Poor Sir Isaac Newton. After he got hit on the head by the infamous gravity-demonstrating apple, he began having problems--four of them to be exact. Since therapists and such had not been invented yet, Newton realized he had to solve his own problems...but he could only do this by making certain assumptions. Even though Isaac Newton has been dead for a long time, his problems still exist (they outlived him) and they still have to be solved, only now by us poor grade 11s. Here is a list of the assumptions Newton had to make and that are necessary to apply to the solutions of these problems.
Problem #1: Equilibrium
Assumptions:
- Acceleration: since ax = 0 and ay = 0, overall a = 0 (there is no acceleration)
- Gravity: g = 9.81 m/s^2
- Positive Axes: a set of positive axes must be set
Example Problem:
For some unimaginable reason, Marco decides that he wants to slide down a frictionless rope across Niagara Falls. However, halfway through he realizes that the maximum tension the rope can support is 500 N. If the angle between the rope and the horizontal is 15 º, help Marco figure out what his mass must be to safely cross the waterfall (hint: right now Marco is hanging on the rope).
Solution:
Given: T1 max = 500 N g = 9.81 m/s2 θ = 15 º Required: m = ?
Analysis: ∑ F = ∑ Fx = 0
∑ Fx = Fg – 2Tsin15 º
Fg = 2Tsin15 º
mg = 2Tsin15 º
Substitution: mmax = 2Tsin15 º / g = 2(500)sin15 º / 9.81 = 26.38 kg
Therefore, he must weigh 26.38 kg or under in order to cross safely.
Problem #2: Inclined Planes
Assumptions (Static or no movement):
- Acceleration: since ax = 0 and ay = 0, there is no acceleration or a = 0
- Axes: the x-axis is now parallel to the direction of friction or to the surface, and the y-axis is now parallel to the normal force or perpendicular to the surface
- Positive X-Axis: the positive x-axis is in the direction of potential motion
- Coefficient of Friction: µ = tanθ
Example Problem: If Marco has a mass of 37.6 kg and is sitting without moving on an incline with a coefficient of friction of 0.42, what is the force of friction that he experiences?
Solution:
Given: µ = 0.42 m = 37.6 kg g = 9.81 m/s2 Required: f = ?
Analysis: ∑F = 0 ∑Fx = ∑Fy = 0 X: ∑Fx = 0 Fgx – f = 0 f = Fgx = mgsinθ
Y: ∑Fy = 0 FN – Fgy = 0 FN = Fgy = mgcosθ
µ = tanθ
f = µFN = mgcosθtanθ
Substitution: θ = tan-1 µ = tan-1 (0.42) = 22.78 º
Either f = mgsinθ = (37.6)(9.81)sin(22.78º) = 142.83 N
Or f = mgcosθtanθ = µmgcosθ = (0.42)(37.6)(9.81)cos(22.78º) = 142.83 N
Therefore, he experiences a frictional force of 142.83 N.
Assumptions (Kinetic or with movement):
- Kinetic friction: fk = µk x FN
- Axes: the x-axis is parallel to the direction of friction or the surface, and the y-axis is parallel to the normal force or perpendicular to the surface
- Positive X-Axis: the positive x-axis is in the direction of motion
- Forces: ∑ F = ∑Fx + ∑Fy , ∑Fy = 0 and ∑Fx ≠ 0
Example Problem: Marco wants to feel the wind in his new hair, so he decides to go sliding. The slide is at an angle of
11º and Marco is travelling with a force of 20.5 N despite 54 N of friction. What is the coefficient of friction of the slide?
Solution:
Given: g = 9.81 m/s2 θ = 11º f = 54 N F = 20.5 N
Required: µ = ? (m = ?)
Analysis: ∑F = 20.5N ∑Fx + ∑Fy = 20.5N ∑Fy = 0 ∑Fx = 20.5 N
X: ∑Fx = 20.5 N Fgx – f = 20.5
Y: ∑Fy = 0 Fgy = FN = mgcosθ
f = µFN µ = f/FN = f/mgcosθ
m = ? Fgx – f = 20.5 mgsinθ – f = F mgsinθ = F + f m = (F + f)/gsinθ
Substitution: m = (F + f)/gsinθ = (20.5 + 54)/(9.81)sin11º = 39.8 kg
µ = f/mgcosθ = 54/(39.8)(9.81)cos11º = 0.14
Therefore, the coefficient of friction on the slide was 0.14.
Problem #3: Pulleys
Assumptions:
- Air resistance: there is none
- Friction: there is no friction when using the pulley
- Multiple Systems: each side of the pulley can be broken into multiple systems
- Tension: the tension in both systems is equal and opposite
- Positive Axis: for each system, the positive axis is in the direction of motion or potential motion
- Acceleration: the acceleration is consistent for both systems
Example Problem: Marco is hanging off one end of a frictionless pulley while Marco’s friend Tony is hanging off the other end. If Tony has twice as much mass as Marco and the tension in the pulley is 78 N, what is Marco’s mass?
Solution:
Given: m1 = Marco m2 = Tony m2 = 2m1 T = 78 N g = 9.81 m/s2
Required: m1 = ?
Analysis: System 1: ∑ F1 = m1a = T - m1g a = (T - m1g)/ m1
System 2: ∑ F2 = m2a = m2g – T a = (m2g – T)/ m2
a = a
(T - m1g)/m1 = (m2g – T)/m2
(T - m1g)/m1 = (2m1g – T)/2m1
(T - m1g)(2m1) = (2m1g – T)(m1)
(T - m1g)(2m1) = (2m1g – T)(m1)
2T – 2m1g = 2m1g – T
3T = 4m1g
3T/ (4g) = m1
Substitution: m1 = 3(78) / (4)(9.81) = 5.96 kg
Therefore, Marco has a mass of 5.96 kg.
Problem #4: Trains
Assumptions:
- Multiple Systems: each car of the train can be separated into its own system
- Air Resistance: there is none
- Positive Axes: the positive axis (generally it is the x-axis that involves motion) is in the direction of motion
- Acceleration: the acceleration is consistent for all the system. For this reason, the train can be treated as one entity in order to find acceleration
Example Problem: In Marco's train (shown below), what is the amount of force opposing the motion of the middle car (hint: the amount of force in the opposite direction of motion for the middle car).
Given: m1 = 20 kg m2 = 30 kg m3 = 40 kg µ = 0.16 FA = 450 N g = 9.81 m/s2
Required: T1 – m2a = x = ? (a = ? T1 = ?)
Analysis: and Substitution: Total System: ∑FT = mTa
∑FT = ∑Fx + ∑Fy
∑Fy = 0
∑Fx = FA – f
= mTa
a = (FA – f)/mT
= (FA - µFN)/mT
= (FA - µmTg)/mT
= [450 – (0.16)(90)(9.81)]/90
= 3.4304 m/s2
System 1: ∑F1 = m1a
∑F1 = ∑Fx + ∑Fy
∑Fy = 0
∑Fx = FA – T1 – f
m1a = FA – T1 – f
= FA – T1 - µFN
= FA – T1 - µm1g
T1 = FA - µm1g – m1a
= 450 – (0.16)(20)(9.81) – (20)(3.4304)
= 350 N
System 2: ∑F2 = m2a
∑F2 = ∑Fx + ∑Fy
∑Fy = 0
∑Fx = T1 – T2 – f
= T1 – x
x = T1 – m2a
= 350 N – (30)(3.4304)
= 247.088 N
Therefore, 247.088 N is the amount of force pulling the middle car back.
Thank you for having the patience to look through this long and incredibly time-consuming to make blog!
