Discussion Class 1
Questions
Q1)
a) In the diagram alongside, if the box is stationary and we
increase the angle 𝜃 between the horizontal and force
𝐹⃗ , do the following quantities increase, decrease or
remain the same: 𝐹𝑥 , 𝑓𝑠 , 𝐹𝑁 , and 𝑓𝑠,𝑚𝑎𝑥
Answer
Justification
𝐹𝑥
Decrease
Since 𝐹𝑥 = 𝐹 cos 𝜃 then as 𝜃 increases then 𝐹𝑥 decreases
𝑓𝑠 ,
Decrease
By Newton’s first law 𝐹𝑛𝑒𝑡,𝑥 = 0 ⇒ 𝐹𝑥 − 𝑓𝑠 = 0 thus as 𝐹𝑥
decreases so must 𝑓𝑠 to compensate
𝐹𝑁
Increase
By Newton’s first law 𝐹𝑛𝑒𝑡,𝑦 = 0 ⇒ 𝐹𝑁 − 𝐹𝑔 − 𝐹 sin 𝜃 = 0 thus as
𝜃 increases then 𝐹 sin 𝜃 increases so must 𝐹𝑁 to compensate
𝑓𝑠,𝑚𝑎𝑥
Increase
Since 𝑓𝑠,𝑚𝑎𝑥 = 𝜇𝑠 𝐹𝑁 then as 𝐹𝑁 increases so must 𝑓𝑠,𝑚𝑎𝑥
b) If instead the box were moving when 𝜃 was increased, does the magnitude of the frictional
force on the box increase, decrease, or remain the same?
Increase. AS explained above 𝐹𝑁 will increase and in the moving case we are dealing with kinetic
friction and 𝑓𝑘 = 𝜇𝑘 𝐹𝑁 thus friction will increase as 𝜃 is increased
Q3) In the diagram alongside, a horizontal force 𝐹⃗1 of magnitude 10N
is applied to a box on a floor, but the box does not slide. Then as the
magnitude of vertical force 𝐹⃗2 is increased from zero, do the following
quantities increase, decrease, or remain the same: 𝑓𝑠 , 𝐹𝑁 , and
𝑓𝑠,𝑚𝑎𝑥 ? Does the box eventually slide?
𝑓𝑠 remains the same: By Newton’s first law 𝐹𝑛𝑒𝑡,𝑥 = 0 ⇒ 𝐹1 − 𝑓𝑠 = 0 thus as 𝐹1 remains the same
so must 𝑓𝑠
𝐹𝑁 increases: By Newton’s first law 𝐹𝑛𝑒𝑡,𝑦 = 0 ⇒ 𝐹𝑁 − 𝐹𝑔 − 𝐹2 = 0 thus as increases so must 𝐹𝑁 to
compensate
𝑓𝑠,𝑚𝑎𝑥 increases: Since 𝑓𝑠,𝑚𝑎𝑥 = 𝜇𝑠 𝐹𝑁 then as 𝐹𝑁 increases so must 𝑓𝑠,𝑚𝑎𝑥
Problems
P11) A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined
15° above the horizontal.
𝐹⃗ = 𝐹 cos 𝜃 𝑖̂ + 𝐹 sin 𝜃 𝑗̂
𝐹⃗𝑁 = 0𝑖̂ + 𝐹𝑁 𝑗̂
𝐹⃗𝑔 = 0𝑖̂ − 𝑚𝑔𝑗̂
𝑓⃗ = −𝑓𝑖̂ + 0𝑗̂
𝐹⃗𝑛𝑒𝑡 = (𝐹 cos 𝜃 −𝑓𝑠 )𝑖̂ + (𝐹 sin 𝜃 + 𝐹𝑁 − 𝑚𝑔)𝑗̂
a) If the coefficient of static friction is 0.65, what is the minimum force magnitude required to
start the crate moving?
By Newton’s 1st law and looking at the key word minimum:
𝐹⃗𝑛𝑒𝑡 = 0𝑖̂ + 0𝑗̂ ⇒ (𝐹 cos 𝜃 −𝜇𝑠 𝐹𝑁 )𝑖̂ + (𝐹 sin 𝜃 + 𝐹𝑁 − 𝑚𝑔)𝑗̂ = 0𝑖̂ + 0𝑗̂
𝐹 sin 𝜃 + 𝐹𝑁 − 𝑚𝑔 = 0 ⇒ 𝐹𝑁 = 𝑚𝑔 − 𝐹 sin 𝜃
𝐹 cos 𝜃 −𝜇𝑠 (𝑚𝑔 − 𝐹 sin 𝜃) = 0 ⇒ 𝐹 =
𝜇𝑠 𝑚𝑔
= 382 𝑁
cos 𝜃 + 𝜇𝑠 sin 𝜃
b) If the coefficient of kinetic friction is 0.35, what is the magnitude of the initial acceleration of
the crate?
