Mechanics

Topic 2.2 Forces and Dynamics

Forces and Free-body Diagrams

To a physicist a force is recognised by the effect or effects that it producesA force is something that can cause an object to Deform (i.e. change its shape) Speed up Slow Down Change direction

The last three of these can be summarised by stating that a force produces a change in velocityOr an acceleration

Free-body Diagrams

A free-body diagram is a diagram in which the forces acting on the body are represented by lines with arrows.The length of the lines represent the relative magnitude of the forces.The lines point in the direction of the force.The forces act from the centre of mass of the bodyThe arrows should come from the centre of mass of the body

Example 1

Normal/Contact Force

Weight/Force due to Gravity

A block resting on a worktop

Example 2A car moving with a constant velocity

Normal/Contact Force

Weight/Force due to Gravity

Motor ForceResistance

Example 3A plane accelerating horizontally

Upthrust/Lift

Weight/Force due to Gravity

Motor ForceAir Resistance

Resolving Forces

Q. A force of 50N is applied to a block on a worktop at an angle of 30o to the horizontal. What are the vertical and horizontal components of this force?

Answer

First we need to draw a free-body diagram

30o

50N

We can then resolve the force into the 2 components

30o

50NVertical = 50 sin 30o

Horizontal = 50 cos 30o

Therefore Vertical = 50 sin 30o = 25N Horizontal = 50 cos 30o = 43.3 = 43N

Determining the Resultant Force

Two forces act on a body P as shown in the diagramFind the resultant force on the body.

30o

50N

30N

Resolve the forces into the vertical and horizontal componenets (where applicable)

Solution

30o

50N

30N

50 sin 30o

50 cos 30o

Add horizontal components and add vertical components.

50 sin 30o = 25N

50 cos 30o – 30N = 13.3N

Now combine these 2 components

25N

13.3N

R

R2 = 252 + 13.32

R = 28.3 = 28N

Finally to Find the Angle25N

13.3N

R

tan = 25/13.3 = 61.987 = 62o

The answer is therefore 28N at 62o upwards from the horizontal to the right

Springs

The extension of a spring which obeys Hooke´s law is directly proportional to the extending tensionA mass m attached to the end of a spring exerts a downward tension mg on it and if it is stretched by an amount x, then if k is the tension required to produce unit extension (called the spring constant and measured in Nm-1) the stretching tension is also kx and so

mg = kx

Spring Diagram

x

Newton´s Laws

The First Law

Every object continues in a state of rest or uniform motion in a straight line unless acted upon by an external force

Examples

Any stationary object!Difficult to find examples of moving objects here on the earth due to frictionPossible example could be a puck on ice where it is a near frictionless surface

Equilibrium

If a body is acted upon by a number of coplanar forces and is in equilibrium ( i.e. there is rest (static equilibrium) or unaccelerated motion (dynamic equilibrium)) then the following condition must applyThe components of the forces in both of any two directions (usually taken at right angles) must balance.

Newton´s Laws

The Second LawThere are 2 versions of this law

Newton´s Second Law

1st version

The rate of change of momentum of a body is proportional to the resultant force and occurs in the direction of the force.F = mv – mu F =

t t

Newton´s Second Law

2nd version

The acceleration of a body is proportional to the resultant force and occurs in the direction of the force.F = ma

Linear Momentum

The momentum p of a body of constant mass m moving with velocity v is, by definition mvMomentum of a body is defined as the mass of the body multiplied by its velocityMomentum = mass x velocityp = mvIt is a vector quantityIts units are kg m s-1 or NsIt is the property of a moving body.

Impulse

From Newtons second lawF = mv – mu F =

t tFt = mv – muThis quantity Ft is called the impulse of the force on the body and it is equal to the change in momentum of a body.It is a vector quantityIts units are kg m s-1or Ns

Law of Conservation of Linear Momentum

The law can be stated thusWhen bodies in a system interact the total momentum remains constant provided no external force acts on the system.

Deriving This Law

To derive this law we apply Newton´s 2nd law to each body and Newton´s 3rd law to the systemi.e. Imagine 2 bodies A and B interactingIf A has a mass of mA and B has a mass mB If A has a velocity change of uA to vA and B has a velocity change of uB to vB during the time of the interaction t

Then the force on A given by Newton 2 isFA = mAvA – mAuA

t

And the force on B isFB = mBvB – mBuB

t

But Newton 3 says that these 2 forces are equal and opposite in direction

ThereforemAvA – mAuA = -(mBvB – mBuB)

t tTherefore

mAvA – mAuA = mBuB – mBvB

Rearranging mAvA + mBvB = mAuA + mBuB

Total Momentum after =Total Momentum before

Newton´s Laws

The Third Law

When two bodies A and B interact, the force that A exerts on B is equal and opposite to the force that B exerts on A.

Example of Newton´s 3rd

Q. According to Newton’s third Law what is the opposite force to your weight?A. As your weight is the pull of the Earth on you, then the opposite is the pull of you on the Earth!

Newton´s 3rd Law

The law is stating that forces never occur singularly but always in pairs as a result of the interaction between two bodies.For example, when you step forward from rest, your foot pushes backwards on the Earth and the Earth exerts an equal and opposite force forward on you. Two bodies and two forces are involved.

Important

The equal and opposite forces do not act on

the same body!