Discussion Class 1
Questions
Q1) Below are the free-body diagrams for four situations in which an object is pulled by several
forces across frictionless floor, as seen from overhead. In which situations does the object’s
acceleration 𝑎⃗ have an x component? In which situations does the object’s acceleration 𝑎⃗ have an y
component? In each situation give the direction of 𝑎⃗ by naming either a quadrant or an axis
𝐹⃗𝑛𝑒𝑡 = 𝑚𝑎⃗ thus if there is a net force in along any
axis there is an acceleration along that axis:
Situation 1:
𝐹⃗𝑛𝑒𝑡 = (5 − 3 − 2)𝑖̂ + (7 − 4)𝑗̂ = 0𝑖̂ + 3𝑗̂
⇒ 𝑎⃗ 𝑎𝑙𝑜𝑛𝑔 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑦 𝑎𝑥𝑖𝑠
Situation 2:
𝐹⃗𝑛𝑒𝑡 = (3 − 2)𝑖̂ + (6 − 4 − 2)𝑗̂ = 1𝑖̂ + 0𝑗̂
⇒ 𝑎⃗ 𝑎𝑙𝑜𝑛𝑔 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑥 𝑎𝑥𝑖𝑠
Situation 3:
𝐹⃗𝑛𝑒𝑡 = (5 − 4)𝑖̂ + (6 − 4 − 3)𝑗̂ = 1𝑖̂ − 1𝑗̂
⇒ 𝑎⃗ 𝑝𝑜𝑖𝑛𝑡𝑠 𝑖𝑛𝑡𝑜 4𝑡ℎ 𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡
Situation 4:
𝐹⃗𝑛𝑒𝑡 = (3 − 5)𝑖̂ + (2 + 3 − 4 − 5)𝑗̂ = −2𝑖̂ − 4𝑗̂
⇒ 𝑎⃗ 𝑝𝑜𝑖𝑛𝑡𝑠 𝑖𝑛𝑡𝑜 3𝑟𝑑 𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡
Q3) In the diagram alongside, forces 𝐹⃗1 and 𝐹⃗2 are applied to
a box as it moves at constant velocity over a frictionless
floor. We manage to decrease the angle 𝜃 without changing
the magnitude of 𝐹⃗1 . For constant velocity to be kept at all
stages should we increase, decrease, or maintain the
magnitude of 𝐹⃗2 ?
According to Newton’s first law: 𝐹𝑛𝑒𝑡,𝑥 = 0 ⇒ 𝐹1 cos 𝜃 − 𝐹2 = 0 ⇒ 𝐹2 = 𝐹1 cos 𝜃 as 𝜃 decreases
then cos 𝜃 increases and thus so must 𝐹⃗2 in order to maintain constant velocity
Problems
P6) In a two-dimensional tug-of-war, Alex, Betty, and Charles pull horizontally
on a car tyre at the angles shown in the overhead view alongside. The tire
remains stationary in spite of the three pulls. Alex pulls with force 𝐹⃗𝐴 of
magnitude 220 N and Charles pulls with force 𝐹⃗𝐶 of magnitude 170 N (note that
the direction of 𝐹⃗𝐶 is not given). What is the magnitude of Betty’s force 𝐹⃗𝐵 ?
𝐹⃗𝐴 = 220 cos 133° 𝑖̂ + 220 sin 133° 𝑗̂
𝐹⃗𝐵 = 0𝑖̂ − 𝐹𝐵 𝑗̂
𝐹⃗𝐴 = 170 cos 𝜃 𝑖̂ + 170 sin 𝜃 𝑗̂
𝐹⃗𝑛𝑒𝑡 = (220 cos 133° + 170 cos 𝜃)𝑖̂ + (220 sin 133° − 𝐹𝐵 + 170 sin 𝜃)𝑗̂ = 0𝑖̂ + 0𝑗̂
220 cos 133° + 170 cos 𝜃 = 0 ⇒ 𝜃 = cos −1 (−
220 cos 133°
) = 28°
170
220 sin 133° − 𝐹𝐵 + 170 sin 𝜃 = 0 ⇒ 𝐹𝐵 = 220 sin 133° + 170 sin 28° = 241𝑁
P17) In the diagram below the mass of the block is 8.5kg and the angle of the incline 𝜃 is 30°.
