1. Newton 2nd Law problems - Atwood Machines -Incline Planes -Tension Problems -Other Object Connected problems

2. Tension

3. Ropes- Assumptions Ignore any frictional effects of the rope
Ignore the mass of the rope
The magnitude of the force exerted along the rope is called the tension
The tension is the same at all points in the rope
Section 4.5

4. Tension Forces A taut rope has a force exerted on it.
If the rope is lightweight and flexible the force is uniform over the entire length.
This force is called tension and points along the rope.
forces on the block
forces on the rope FT FN
Frope
-FT
Ffr m m Fg
Frope = -FT by the law of reaction

5. Tension or Normal Force
Tension and normal forces are different.
A pull on an object - tension
A push from a surface - normal force
Either one or both may be present.
Normal force
Tension force FN FT
Ffr m Fg

6. Coupled Motion Objects linked by tension move together
v1 = v2
FT2
FT2
FT1 m2 m1
Ffr2
Ffr1
Objects linked by tension move together
Same velocity and acceleration
Tension may not be the same on two ropes
FT2 = Ffr2
FT1 = Ffr1 + FT2 = Ffr1 + Ffr2

7. Pulley A pulley uses tension to transfer a force to another direction.FT
forces on the rope m1 FT m2
forces on block 1
forces on block 2
Ffr
Frope m1 m1
Frope m2 m2 Fg

8. Tension is equally distributed in a rope and can be bidirectional
Draw FBD block Draw FBD hand
30N
Tension does not always equal weight:

9. Derive a general formula for the acceleration of the system and tension in the ropes

10. Derive a general formula for the acceleration of the system and tension in the ropes
Draw FBDs and set your sign conventions: Up is positive down is negative m1 m2 m1 m2

11. Derive a general formula for the acceleration of the system and tension in the ropes
Draw FBDs and set your sign conventions: If up is positive down is negative on the right, the opposite is happening on the other side of the pulley. -T
+m1g +T
-m2g m1 m2 m1 m2
Then, do ΣF= ma to solve for a if possible.

12. Derive a general formula for the acceleration of the system and tension in the ropes
+m1g +T
-m2g m1 m2 m1 m2
Σ F2 = T – m2g = m2 a
Σ F1 = - T + m1g = m1 a
To find acceleration I must eliminate T: so solve one equation for T and then substitute it into the other equation (This is called “solving simultaneous equations” )

13. Σ F1 = - T + m1g = m1 a
Σ Fnet = m1 a+ m2 a = - T + m1g + T – m2g
-T and T Cancel !!!!!
Σ Fnet =m1 a+ m2 a = m1g – m2g
Now get all the a variables on one side and solve for a. Solve!!!!!
Σ F2 = T – m2g = m2 a

14. m1g - m1 a – m2g = m2 a
m1g – m2g = m2 a + m1 a
Factor out g on left and a on right
g(m1– m2) = a (m2 + m1)
Divide this away
(m1– m2)
g = a
(m2 + m1)

15. Now you can find the tension formula by eliminating a
Now you can find the tension formula by eliminating a. Just plug the new a formula either one of the original force equations.
Σ F2 = T – m2g = m1 a
Σ F1 = - T + m1g = m1 a -T
+m1g m1 m1 m2

16. Pulley Acceleration
The normal force on m1 equals the force of gravity.
The force of gravity is the only force on m2.
Both masses must accelerate together.
Consider two masses linked by a pulley
m2 is pulled by gravity
m1 is pulled by tension
frictionless surface
Ffr
Frope m1 m2

17. Atwood’s Machine Problem 1
In an Atwood machine both masses are pulled by gravity, but the force is unequal.
The heavy weight will move downward at
( kg)(9.8 m/s2)/( kg) = 1.8 m/s2.
Using y = (1/2)at2, it will take t2 = 2(1.80 m)/(1.8 m/s2)
t = 1.4 s.