𝐹𝑁 will not change in this instance since the object is not moving through the surface thus 𝐹𝑛𝑒𝑡,𝑦 = 0
However according to Newton’s 2nd law for any force infinitesimally greater than the force in (a):
𝐹𝑛𝑒𝑡,𝑥 = 𝑚𝑎 ⇒ 𝑎 =
𝐹 cos 𝜃 −𝜇𝑘 𝐹𝑁 𝐹 cos 𝜃 −𝜇𝑘 (𝑚𝑔 − 𝐹 sin 𝜃)
=
= 2.50 𝑚/𝑠 2
𝑚
𝑚
P23) When the three blocks in the diagram below are released from rest, they accelerate with
magnitude of 0.500 𝑚/𝑠 2 . Block 1 has mass M, block 2 has mass 2M, and block 3 has mass 2M.
What is the coefficient of kinetic friction between block 2 and the table?
Block 3 is heavier than block 1 thus we assume that motion (if any) would be to the right for block 2
For Block 1:
𝐹𝑛𝑒𝑡,1 = 𝑀𝑎 ⇒ 𝑇1 − 𝑀𝑔 = 𝑀𝑎
(1)
For Block 3:
𝐹𝑛𝑒𝑡,3 = 2𝑀𝑎 ⇒ 𝑇2 − 2𝑀𝑔 = 2𝑀(−𝑎)
(2)
For Block 2:
𝐹⃗𝑛𝑒𝑡,2 = 2𝑀𝑎𝑖̂ ⇒ (𝑇2 − 𝑇1 − 𝜇𝑘 𝐹𝑁 )𝑖̂ + (𝐹𝑁 − 2𝑀𝑔)𝑗̂ = 2𝑀𝑎𝑖̂
⇒ 𝐹𝑁 − 2𝑀𝑔 = 0
⇒ 𝑇2 − 𝑇1 − 𝜇𝑘 2𝑀𝑔 = 2𝑀𝑎
(3)
(1) − (2) + (3) ⇒ −𝑀𝑔 + 2𝑀𝑔 − 𝜇𝑘 2𝑀𝑔 = 𝑀𝑎 + 2𝑀𝑎 + 2𝑀𝑎 = 5𝑀𝑎
⇒ 𝜇𝑘 =
𝑀𝑔 − 5𝑀𝑎 𝑔 − 5𝑎
=
= 0.372
2𝑀𝑔
2𝑔
Discussion Class 2
Problems
P16) A loaded penguin sled weighing 80 N rests on a plane inclined at angle 𝜃 = 20° to the
horizontal. Between the sled and the plane, the coefficient of static friction is 0.25, and the
coefficient of kinetic friction is 0.15. A force 𝐹⃗ is applied to the sled that is parallel to the plane and
directed up the plane.
𝐹⃗ = 𝐹𝑖̂ + 0𝑗̂
𝐹⃗𝑁 = 0𝑖̂ + 𝐹𝑁 𝑗̂
𝐹⃗𝑔 = 𝑊 cos 250° 𝑖̂ + 𝑊 sin 250° 𝑗̂
𝑓⃗ = ±𝑓𝑖̂ + 0𝑗̂
𝐹⃗𝑛𝑒𝑡 = (𝐹 + 𝑊 cos 250° ± 𝑓)𝑖̂ + (𝐹𝑁 + 𝑊 sin 250°)𝑗̂
a) What is the minimum magnitude of 𝐹⃗ is required so that the sled does not slide down the
plane?
𝐹⃗𝑛𝑒𝑡 = 0𝑖̂ + 0𝑗̂ ⇒ (𝐹 + 𝑊 cos 250° + 𝜇𝑠 𝐹𝑁 )𝑖̂ + (𝐹𝑁 + 𝑊 sin 250°)𝑗̂ = 0𝑖̂ + 0𝑗̂
⇒ 𝐹𝑁 = −𝑊 sin 250°
⇒ 𝐹 = 𝜇𝑠 𝑊 sin 250° − 𝑊 cos 250° = 8.57 𝑁
b) What is the minimum magnitude of 𝐹⃗ is required so that the sled will start to move up the
plane?