⃗⃗ = 𝑇𝑖̂ + 0𝑗̂
𝑇
𝐹⃗𝑁 = 0𝑖̂ + 𝐹𝑁 𝑗̂
𝐹⃗𝑔 = 𝑚𝑔 cos 240° 𝑖̂ + 𝑚𝑔 sin 240° 𝑗̂
𝐹⃗𝑛𝑒𝑡 = (𝑇 + 𝑚𝑔 cos 240°)𝑖̂ + (𝐹𝑁 + 𝑚𝑔 sin 240°)𝑗̂
a) Find the tension in the cord.
By Newton’s First Law: 𝐹𝑛𝑒𝑡,𝑥 = 0 ⇒ 𝑇 + 𝑚𝑔 cos 240° = 0 ⇒ 𝑇 = −𝑚𝑔 cos 240° = 41.65 𝑁
b) Find the normal force acting on the block.
By Newton’s First Law: 𝐹𝑛𝑒𝑡,𝑦 = 0 ⇒ 𝐹𝑁 + 𝑚𝑔 sin 240° = 0 ⇒ 𝐹𝑁 = −𝑚𝑔 sin 240° = 72.14 𝑁
c) If the cord is cut find the magnitude of the acceleration experienced by the block.
If cord is cut then there is no tension and 𝐹𝑛𝑒𝑡,𝑥 = 𝑚(−𝑎) ⇒ 𝑚𝑔 cos 240° = 𝑚(−𝑎)
𝑎 = −𝑔 cos 240° = 4.9 𝑚/𝑠 2
Discussion Class 2
Questions
Q10) The diagram shows three blocks being pushed across a horizontal floor by horizontal force 𝐹⃗ .
What is the total mass accelerated to the right by force 𝐹⃗ ?
The presence of 𝐹⃗ means all the blocks move and it accelerates the entire system i.e. 17 kg
a) What is the total mass accelerated to the right by force 𝐹⃗12 (force of block 1 on 2)?
𝐹⃗12 accelerates all masses to the right of it hence mass is 12 kg
b) What is the total mass accelerated to the right by force 𝐹⃗23 (force of block 2 on 3)?
𝐹⃗23 accelerates all masses to the right of it hence mass is 10 kg
c) Rank the blocks according to the magnitude of acceleration experienced by each (greatest
first)
All tied. System moves as one unit thus meaning it is impossible for any piece of the system to have
different accelerations
d) Rank forces 𝐹⃗ , 𝐹⃗12 and 𝐹⃗23 according to magnitude (greatest first)
𝐹⃗ > 𝐹⃗12 > 𝐹⃗23. 𝐹⃗𝑛𝑒𝑡 = 𝑚𝑎⃗ by Newton’s second law thus by considering forces and the
corresponding total masses accelerated by each force then we see that the force is proportional to
these masses since acceleration is the same
Q11) A vertical force 𝐹⃗ is applied to a block of mass m that lies on a floor.
a) What happens to the magnitude of the normal force 𝐹⃗𝑁 on the block from the floor as
magnitude F is increased from zero if force 𝐹⃗ is downwards?
By Newton’s first law since the object does not move through the surface
𝐹𝑛𝑒𝑡 = 0 ⇒ 𝐹𝑁 − 𝑚𝑔 − 𝐹 = 0 ⇒ 𝐹𝑁 = 𝑚𝑔 + 𝐹
Thus as 𝐹 increases then so does 𝐹𝑁
b) What would happen if force 𝐹⃗ was directed upwards instead?
Initially since 𝐹⃗ is very small then it is not bigger than gravitational force hence Newton’s first law
still applies and thus
𝐹𝑛𝑒𝑡 = 0 ⇒ 𝐹𝑁 − 𝑚𝑔 + 𝐹 = 0 ⇒ 𝐹𝑁 = 𝑚𝑔 − 𝐹
Thus as 𝐹 increases then 𝐹𝑁 decreases until it reaches zero after which it lifts off the surface
Problems
P49) A block of mass 𝑚 = 5.00 𝑘𝑔 is pulled along a horizontal frictionless floor by a force of
magnitude 𝐹 = 12.0 𝑁 at an angle of 𝜃 = 25.0°.
