about a point must equal the sum of moments tending to rotate the body counterclockwise about the same point: ΣM = 0

Sample Problem:
1. A boy weighing 225N sits at one end of a seesaw 4m long. If a girl sits opposite him 1.5m away from the fulcrum, they balance each other. Find the weight of girl? (Disregard the weight of the seesaw)

----------------------------------------M girl = M boy
------------------------------------G X 1.5 m = 225N X 2m
---------------------------------------------G =300N

2. A rigid bar 2.8m long is pivoted at a point O, 1.2m from one end where a 20N weight is hung. Find the weight at the opposite end of the bar will produce equilibrium and the force exerted at the pivot point.

----------------------------------------M of W2 = M of W1
-------------------------------------W2 X 1.6 m = 20N X 1.2m
----------------------------------------------W2 = 15N

----------------------------------------P = W1 + W2
----------------------------------------P = 20N + 15N
----------------------------------------P = 35N