𝐹⃗𝑛𝑒𝑡 = 0𝑖̂ + 0𝑗̂ ⇒ (𝐹 + 𝑊 cos 250° − 𝜇𝑠 𝐹𝑁 )𝑖̂ + (𝐹𝑁 + 𝑊 sin 250°)𝑗̂ = 0𝑖̂ + 0𝑗̂
⇒ 𝐹𝑁 = −𝑊 sin 250°
⇒ 𝐹 = −𝜇𝑠 𝑊 sin 250° − 𝑊 cos 250° = 46.2 𝑁
c) What magnitude of 𝐹⃗ is required so that the sled maintains constant velocity once it starts
moving up the plane?
𝐹⃗𝑛𝑒𝑡 = 0𝑖̂ + 0𝑗̂ ⇒ (𝐹 + 𝑊 cos 250° − 𝜇𝐾 𝐹𝑁 )𝑖̂ + (𝐹𝑁 + 𝑊 sin 250°)𝑗̂ = 0𝑖̂ + 0𝑗̂
⇒ 𝐹𝑁 = −𝑊 sin 250°
⇒ 𝐹 = −𝜇𝐾 𝑊 sin 250° − 𝑊 cos 250° = 38.6 𝑁
P25) Block B in the diagram alongside weighs 750N. The coefficient of static friction between the
block and table is 0.25; angle 𝜃 is 30°; assume the cord between B and the knot is horizontal. Find
the maximum weight of block A for which the system will be stationary.
For Block B:
𝐹⃗𝑛𝑒𝑡 = 0𝑖̂ + 0𝑗̂ ⇒ (𝑇𝐵 − 𝜇𝑠 𝐹𝑁 )𝑖̂ + (𝐹𝑁 − 𝐹𝑔 )𝑗̂ = 0𝑖̂ + 0𝑗̂
⇒ 𝐹𝑁 = 750
⇒ 𝑇𝐵 = 𝜇𝑠 𝐹𝑁 = 187.5 𝑁
For Knot:
𝐹⃗𝑛𝑒𝑡 = 0𝑖̂ + 0𝑗̂ ⇒ (𝑇𝐵 − 𝑇𝐶 cos 𝜃)𝑖̂ + (𝑇𝐶 sin 𝜃 − 𝑇𝐴 )𝑗̂ = 0𝑖̂ + 0𝑗̂
⇒ 𝑇𝐶 =
⇒ 𝑇𝐴 =
𝑇𝐵
cos 𝜃
𝑇𝐵
sin 𝜃 = 𝑇𝐵 tan 𝜃
cos 𝜃
For Block A:
𝐹𝑛𝑒𝑡 = 0 ⇒ 𝑇𝐴 − 𝑊𝐴 = 0
⇒ 𝑊𝐴 = 𝑇𝐵 tan 𝜃 = 108 𝑁
P28) Body A weighs 102 N, and body B weighs 32 N. The coefficients of friction between A and the
incline are 𝜇𝑠 = 0.56 and 𝜇𝐾 = 0.25. Angle 𝜃 is 40°. Let the positive direction of the x axis be up the
incline.
For Block A
⃗⃗ = 𝑇𝑖̂ + 0𝑗̂
𝑇
𝐹⃗𝑁 = 0𝑖̂ + 𝐹𝑁 𝑗̂
𝐹⃗𝑔 = 𝑊𝐴 cos 230° 𝑖̂ + 𝑊𝐴 sin 230° 𝑗̂
𝑓⃗ = ∓𝑓𝑖̂ + 0𝑗̂
𝐹⃗𝑛𝑒𝑡 = (𝑇 + 𝑊𝐴 cos 230° ∓ 𝑓)𝑖̂ + (𝐹𝑁 + 𝑊𝐴 sin 230°)𝑗̂
For Block B
⃗⃗ = 0𝑖̂ + 𝑇𝑗̂
𝑇
𝐹⃗𝑔 = 0𝑖̂ − 𝑊𝐵 𝑗̂
𝐹⃗𝑛𝑒𝑡 = 0𝑖̂ + (𝑇 − 𝑊𝐵 )𝑗̂
a) What is the acceleration of A if the system is initially at rest?