𝐹⃗ = 𝐹 cos 𝜃 𝑖̂ + 𝐹 sin 𝜃 𝑗̂
𝐹⃗𝑁 = 0𝑖̂ + 𝐹𝑁 𝑗̂
𝐹⃗𝑔 = 0𝑖̂ − 𝑚𝑔𝑗̂
𝐹⃗𝑛𝑒𝑡 = (𝐹 cos 𝜃)𝑖̂ + (𝐹 sin 𝜃 + 𝐹𝑁 − 𝑚𝑔)𝑗̂
a) What is the magnitude of the block’s acceleration?
By Newton’s 2nd Law 𝐹𝑛𝑒𝑡,𝑥 = 𝑚𝑎 ⇒ 𝐹 cos 𝜃 = 𝑚𝑎 ⇒ 𝑎 =
𝐹 cos 𝜃
𝑚
= 2.18 𝑚/𝑠 2
b) The force magnitude 𝐹 is slowly increased. What is its magnitude just before the block lifts
off the floor?
At point when block is about to leave floor then 𝐹𝑁 = 0 thus by Newton’s First Law:
𝐹𝑛𝑒𝑡,𝑦 = 0 ⇒ 𝐹 sin 𝜃 − 𝑚𝑔 = 0 ⇒ 𝐹 =
𝑚𝑔
= 116 𝑁
sin 𝜃
c) What is the acceleration of the block just before the block is lifted off the floor?
As per (a) we have 𝑎 =
𝐹 cos 𝜃
𝑚
= 21.0 𝑚/𝑠 2
P50) Three ballot boxes are connected by cords, one which wraps over a pulley having negligible
friction on its axle and negligible mass. The three masses are 𝑚𝐴 = 30.0 𝑘𝑔, 𝑚𝐵 = 30.0 𝑘𝑔 and
𝑚𝐶 = 10.0 𝑘𝑔.
a) When the assembly is released from rest, what is the tension in the cord connecting B and
C?
𝐹𝑛𝑒𝑡,𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑚𝑡𝑜𝑡𝑎𝑙 𝑎
𝑚𝐵 𝑔 + 𝑚𝐶 𝑔 = (𝑚𝐴 + 𝑚𝐵 + 𝑚𝐶 )𝑎
𝑎=
𝑚𝐵 𝑔 + 𝑚𝐶 𝑔
= 5.6 𝑚/𝑠 2
𝑚𝐴 + 𝑚𝐵 + 𝑚𝐶
𝐹𝑛𝑒𝑡,𝐶 = 𝑚𝐶 (−𝑎)
𝑇 − 𝑚𝐶 𝑔 = −𝑚𝐶 𝑎
𝑇 = 𝑚𝐶 𝑔 − 𝑚𝐶 𝑎 = 42𝑁
b) How far does A move in the first 0.250 s (assuming it does not reach the pully)
1
1
∆𝑥 = 𝑣0 𝑡 + 𝑎𝑡 2 = 0(0.25) + (5.6)(0.25)2 = 0.175 𝑚
2
2
Discussion Class 3
Problems
P55) Two blocks are in contact on a frictionless table. A horizontal force is applied to the larger
block. If 𝑚1 = 2.3 𝑘𝑔, 𝑚2 = 1.2 𝑘𝑔 and 𝐹 = 3.2 𝑁.
a) Find the magnitude of the force between the two blocks
𝐹𝑛𝑒𝑡,𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑚𝑡𝑜𝑡𝑎𝑙 𝑎
𝐹
3.2
𝐹 = (𝑚1 + 𝑚2 )𝑎 ⇒ 𝑎 =
=
𝑚1 + 𝑚2 3.5
3.2
𝐹𝑛𝑒𝑡,2 = 𝑚2 𝑎 ⇒ 𝐹12 = 1.2 ( ) = 1.10𝑁
3.5
b) Show that if the force 𝐹 is instead applied to the smaller block in the opposite direction to
before, then the force between the two blocks is 2.1 N (Note this value is not the same as in
(a))
As before 𝑎 = 𝑚
𝐹
+𝑚
1
2
3.2
= 3.5 but now there is only one force on 𝑚1 thus
3.2
𝐹𝑛𝑒𝑡1 = 𝑚1 𝑎 ⇒ 𝐹21 = 2.3 ( ) = 2.10𝑁
3.5
P57) A block of mass 𝑚1 = 3.70 𝑘𝑔 on a frictionless plane inclined at 𝜃 = 30° is connected by a cord
over a massless, frictionless pulley to a second block of mass 𝑚2 = 2.30 𝑘𝑔.
