UNIT 4
Conditions for Static Equilibrium
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Mechanical Properties of Solids
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6

 

 

Conditions for Static Equilibrium

Objects in Equilibrum

— The linear momentum (momentum)of its center of mass is zero.

— Its angular momentum (momentum_angular)about its center of mass, or about any other point, is also zero.

Recall Newton's first law about stationary objects ⇒ if the object is stationary (at rest) all forces on that object are balanced.

Sum of the forces = 0

Sum of the torques = 0

Which of the following are in Static Equilibrum?

— A vase resting on a desk
— A ball rolling on a horizontal floor
— The propellor of a plane rotating
— A wheel spinning in a machine

Stable Equilibrum

— Object returns to a state of static equilibrium after it has been displaced. e.g. stable

Unstable Equilibrum

— Static equilibrium of object ends by a small force. e.g. unstable

Conditions for Equilibrium

Newton's second Law of motion can be written as f = dpdt

For translational equibibrum f = 0
so dP = 0 as P is constant.

For rotational equilibrium taunet = 0
so dLdt = 0 as P is constant.

To summarize: For a body to be in equilibrium

1. The vector sum of all the external forces that act on the body must be zero.
2. The vector sum of all external torques that act on the body, measured about any possible point, must also be zero.
3. The linear momentum of the body must be zero.
4. angular momentum about its center of mass, or about any other point, is also constant

Solving Problems involving Equilibrium.

Motion can be:

— Translation, i.e displacement from on point to another. E.g. a car driving on a highway.
— Spin - i.e. rotation about a axis that is within the body. E.g. a fan blade.
— Orbital, i.e. rotation about a axis that is not within the body. E.g. the earth going around the sun.

Equilibrium can be of two kinds of equilibrium - Static and Dynamic (or Kinetic):

For static equilibrium there is no motion with respect to the observer so all forces must be balanced and there must be no torque. E.g. a painting hanging on the wall.
— For dynamic equilibrium there can be motion with respect to the observer but all all forces must still be balanced and there must be no torque. This can only happen when the object has constant velocity.

Since there is no resultant force on the object it's acceleration must be ZERO.

Some Pointers for Solving Problems.

— When there are three coplanar forces, the lines of action of the three forces must all pass through the same point. This is because of the condition that there be no resultant or net torque. This only holds for three coplanar forces.
— The principle of moments: the sum of all the clock-wise moments about any point must have the same magnitude as the sum of all the anti-clockwise moments about the same point. Remember anticlockwise moments are positive and clocwise moments are considered to be negative.
— Identify two perpendicular planes (usually vertical and horizontal) and resolve vectors in these planes. They can then be treated independentally as a vector can have no component in a perpendicular plane so then the sum of vectors must be zero in each plane . In inclined planes it may be easier to use the plane of the incline and a plane perpendicular to this.

Consider a a uniform ladder leaning against a smooth vertical wall.
static eq static eq static eq

Inspection of the force diagram may give some indication of the way to proceed.

In (ii) we can see that Ff= W so as W is known we can use this approach to get Ff and then Fg.
Form here it is easy to calculate Fw.

sineOn the othere hand if we use (iii) we can use the law of Sines to calculate Ff and Fg if θ is known.

 

Another approach might be to take moments about point B (point where Fg originates). Then we would equate two torques so if the perpendicular distance from these forces are known and one of the forces is know we can find the other.

So in the final analysis the ease of the solution depends on what data is given.


The forces on the ladder are:
1. Fw at the top (there is only a normal reaction from the wall as there is no friction on the wall.
2. W, the weight acting vertically downwards and at the centre of the ladder as it is uniform.
3. Fg at the ground level. This is really the combination of two forces, friction and the normal reaction from the floor.

The forces Fgn and Ff are components of Fg and as such:
Fgn = Fg Sin θ and
Ff= Fg Cos θ

 

 

 


In (i) we can treat the ladder as a point object and draw this force diagram.
In (ii) we also treat the ladder as a point object and draw this force diagram.
In (iii) we apply the principle that the sum of all the forces is zero so they form a closed polygon (in this case a triangle) when added, i.e. drawn connecting head to tail.
   
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Concept by Kishore Lal. Programmed by Kishore Lal... Copyright © 2015 Kishore Lal. All rights reserved.