Since object is initially at rest we first need to check whether or not it will move so assume it is
stationary so assume attempted motion is up the slope:
For Block B 𝐹⃗𝑛𝑒𝑡 = 0𝑖̂ + 0𝑗̂ ⇒ 𝑇 = 𝑊𝐵
Thus for Block A: 𝐹⃗𝑛𝑒𝑡 = 0𝑖̂ + 0𝑗̂ ⇒ 𝑇 + 𝑊𝐴 cos 230° − 𝑓 = 0
⇒ 𝑓 = −𝑊𝐴 cos 230° − 𝑊𝐵 = 33.56 𝑁
And 𝐹𝑁 + 𝑊𝐴 sin 230° = 0 ⇒ 𝐹𝑁 = −𝑊𝐴 sin 230° = 78.14 𝑁 ⇒ 𝑓𝑠,𝑚𝑎𝑥 = 43.76 𝑁
Thus 𝑓 < 𝑓𝑠,𝑚𝑎𝑥 and system is still stationary thus acceleration is zero
b) What is the acceleration of A if A is initially moving up the incline?
For Block B 𝐹⃗𝑛𝑒𝑡 = 0𝑖̂ − 𝑚𝐵 𝑎𝑗̂ ⇒ 𝑇 = 𝑊𝐵 − 𝑚𝐵 𝑎
Thus for Block A: 𝐹⃗𝑛𝑒𝑡 = 𝑚𝐴 𝑎𝑖̂ + 0𝑗̂
Thus as before 𝐹𝑁 = −𝑊𝐴 sin 230° = 78.14 𝑁 but
⇒ 𝑇 + 𝑊𝐴 cos 230° − 𝜇𝐾 𝐹𝑁 = 𝑚𝐴 𝑎
𝑚𝐴 𝑎 = 𝑊𝐵 − 𝑚𝐵 𝑎 + 𝑊𝐴 cos 230° − 𝜇𝐾 𝐹𝑁
⇒𝑎=
𝑊𝐵 + 𝑊𝐴 cos 230° − 𝜇𝐾 𝐹𝑁
= −3.88 𝑚/𝑠 2
𝑚𝐴 + 𝑚𝐵
c) What is the acceleration of A if A is initially moving down the incline?
For Block B 𝐹⃗𝑛𝑒𝑡 = 0𝑖̂ + 𝑚𝐵 𝑎𝑗̂ ⇒ 𝑇 = 𝑊𝐵 + 𝑚𝐵 𝑎
Thus for Block A: 𝐹⃗𝑛𝑒𝑡 = −𝑚𝐴 𝑎𝑖̂ + 0𝑗̂
Thus as before 𝐹𝑁 = −𝑊𝐴 sin 230° = 78.14 𝑁 but
⇒ 𝑇 + 𝑊𝐴 cos 230° + 𝜇𝐾 𝐹𝑁 = −𝑚𝐴 𝑎
−𝑚𝐴 𝑎 = 𝑊𝐵 + 𝑚𝐵 𝑎 + 𝑊𝐴 cos 230° + 𝜇𝐾 𝐹𝑁
𝑊𝐵 + 𝑊𝐴 cos 230° + 𝜇𝐾 𝐹𝑁
⇒𝑎=
= 1.03 𝑚/𝑠 2
−𝑚𝐴 − 𝑚𝐵
Discussion Class 3
Questions
Q9) The path of a park ride that travels at constant speed is shown below. The ride goes through five
circular arcs of radii 𝑅0 , 2𝑅0 and 3𝑅0 . Rank the arcs according to the magnitude of the centripetal
force on a rider travelling in the arcs greatest first
4,3,1=2=5
𝐹𝑐𝑒𝑛𝑡𝑟𝑖𝑝𝑒𝑡𝑎𝑙 = 𝑚
𝑣2
𝑟
thus greater radius of curve means smaller 𝐹𝑐𝑒𝑛𝑡𝑟𝑖𝑝𝑒𝑡𝑎𝑙
Q11) A person riding a Ferris wheel moves through position at (1) the top, (2) the bottom, and (3)
midheight. If the wheel rotates at a constant rate, rank (greatest first) these three positions
according to the magnitude of the person’s centripetal acceleration, net centripetal force, and
normal force.
