Mechanics Unit 5: Motion and Forces 5.5 Mechanical Equilibrium

Mechanics

Unit 5: Motion and Forces

5.5 Mechanical Equilibrium

...

Fundamentals of physics - Mechanics Mechanical Equilibrium

---------------------------------------------------------------------------------------- The two masses on the spring balance on the right have weights of 20N each, and the hanger 40N. Since weight always act towards the centre of the Earth, the net force on the spring balance is 20N + 20N + 40N = 80N When the spring balance settles down the masses are in equilibrium (no movement).To obey Newton’s First law of motion (inertia) then the weight is balanced by a force pointing away from the center of the Earth of the same magnitude called a tension on the balance.


----------------------------------------------------------------------------------------

So on the masses there are two forces which are equal in magnitude but opposite in direction Newton's first law in a then about mechanical equilibrium. We can write this as F = 0, where the stands for the sum of...


----------------------------------------------------------------------------------------You have already looked at two cases where horizontal forces were in equilibrium! Look around you! All the stationary objects are in mechanical equilibrium so that all the forces on them

are balanced or no net force (F = 0)


----------------------------------------------------------------------------------------

So now let’s investigate cases where there are vertical forces that are balanced or no net force

(Fv = 0). Remember the spring balance Wasn’t Fup = Fdown?So you have actually looked at one case so far... Look at the person hanging onto the rope and answer the following questions. Assume he/she is not moving (in equilibrium).


----------------------------------------------------------------------------------------What are the forces on the person? Fw = his weight or the force with which the earth pulls him towards its centre. Ft = the tension on the rope pulling him upwards.Since he is in Equilibrium by Newton’s First Law:

F = 0

Fw + Ft = 0

Fw = - Ft They are of the same

magnitude but opposite directions.


----------------------------------------------------------------------------------------

Look at the girl hanging by her hands from the cross bar.What are the forces on the girl?Fw = her weight or the force with which the earth pulls her towards its centre.Fleft = the reaction of the crossbar on her left hand pushing her upwards. Fright = the reaction of the crossbar on her left hand pushing her upwards.Since she is in Equilibrium by

Newton’s First Law: F = 0


----------------------------------------------------------------------------------------

Fleft + Fright + Ft = 0

Fw = - (Fleft + Fright) The total forces on the hands are of the same magnitude but opposite direction to her weight.

Question: Can you make any further statement about the forces on the hands?

There is no guarantee that both hands experience equal forces. The most we can say is that the sum of the magnitudes is equal to the magnitude of her weight


----------------------------------------------------------------------------------------The window washer on the left is on a platform supported by two cables which have spring balances. Ignore the weight of the plank. The man has a weight of 1000N.(a) If the spring balance on the left of reads 0N, what is value of F1up?(b) When the man stands in the middle of the platform what is the reading F2up?(c) If the left spring balance reads 170N, What is the value of F3up?


----------------------------------------------------------------------------------------The window washer on the left is on a platform supported by two cables which have spring balances. Ignore the weight of the plank. The man has a weight of 1000N.(d) Make a prediction of the readings Fl and Fr if BC is 3 times the distance AB.(e) Redo all the calculations in the previous setion, if the plank weighs 300N


----------------------------------------------------------------------------------------

Did you get these answers?(a) 1000N (clearly if the plank had no weight it would ‘flip over’..(b) 500N(c) 830N(d) Fl = 250N; Fr = 750N


----------------------------------------------------------------------------------------Did you get these answers?On accounting for the plank’s weight:(a) 1300N(b) 800N (clearly he cannot be standing in the middle)(c) 1130N (d) 400N; 900N (the weight of the plank is distributed equally to both cables)...

Fundamentals of physics - Mechanics Equilibrium Support Forces

----------------------------------------------------------------------------------------

The Earth is constantly pulling on all of objects within gravity field and this force is called weight. Objects would move towards the centre of the Earth if this force (weight) was not balanced. Consider the book on the table. The forces on the book are the weight (acts towards the centre of the Earth) and the normal reaction that the table pushes on the book.

Fr = -Fw Fr + Fw = 0

F = 0


----------------------------------------------------------------------------------------

These are equal in magnitude but opposite in direction so the book remains in equilibrium.

Fr = -Fw; F = 0.

Suppose the book weighs 5N. Then the normal reaction is also 5N.Fr + Fw = 5N + (–5N)

F = 0

Fr = -Fw Fr + Fw = 0

F = 0


----------------------------------------------------------------------------------------

How you know you have weight! If you are standing still the floor pushes on your feet with an equal but opposite reaction to oppose the tendency of your weight to pull you towards the centre of the earth. When sitting it is the chair which provides this normal reaction.Since on earth we normally feel this reaction most of the time we are often startled when it increases or decreases.


