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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!