Transcript Physics_AS_Unit2_23_FmaInPractice

Learning Objective :
1. To look at the how F=ma presents itself
in questions.....
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Towing things
Rockets
Lifts
Pulleys
Sliding down a slope
Book Reference : Pages 135-137
T
Car 1000kg
Caravan
800kg
8000N from the Engine
3000N Resistance
8000N from the Engine
T
Car 1000kg
Caravan
800kg
3000N Resistance
8000N from the Engine
T
Car 1000kg
Caravan
800kg
3000N Resistance
For a rocket of mass m
with thrust T at launch
Weight of the Rocket
mg
Thrust T from the
rocket motors
At launch when the rocket is
accelerating the unbalanced
force is T – mg and hence we
can apply this to “f=ma”...
T - mg = ma
Tension T in
the cable
Like before the unbalanced resultant
force is applied to “f=ma”
T – mg = ma
But there are a number of different
scenarios based upon the motion of the
lift (assuming upwards is positive)
Weight of the lift mg
1. The lift is moving up and accelerating
(a>0) T = mg + ma (T > mg)
2. The lift is moving up at constant
velocity (a=0) T=mg
Tension T in
the cable
3. The lift is moving up and decelerating
(a<0) T = mg + ma (T < mg since ma is -ve)
4. The lift is accelerating downwards (a<0) T
= mg + ma (T < mg since ma is -ve)
5. The lift is moving down but decelerating
(a>0) T = mg + ma (T > mg)
The Cable tension is more than the lift weight if:
The lift is moving up and accelerating (v>0, a>0)
The lift is moving down and decelerating (v<0, a>0)
Weight of the lift mg
The Cable tension is less than the lift weight if:
The lift is moving up and decelerating (v>0, a<0)
The lift is moving down and accelerating (v<0, a<0)
Looking at the resultant
forces on both Masses:
Tension T in
the cable
Tension T in
the cable
m
T – mg = ma
mg
Acceleration
of m
For m:
(1)
M
Mg
Acceleration
of M
For M:
Mg - T = Ma
(2)
Since we have two equations we can remove T by
substitution: (from 1 T = ma + mg, substitute into 2)
Mg – (ma + mg) = Ma
 (M - m)g = (M + m)a
°
This time it is no longer a static system in
Equilibrium, the object is accelerating down
the slope
°
The resultant force down the slope providing
the acceleration is:
mg sin  - F = ma
The problem on the previous page can be
extended to consider a vehicle with an
engine which is providing an engine force FE
This is simply considered when balancing all
the forces which ultimately provide the
acceleration
FE + mg sin  - F = ma
We have applied “F=ma” to a number of different
problems....
In each case it has been a matter combining the forces in
the system to find the unbalanced resultant force which
is causing the acceleration. Effectively this gives the “F”
part of “F=ma” which can then be equated to the “ma”
part
The equations derived here should not be learnt, rather
this lesson should be considered a collection of worked
examples & eventually you should be able to carry out
similar derivations when faced with new problems