Download Newton`s 2: Complicated Forces

Newton’s Laws, Part 2:
Pre AP Physics
Remember:
1. Draw FBD with forces and acceleration drawn and labeled
2. Write separate x and y net force equations
3. Fg does not always equal FN
Systems of objects accelerating:
1. Treat each object in the system separately
a. Its own individual FBD
b. Its own net force equation
2. Acceleration of the whole system is the same
a = Fnet/m
this value is the same for each object in the system
3. The tension in a rope is the same throughout the rope
1
2
T1
T2
Both tensions are the same value (magnitude) but they act in opposite directions
1
These tensions are the same as long as the pulley is a frictionless
pulley
T1
T2
2
Tpull
1
T1
T2
2
4. Many times you will be dealing with 3 or more forces in one direction on an object
FN
Fnet x = Fa – T- FFR OR Fnet = Fa – (T + FFR)
Fnet y = FN – Fg = 0
T1
Fa
FFR
Fg
1
5. You will be dealing with systems of equations
a. However many unknowns you have, you will need that many equations
2 unknowns = 2 equations; 3 unknowns = 3 equations
b. Will solve by substitution or elimination
Forces at an angle:
Either push object or pull object and angle is given relative to the horizontal
Must break force into components on the FBD – draw components as dotted lines and label
Include the components in the appropriate net force equation.
When you pull on an object, you always pull up relative to the horizontal
a
Fpull(sinΘ)
FN
Fpull
Fpull(cosΘ)
Fg
Net force equations:
X: Fpull(cosΘ) = ma
Y: FN + Fpull(sinΘ) = Fg
When you push on an object, you always push down relative to the horizontal
Fpush(cosΘ) FN
Fpush
a
Fpush(sinΘ)
Fg
Y: FN = Fpush(sinΘ) + Fg
Net force equations: X: Fpush(cosΘ) = ma
Following principles still hold true:
2
1. Only net unbalanced forces cause acceleration
2. Horizontal forces cause horizontal acceleration; Vertical forces cause vertical
acceleration
3. ALWAYS need FBD with ALL forces drawn
Inclined planes:
If an object is sitting on an inclined plane, Θ is from the horizontal
y
x
Θ
We still have force of gravity, normal force (since on a surface) and if surface has friction,
friction force
X and Y axes are now tilted
Force of gravity always acts straight down towards center of earth, so Fg is drawn:
y
x
Θ
Fg
Θ
But this doesn’t fit into our x & y, so we need to break Fg into its components: x = mgsinΘ and
y = mgcosΘ view this video for explaination of x and y components
http://www.khanacademy.org/video/inclined-plane-force-components?playlist=Physics
1. Remember, FN is force a surface exerts to balance weight of an object and it acts
perpendicular to the surface
2. mgsinΘ is component of gravity that pulls the mass down the incline (helpful? Slide =
sine down the incline)
3. Force of friction opposes the motion or apparent motion
a. Sliding down, then frictional force acts up incline
b. Sliding up, then frictional force acts down incline
c. Sitting on incline (a = 0), wants to slide down therefore friction acts up the incline
4. If change angle of incline, component of weight acting down the incline changes
a. If increase incline, mgsinΘ is more
b. If decrease incline, mgsinΘ is less
3
Equilibrium:
When net force on an object is zero, we can say that the object is at equilibrium.
We know that there are 2 cases when an object is at equilibrium
1. Object is at rest
2. Object is moving with constant velocity
In both cases there is no acceleration therefore the net force is 0 N.
Right now we will focus on the 1st case, static equilibrium, objects at rest
When an object is at equilibrium all components of forces are balanced  left = right & up =
down
Draw FBD for knot (where ropes connect) and FBD for any hanging objects.
Ex. Man on 1 scale  FN = Fg = 500 N
Man on 2 scales  what does each scale read? 250 N
Ex 2: hanging a picture
Find the tension in the cable
Fg = 50 N
Θ = 30o
Θ
1. Have to look where cable is attached to object & find those angles (150/2) = 75o
2. Break into components and solve
T = 25.9 N
B: Θ = 100o
T = 38.8 N
C: Θ = 150o
T = 96.5 N
Notice: only vertical components can support vertical weight. The smaller the angle with the
horizontal, the less vertical component there is so the tension must be greater
What arrangement would have the least tension?
When both cables are vertical because then 100% support of vertical weight
Discussion:
1. swing sets -
not
Why?
2. power lines in winter – small mass of ice breaks them, why?
3. support cables of suspension bridge are designed to sag
4