18. Atwood’s Machine (simple)

19. Forces of Friction
When an object is in motion on a surface or through a viscous medium, there will be a resistance to the motion
This is due to the interactions between the object and its environment
This is resistance is called friction
Section 4.6

20. More About Friction Friction is proportional to the normal force
The force of static friction is generally greater than the force of kinetic friction
The coefficient of friction (µ) depends on the surfaces in contact
The direction of the frictional force is opposite the direction of motion
The coefficients of friction are nearly independent of the area of contact
Section 4.5

21. Static Friction, ƒs
Static friction acts to keep the object from moving
If F increases, so does ƒs
If F decreases, so does ƒs
ƒs  µs n
Use = sign for impending motion only
Section 4.5

22. Kinetic Friction, ƒk
The force of kinetic friction acts when the object is in motion
ƒk = µk n
Variations of the coefficient with speed will be ignored
Section 4.5

23. Friction, final Adjust the force and see where you are on the graph
Note especially where ƒ = Fs
Section 4.5

24. Some Coefficients of Friction
Section 4.5

25. Other Types of Friction
Friction between the moving car’s wheels and the road is static friction
Unless the car is skidding
Also have the air resistance,
Section 4.5

26. Inclined Planes

27. Axes for Inclined Planes
X axis is parallel to the inclined plane
Y axis is perpendicular to the inclined plane
Friction force (f or Ff) is on the x axis opposite the motion
Weight or Force of Gravity (Fg) is straight down and must be split into x and y components.
Normal force (FN) is upward along the y axis
Applied force (Fa) may be in either direction on the x axis or at angle (which would require it to be split into x and y components)

28. Forces Acting on the Objecty
Note: The applied force and the force of friction can be in either direction as long as the friction force is opposite to the motion. FN Fa f
Fgx
Fgy  Fg
Fgx x 
Remember: Weight or force of gravity = F g= mg (some may use different symbols such as W or FW for weight)

29. Useful Equations for Inclined Plane Problems
Formulas
Fnet,x = Fx = max
Fnet,y = Fy = may
Fg = mg
Fgx = mg sin θ
Fgy = mg cos θ
Ff= FN
* Assumes that Ɵ is as shown on previous diagram.
Explanation of Symbols
Fnet = net force
m = mass
Fg = weight (force of gravity)
Ff or f = friction force
 = coeff. of friction
FN = normal force
Fa = applied force

30. Problems With No Acceleration
When the object on the plane does not have any acceleration, then the forces acting on the object in both x and y must be equal to zero. y
Fx = Fa - f - Fgx = max = 0 Fa
Fy= FN - Fgy= may = 0 FN
Note: There might not be an applied force or it may be down the incline (so f would be up the incline). f
Fgy
Fgx Fg x
Fgx 

31. Problems With Acceleration
When the object on the plane has acceleration, there is a net force along the incline (x) but the net force perpendicular to incline (y) is still zero. y
Fx = Fa - f - Fgx = max Fa
Fy= FN - Fgy= may =0 FN
Note: There might not be an applied force or it may be down the incline (so f would be up the incline). f
Fgy
Fgx Fg x
Fgx 

32. Solving an Inclined Plane Problem # 1 A box slides down an plane 8 meters long inclined at 300 with a coefficient of friction of (a) what is the acceleration of the box? (b) what is its velocity at the bottom of the incline?
FParallel = W sin 
FNormal = W cos  
Normal force
Force parallel
Upward force of plane
Friction
force
weight
Center of Mass
FParallel = w sin 300
Ffriction =  Fnormal
Ffriction = 0.25 x w cos 300
Fnet = P – Ffriction
Fnet = 0.5 w – 0.17 w
(a) Fnet = ma, w = mg , m = w /g , Fnet = (w/g) a
0.5 w – 0.17 w = (w/9.8) a , a = 2.77 m/s2
(b) Vo = 0 (the box starts at rest), a = (Vi2 – Vo2) / 2 S
Vi =( 2 S a + Vo)1/2 , Vi = ( 2 x 8 x )1/2 , Vi = 6.7 m/s

33. Inline Plane Problems # 2

34. Connected Objects Apply Newton’s Laws separately to each object
The magnitude of the acceleration of both objects will be the same
The tension is the same in each diagram
Solve the simultaneous equations
Section 4.5

35. Modified Atwood’s Machine Problem 2
A 15 kg cart is attached to a hanging 25 kg mass. Friction is negligible. What is the acceleration of the 15 kg cart?

36. What is the acceleration of the system
What is the acceleration of the system? What is the tension developed in the connector?