----------------------------------------------------------------------------------------The normal reaction decrease to zero if we are falling freely. This condition where ‘weightlessness’ is simulated by having an aircraft fall freely so that the seat does not need to provide a normal reaction when both the seat and person are falling at the same rate. The person is moving towards the centre of the earth but doesn’t notice it because he is enclosed in the aircraft which is falling freely. A person in another aircraft can easily notice that the person and aircraft are falling.Of course the opposite happens if the aircraft climbs steeply.

Fundamentals of physics - Mechanics Dynamic Equilibrium in Movement: Elevators

----------------------------------------------------------------------------------------Most people would be more familiar with a reduced effect on a rollercoaster or an elevator.

Let us examine riding in a very fast elevator (turbo lift).

Forces on you are Weight (W) and normal reaction of the floor (Rn)

Elevator Stop


----------------------------------------------------------------------------------------Forces on you are Weight (W) and normal reaction of the floor (Rn)

When the elevator has stopped at floor level we can easily walk into it and the normal reaction of the floor feels pretty ordinary... (W = Rn)

No net force...

Elevator Stop


----------------------------------------------------------------------------------------However when the elevator begins to move downward (with you in it) then there must be an net force downward force on you!You weight (W)has not changed (the earth is still pulling on you with the same force) so it is clear that the normal reaction (Rn) would have decreased!

(W > Rn)Net force down...

Elevator starting to go down

Fundamentals of physics - Mechanics Equilibrium in Movement: Elevators

----------------------------------------------------------------------------------------No what happens when the elevator begins to slow down?Well you must now have a net upward force (upwards) on you to slow you down to a stop. The normal reaction increases to more than your weight so there is a net upward force…

(W < Rn)

Elevator (moving downward) stopping.


----------------------------------------------------------------------------------------No what happens when the elevator begins to slow down?Well you must now have a net upward force (upwards) on you to slow you down to a stop. The normal reaction increases to more than your weight so there is a net upward force…

(W < Rn)



----------------------------------------------------------------------------------------Questions: Does your weight change during the journey?Is the normal reaction always the same?



----------------------------------------------------------------------------------------Can you work out the forces on you when the elevator begins to move upwards and when it comes to a stop?I have given you the first one: Elevator starting to move upward:

(W < Rn)Work out these others:

(W > Rn)(W = Rn)

Elevator starting to move upward.

Fundamentals of physics - Mechanics Dynamic Equilibrium in Movement: Airplanes

----------------------------------------------------------------------------------------Consider the airplane climbing upward…There are for forces (a) the weight (Fw) pointing towards the centre of the earth, (b) The lift (Fl) from the wings pointing upwards, (c) the thrust (Ft) from the engines pointing horizontally forward and the drag or air resistance (Fr) from air resistance pointing horizontally backward.

Airplane climbing upward while increasing forward

velocity.(Fl > Fw) accelerating upward(F1 = Fw) moving upward with constant upward velocity…(Fl < Fw) decelerating upward(Ft > Fr) accelerating forward(Ft = Fr) moving forward with constant velocity…(Ft < Fr) decelerating forward


----------------------------------------------------------------------------------------When the plane reaches constant forward velocity (Ft = Fr)

FYI trivia... Drag increases with velocity so most of the energy from the fuel is in overcoming air resistance...

Airplane climbing upward with constant forward velocity.

(Fl > Fw) accelerating upward(F1 = Fw) moving upward with constant upward velocity…(Fl < Fw) decelerating upward

(Ft = Fr) moving forward with constant velocity…


----------------------------------------------------------------------------------------Consider the airplanes...There are for forces (a) the weight (Fw) pointing towards the centre of the earth, (b) The lift (Fl) from the wings pointing upwards, (c) the thrust (Ft) from the engines pointing horizontally forward and the drag or air resistance(Fr) from air resistance pointing horizontally backward.

(a) Airplane flying level at constant velocity.

(b) Airplane flying level and increasing velocity

(c) Airplane flying level and decreasing velocity


----------------------------------------------------------------------------------------(a) Airplane flying level at

constant velocity.

(b) Airplane flying level and increasing velocity

(c) Airplane flying level and decreasing velocity

Write the relationships between each pair of forces in each case. Also draw the forces in the diagrams.

(a) (Fl Fw) and (Ft Fr)

(b) (Fl Fw) and (Ft Fr)

(c) (Fl Fw) and (Ft Fr)


----------------------------------------------------------------------------------------(a) Airplane descending at

constant velocity.

(b) Airplane descending with increasing velocity

(c) Airplane descending with decreasing velocity

Consider the airplanes...There are for forces (a) the weight (Fw) pointing towards the centre of the earth, (b) The lift (Fl) from the wings pointing upwards, (c) the thrust (Ft) from the engines pointing horizontally forward and the drag or air resistance(Fr) from air resistance pointing horizontally backward.