Centripetal acceleration
Ranking
All same (𝑣 and 𝑟 are all the same and acceleration is given by 𝑣 2 /𝑟)
Net Centripetal Force
All same (𝐹𝑐𝑒𝑛𝑡𝑟𝑖𝑝𝑒𝑡𝑎𝑙 = 𝑚𝑎𝑐𝑒𝑛𝑡𝑟𝑖𝑝𝑒𝑡𝑎𝑙 )
Normal Force
2,3,1 (𝐹𝑁 and 𝐹𝑔 combined is the force in responsible for acceleration
towards the centre of the circle at all times. At the bottom 𝐹𝑁 it must
oppose gravity but at the top it is aidied by gravity to produce the
net force towards the centre)
Problems
P57) A puck of mass 𝑚 = 1.50 𝑘𝑔 slides in a circle of radius 𝑟 = 0.20 𝑚 on
a frictionless table while attached to a hanging cylinder of mass 𝑀 =
2.50 𝑘𝑔 by a cord through a hole in the table. Show (including all working
and diagrams) that the speed of the puck be 1.81 𝑚. 𝑠 −1
For the cylinder:
Since it is stationary then by Newton’s 1st law
𝐹𝑛𝑒𝑡 = 0 ⇒ 𝑇 − 𝑀𝑔 = 0 ⇒ 𝑇 = 𝑀𝑔
For the puck:
𝐹𝑐𝑒𝑛𝑡𝑟𝑖𝑝𝑒𝑡𝑎𝑙 = 𝑚
𝑣2
𝑣2
𝑇𝑟
𝑀𝑔𝑟
⇒𝑇=𝑚
⇒𝑣=√ =√
= 1.81 𝑚. 𝑠 −1
𝑟
𝑟
𝑚
𝑚
P68) If a car goes through a curve too fast, the car tends to slide out of the curve. For a banked curve
with friction, a frictional force acts on the car to oppose the tendency to slide out of the curve. The
force is directed down the bank. Consider a circular curve of radius R=250 m and bank angle 𝜃,
where the coefficient of static friction between tyres and tar is 𝜇𝑠 . A car (without negative lift) is
driven around the curve as shown below.
𝐹⃗𝑁 = 𝐹𝑁 sin 𝜃 𝑖̂ + 𝐹𝑁 cos 𝜃 𝑗̂
𝐹⃗𝑔 = 0𝑖̂ − 𝑚𝑔𝑗̂
𝑓⃗𝑠 = 𝑓𝑠 cos 𝜃 𝑖̂ − 𝑓𝑠 sin 𝜃 𝑗̂
a) Find an expression for the max speed 𝑣𝑚𝑎𝑥 that puts the car on the verve of sliding out.
Since we have max speed then we know 𝑓𝑠 = 𝜇𝑠 𝐹𝑁 hence
𝐹𝑐𝑒𝑛𝑡𝑟𝑖𝑝𝑒𝑡𝑎𝑙 = 𝑚
𝑣𝑚𝑎𝑥 2
𝑣𝑚𝑎𝑥 2
⇒ 𝐹𝑁 sin 𝜃 + 𝜇𝑠 𝐹𝑁 cos 𝜃 = 𝑚
𝑟
𝑟
But also 𝐹𝑛𝑒𝑡,𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 = 0 ⇒ 𝐹𝑁 − 𝑚𝑔 cos 𝜃 = 0 ⇒ 𝐹𝑁 = 𝑚𝑔 cos 𝜃
Thus
𝑚𝑔 cos 𝜃 sin 𝜃 + 𝜇𝑠 𝑚𝑔 cos 2 𝜃 = 𝑚
𝑣𝑚𝑎𝑥 2
𝑟
𝑣𝑚𝑎𝑥 = √𝑔𝑟 cos 𝜃 sin 𝜃 + 𝜇𝑠 𝑔𝑟 cos 2 𝜃
b) Calculate 𝑣𝑚𝑎𝑥 in kilometres per hour for a bank of angle 𝜃 = 10° in dry conditions (𝜇𝑠 =
0.60) and again in wet conditions (𝜇𝑠 = 0.050). (Now you know why there are so many
accidents on freeways under wet conditions)
Using formula above for dry conditions: 𝑣𝑚𝑎𝑥 = 42.94 𝑚/𝑠 = 155𝑘𝑚/ℎ
for wet conditions: 𝑣𝑚𝑎𝑥 = 23.19 𝑚/𝑠 = 83.5 𝑘𝑚/ℎ
Discussion Class 4
Problems
Coming Soon
P77) What is the terminal speed of a 6.00 kg sphere that has a radius of 3.50 cm and a drag
coefficient of 1.60? The density of the air through which it falls is 1.20 kg/m3
P80) Calculate the magnitude of the drag force oj a missile 60 cm in diameter, cruising at 250 m/s at
low altitude, where the density of air is 1.20 kg/m3. Assume C is 0.75.
ST2 - 2013 1 e) A car, with mass 900 𝑘𝑔, has a drag coefficient of 0.3 and the density of air is
1.3 𝑘𝑔/𝑚3. If the car coasts down a hill (in neutral) with an incline of 15.5°. Determine the terminal
velocity of the car if it is 2.5 𝑚 wide and 1.12 𝑚 high. The coefficient of kinetic friction between the
tyres and road is 𝜇𝑘 = 0.27.