37. Tension Problems

38. What is Tension?
Tension is defined as a force transmitted along a rope, chain, or wire.
Tension will remain constant throughout the length of the rope.
Tension is treated as a force in force diagrams and calculations
Tension is measured in force units. (Newtons, dynes, or pounds)

39. Weight or mass? What’s the difference?
Mass is the amount of matter that an object is made up of, measured in mass units (g, kg, slugs)
The mass of an object remains constant at any location in the universe unless matter is added or removed
Weight is the force of gravity acting on an object, measured in force units (N, dynes, or pounds)
Weight depends on the acceleration due to gravity for the location of the object.
Weight (w) = mg

40. Problems in Equilibrium
Equilibrium means that the object has a net force and acceleration of zero.
Since Fnet,x or Fx= 0, the x-components of tension should cancel each other.
(Fright = Fleft)
Since Fnet, y or Fy= 0, the y-components of tension should cancel the objects weight.
(Fup = Fdown )

41. Force Diagram (Free-Body Diagram)1 2 T1 T2
T2y=T2sin2
T1y=T1sin1 1 2
T1x=T1cos1
T2x=T2cos2
Fg = mg

42. Applying Newton’s 2nd Law
Since the sign is in equilibrium,
the net force must be equal to zero.
Fy = T1 sin θ1+T2 sin θ2 – mg = 0
Fx =T1 cos θ1 –T2 cos θ2 = 0 θ1 θ2 T1 T2
Fg = mg

43. Example 4.4
Given that Mlight = 25 kg, find all three tensions
T3 = N, T1 = N, T2 = N

44. Other objects connected

45. Objects connected
Example: Three objects, m1, m2, and m3 are tied together by a rope and pulled along a level surface by an applied force Fa. All three objects have the same coefficient of friction, μ. Find the tensions in the ropes connecting the masses. m1 m2 m3
Some real life situations that this model applies to include: train cars connected by couplings, find the tension in the couplings; pulling sleds tied together; pulling a trailer (only 2 objects)…

46. Objects connected Drawing a force diagram
Fg1=m1g
Fg3=m3g
Fg2=m2g Fa
FN2
FN1 T2 T1
f1=μFN1
f2=μFN2
f3=μFN3
FN3

47. Objects connected Write the force equations for each object.m1 m2 m3
Fg1=m1g
Fg3=m3g
Fg2=m2g Fa
FN2
FN1 T2 T1
f1=μFN1
f2=μFN2
f3=μFN3
FN3
For mass m2:
For mass m1:
For mass m3:

48. Case 1: Constant velocity
Objects connected
Case 1: Constant velocity

49. Objects connected constant velocity, summary
So we started with a force diagram, then wrote equations for each mass. We did not need to evaluate the equations for the last mass because we had solved for both tensions already, so we were done.
Note: We could have started at mass 3 and worked back to mass 1. You could solve this problem for applied force to keep the object moving at constant velocity in this method. Applied force did not need to be given.

50. Case 2: Accelerated motion
Objects connected
Case 2: Accelerated motion

51. Objects connected Write the force equations for each object.m1 m2 m3
Fg1=m1g
Fg3=m3g
Fg2=m2g Fa
FN2
FN1 T2 T1
f1=μFN1
f2=μFN2
f3=μFN3
FN3
For mass m2:
For mass m1:
For mass m3:

52. Objects connected with constant velocity, solving T2
Let m1 = 10 kg, m2 = 20 kg, m3 = 15 kg, Fa = 110, μ = 0.25
Tension between masses m2 and m3.

53. Objects connected constant velocity, solving mass T1
Let m1 = 10 kg, m2 = 20 kg, m3 = 15 kg, Fa = 110, μ = 0.25
T 2 = 73.2 N
Tension in the string between masses m1 and m2.

54. Objects connected accelerated motion, solving mass 1
On the last example we started with mass 3 and worked toward mass 1. This time I am beginning at mass 1 and moving toward mass 3 for no other reason than to show that you can work these problems either way.
Let m1 = 10 kg,
m2 = 20 kg,
m3 = 15 kg,
Fa = 150N,
μ = 0.25
a = 0.88 m/s2

55. Objects connected accelerated motion, solving mass 2
Let m1 = 10 kg,
m2 = 20 kg,
m3 = 15 kg,
Fa = 150N,
μ = 0.25
a = 0.88 m/s2
T1 = 33.3 N

56. Objects connected accelerated motion, must find acceleration first
To find acceleration to use for this problem you can treat the whole system together as one object being accelerated under the influence of the applied force.
m1 + m2 + m3 FN Fa
Fg = (m1 + m2 + m3)*g f
Now we have acceleration of each mass to use as we solve for tensions.