----------------------------------------------------------------------------------------

Write the relationships between each pair of forces in each case. Also draw the forces in the diagrams.

(a) (Fl Fw) and (Ft Fr)

(b) (Fl Fw) and (Ft Fr)

(c) (Fl Fw) and (Ft Fr)

(a) Airplane descending at constant velocity.

(b) Airplane descending with increasing velocity

(c) Airplane descending with decreasing velocity


----------------------------------------------------------------------------------------Consider the Blackhawk helicopter. When moving there are for forces (a) the weight (Fw) pointing towards the centre of the earth, (b) The lift (Fl) from the rotor pointing upwards, (c) the thrust (Ft) from the engines pointing horizontally forward and the drag or air resistance(Fr) from air resistance pointing horizontally backward

(a) Blackhawk helicopter hovering over a field.


----------------------------------------------------------------------------------------If it is hovering over a field... What would be the value of Fr?Write the relationships between each pair of forces in each case. Also draw the forces in the diagram.

(a) (Fl Fw) and (Ft Fr) (a) Blackhawk helicopter

hovering over a field.

Fundamentals of physics - Mechanics The Earth moves!

----------------------------------------------------------------------------------------Can you explain this?My favourite sceptic says that it is impossible for the earth to be moving by this simple experiment...

From calculations the Earth rotates from west to east at about 1500kmph near the equator and goes around the sun because it moves even faster at 107000 kmph...When I drop a stone off a cliff it falls straight down. the stone would fall to the west of the cliff due to the Earths rotation on its axis. The stone should also drift away from the direction the Earth is rotating around the sun...

Fundamentals of physics - Mechanics Some Practice Exercises!

----------------------------------------------------------------------------------------Can you figure out this?If you let go of a coin in a moving car how will it fall down?

(a) Straight down(b) Towards the front of the car.(c) Towards the back of the car...


----------------------------------------------------------------------------------------

Does it matter if the car is travelling with:(a) Constant velocity(b) Accelerating forward(c) Decelerating


----------------------------------------------------------------------------------------Can you figure out this?When you flip a coin in a high speed airplane at constant velocity, it behaves as if the airplane were at rest. (a) Why does the coin keep up with your motion?(b) If the airplane was accelerating or decelerating would the coin behave differently?

Fundamentals of physics - Mechanics Summary

----------------------------------------------------------------------------------------

Newton's first law of motion (the law of inertia). Every object continues in its state of rest, or in uniform motion in a straight line, unless acted upon by an eternal force.Inertia: The property of objects to resist changes in motion (laziness).(a) If at rest start moving.(b) If moving slow down, speed up, stop or change direction.Note: movement may be translation (change of position) or rotation.

Fundamentals of physics - Mechanics Summary

----------------------------------------------------------------------------------------

Force is a push or a pull.Net force: The vector sum of forces that act on an object.Mechanical equilibrium: The state of an object or system of objects for which there are no changes in motion. In accord with Newton's first law, if at rest, the state of rest persists. If moving, motion continues withoutChange (slow down, speed up, stop or change direction).Mechanical Equilibrium Rule: For any object or system of objects in equilibrium, the sum of the forces acting

equals zero. In equation form, F = O.

Fundamentals of physics - Mechanics Review Exercises - Net Force

----------------------------------------------------------------------------------------

1. Two children pull on a teddy bear. What is the net force on the teddy bear if one child pulls to the right with 60N and the other to the left with 100N?

Answer:Taking right to be +veFn = Fr + FlFn = 60N + (-100N)Fn = - 40N40 N to the left.


----------------------------------------------------------------------------------------

(2) This chain hoist in the diagram is used to lift the engine which is stationary at the moment.The forces on the hook are (i) the weight of the engine (Fw) and (ii) the tension on the chain (Ft).(a) If the engine is not moving what is the relationship between Fw and Ft?(b) What is the net force on the hook if the engine weighs 2000N and the tension on the chain is 3000N?(c) In what direction will the engine move in part (b)?


----------------------------------------------------------------------------------------Answer: (a) Fw + Ft = 0 (engine is not moving)(b) Taking up as +ve.(-2000N) + 3000N = -1000N1000N upwards.(c) The engine will move up.


----------------------------------------------------------------------------------------

(2) To keep a shopping cart moving with constant velocity a woman pushes with a force of 100N. How much is the friction is acting on the cart?


----------------------------------------------------------------------------------------

Answer: Since the cart is moving with constant velocity then the net force is 0.The forces acting on the cart are the force with which the woman pushes (Fw) and the force of friction (Ff) opposing the motion of the cartFn = Fw + Ff0N = 100N + Ff (Taking right to be +ve) Ff = -100N Ff = 100N to the left.

Documents

Mechanics Unit 5: Motion and Forces 5.5 Mechanical Equilibrium