phyForces.htm

More on Forces and Applications topics basics of Newton's Laws

Applying Newton's Laws to situations relies on

creating a good visualisation of the situation

A good diagram large enough to sketch on and identify angles is needed

confidently identifying the physical "things"affecting the object under study

"things" include -
                           "normal" forces, - the ground pushes up
                            forces propelling objects via the drive wheels or feet
                                          (Newton's Third Law is the key here - feet push back on ground, ground pushes forwards on feet)
                            frictional effects including air resistance, rolling resistance, dragging resistance
                            the weight
                             attached ropes giving tensional forces along their lengths - they always pull directly along their length
                            "pushes" from people, or jets out of the back of engines
                             Electric,Gravitational or Magnetic Forces

RARELY ARE THERE MORE THAN THREE OR FOUR FORCES AT THIS LEVEL.

turning these "things" into vector forces - making the diagrams above

deciding whether momentum is changing or not - this is CRUCIAL - it decides whether the sum of the vectors is ZERO or if it comes to "Δmv/Δt"

BALANCED OR UNBALANCED ?

The key to understanding and utilising Newton's Second Law

Example

The skier, mass 70 kg, above is accelerating down the 40⁰ slope at 6ms^-2. Find the drag force acting on her.

Solution:

Is she accelerating? - sure is - forces on her therefore add (vectorially) to "ma".

Unbalanced Force ( by Newton's Second Law ) = ma = 70 x 6 newton = 420 N down the hill in this case

Weight = mg = 70 x 9.8 N = 686N vertically down

Next we must sort out the angles so we can work on the triangle.

We have a right angled triangle with a 40⁰ apex at the top.

Using simple trig, sin40⁰ = Opp / Hyp = unknown / mg thus unknown = 686 x sin40⁰

unknown = frictional force + ma = 440

thus frictional force = 440 - 420 = 20 N up hill.

Questions

Can we always use simple trig?

When must we use cos rule etc?

Can you see an alternative way of calculating the friction using components only?

Could you calculate the "Normal Force"?

What generally happens to the "Normal Force" as the slope gets steeper?