For box 1:
𝑖̂
𝑚1 𝑔 cos 240°
𝑗̂
𝑚1 𝑔 sin 240°
𝐹𝑁,1
0
𝑖̂
0
𝑗̂
−𝑚2 𝑔
0
𝑇
⃗⃗⃗⃗
𝐹𝑔
⃗⃗⃗⃗⃗
0
𝐹𝑁
⃗⃗
𝑇
𝑇
𝐹⃗𝑛𝑒𝑡 1 = (𝑇 + 𝑚1 𝑔 cos 240°)𝑖̂ + (𝐹𝑁,1 + 𝑚1 𝑔 sin 240°)𝑗̂ = 𝑚1 𝑎𝑖̂
For box 2:
⃗⃗⃗⃗
𝐹𝑔
⃗⃗
𝑇
𝐹⃗𝑛𝑒𝑡 2 = 0𝑖̂ + (𝑇 − 𝑚2 𝑔)𝑗̂ = −𝑚2 𝑎𝑖̂
a) What is the magnitude of the acceleration of each block?
𝑚1 𝑎 + 𝑚2 𝑎 = 𝐹𝑛𝑒𝑡 1,𝑥 − 𝐹𝑛𝑒𝑡 2,𝑦 = 𝑚1 𝑔 cos 240° + 𝑚2 𝑔
⇒𝑎=
𝑚1 𝑔 cos 240° + 𝑚2 𝑔
= 0.735 𝑚/𝑠2
𝑚1 + 𝑚2
b) What is the direction of the acceleration of the hanging block?
Downwards
c) What is the tension in the cord?
𝐹𝑛𝑒𝑡 2,𝑦 = −𝑚2 𝑎 ⇒ 𝑇 = 𝑚2 𝑔 − 𝑚2 𝑎 = 20.85 𝑁
P71) The diagram shows a block of dirty money (𝑚1 = 3.0 𝑘𝑔) on a frictionless inclined plane at
angle 𝜃1 = 30°. The box is connected via a cord of negligible mass to a box of laundered money
(𝑚2 = 2.0 𝑘𝑔) on a frictionless plane inclined at angle 𝜃2 = 60°. The pulley is frictionless and has
negligible mass. What is the tension in the cord?
acceleration of each block
For box 1:
𝑖̂
𝑗̂
⃗⃗⃗⃗
𝑚1 𝑔 cos 240°
𝑚1 𝑔 sin 240°
𝐹𝑔
⃗⃗⃗⃗⃗
𝐹𝑁,1
0
𝐹𝑁
⃗⃗
𝑇
0
𝑇
𝐹⃗𝑛𝑒𝑡,1 = (𝑇 + 𝑚1 𝑔 cos 240°)𝑖̂ + (𝐹𝑁,1 + 𝑚1 𝑔 sin 240°)𝑗̂ = 𝑚1 𝑎𝑖̂
For box 2:
𝑖̂
𝑗̂
⃗⃗⃗⃗
𝑚
𝑔
cos
210°
𝑚
𝑔
sin
210°
𝐹𝑔
2
2
⃗⃗⃗⃗⃗
𝐹𝑁,2
0
𝐹𝑁
⃗⃗
𝑇
0
𝑇
𝐹⃗𝑛𝑒𝑡,2 = (𝑇 + 𝑚2 𝑔 cos 210°)𝑖̂ + (𝐹𝑁,2 + 𝑚2 𝑔 sin 210°)𝑗̂ = −𝑚2 𝑎𝑖̂
Hence: 𝑚1 𝑎 + 𝑚2 𝑎 = (𝐹⃗𝑛𝑒𝑡,1 − 𝐹⃗𝑛𝑒𝑡,2 )𝑥 = 𝑚1 𝑔 cos 240° − 𝑚2 𝑔 cos 210°
𝑎=
𝑚1 𝑔 cos 240° − 𝑚2 𝑔 cos 210°
= 0.455 𝑚/𝑠 2
𝑚1 + 𝑚2
From box 1: 𝑇 = −𝑚1 𝑔 cos 240° + 𝑚1 𝑎 = 16.1 𝑁