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02_faq

Page history last edited by johnlittlephysics 10 years, 12 months ago

02 Dynamics FAQ


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Comments (Show all 41)

Pang Seok Mun said

at 9:33 pm on Apr 5, 2013

Mr Ang, can you explain the difference between rising in a parabolic path and parabolic trajectory?

Why isn't there air resistance acting on an object rising in a parabolic path?

johnlittlephysics said

at 5:42 pm on Apr 6, 2013

You are referring to WS 2.1 Exercise E.
Basically these two terms mean the same thing.
In Question 2 (e) and (f), air resistance was assumed negligible.
Otherwise, there should be force of air resistance opposite to its velocity at that instant, in (e) arrow point down slant towards left, and in (f), horizontal arrow towards left.

johnlittlephysics said

at 5:31 pm on Apr 6, 2013

Why two bodies (e.g. a box lying on flat ground) are in contact, they exert forces on each other. At the microscopic level, their molecules at the contact surface exert forces on each other.
E.g. the molecules of the ground repel the molecules of the box above it, so the molecules of the box (at its bottom) also repel the molecules of the ground. We describe the ground as pushing up on the box (exerting an upward normal contact force), otherwise, the box would fall vertically down!
Similarly the box is also pushing down on the ground. This is a pair of action-reaction forces (by Newton's 3rd law of motion). If the ground was replaced by a weighing scale, it would show a reading, due to the downward force from the box.

Tam Li June said

at 9:53 pm on Apr 8, 2013

Hi Mr Ang,


Is it true that the gravitational acceleration will always be negative, -10, when an object is going up and it is positive,10, when an object is coming down?

johnlittlephysics said

at 9:59 pm on Apr 8, 2013

1. The acceleration due to gravity (g) is caused by the weight (W) where W = mg. This is just like F = ma, where force F caused acceleration of a body of mass m).
2. Since W act downwards, the direction of acceleration g of a body is downwards (both are vectors in same direction).
Hence, a body moving (or falling) downwards will accelerate or move faster.
On the other hand, a body moving moving upwards will deceleration or slow down (since W and g are opposite to its motion).

3. Whether the sign of the acceleration a is positive or negative depends on the "sign convention" you choose.
a. If you choose upwards is positive, acceleration due to gravity a = -10 m/s^2 (since a is directed downwards, opposite to your positive sign convention).
b. Similarly, if you choose downwards is positive, acceleration due to gravity a = +10 m/s^2 (since a is directed downwards!)
- the sign would be important when applying the equations of motion, or for velocity-time graph.

johnlittlephysics said

at 9:16 pm on Apr 15, 2013

From Li June: May I know what is thrust? When does it exist? For up thrust, I understand that it is an a upward force exerted by a fluid on an object in the fluid that opposes the object's weight, but I am not sure about upthrust. Is it the a force present when an object is moving?

johnlittlephysics said

at 9:20 pm on Apr 15, 2013

Upthrust is a single word: an upward force acting on an object when it is immersed (partially or fully) in a liquid , regards of whether it is at rest or moving.
Note: This force is present when an object is in a gas (e.g. air), however this is normally neglected because it is a very small force due to the low density of gas.

johnlittlephysics said

at 9:23 pm on Apr 15, 2013

Thrust is usually used to label a force exerted on a machine or vehicle (e.g. a plane, spacecraft or a rocket) to propel it in a direction in which it is supposed to move.

Pang Seok Mun said

at 5:20 pm on Apr 17, 2013

Hi Mr Ang! Is weight and normal contact force acting on an object a pair of action reaction forces?

johnlittlephysics said

at 10:20 pm on Apr 17, 2013

You can refer to WS2.4 on Newton's 3rd law. A pair of action-reaction forces must be of same type of forces.
In AS2.1 when u describe a force by A on B, and the force by B on A, these represent such a pair of forces.
E.g. of a pair of action-reaction forces:
1. Weight of stone (or gravitational force by Earth on stone)
2. Gravitational force on stone by Earth

johnlittlephysics said

at 1:39 pm on Apr 21, 2013

312 Yu Kee:
1. free body diagrams, we rarely have to label up thrust or air resistance right? Then in which cases do we need to include them?
2. Does the weight an object affect the amount of friction acting on it? I mean basically do they have any relation? I'm not sure because they're both acting in different directions, but they still seem to be related.

johnlittlephysics said

at 1:40 pm on Apr 21, 2013

1. Upthrust: body is placed in a liquid; direction is upwards
Air resistance: body moves through air, when air resistance is not neglected; direction is opposite to velocity of body
2. The greater the weight, the larger the normal contact force between the body & the surface,
the larger the maximum friction that can act horizontally by the surface on the body.

johnlittlephysics said

at 10:02 pm on Jul 22, 2013

302 Li June: May I check when does friction exist? Why does it not exist in the case of AS 2.1 4b) between the man and the plank?

johnlittlephysics said

at 10:05 pm on Jul 22, 2013

1. Does friction exist in a system?
Friction can exist between any 2 surfaces (of any states of matter) in contact. E.g. it can exist between the man and the plank in AS2.1 Q4.

2. Does friction actually act (or is exerted on a body) in a system in a specific state?
For 2 solids, e.g. man's feet and plank, friction only acts "along" (or parallel) to the plank's surface. Normal contact force acts "perpendicular" (or normal) to the surface of plank.
In AS 2.1, Q4, since all objects in the system are at rest, all the forces are acting in the vertical direction. Hence there is no horizontal forces involved and so friction (horizontal force along plank) does not act in this state.

Friction may act on the man (and on the plank) if
a. the man starts to walk, as friction acts horizontally on his feet (by the plank) as friction is involved in helping him to walk (along plank);
b. the man stands with his feet wide apart, friction acts horizontally on each feet in opposite directions (towards each other) to keep him at rest. Try to sketch this!

johnlittlephysics said

at 5:49 pm on Jul 26, 2013

305 Elizabeth: I am unable to solve the question attached. The answer is option D, but I do not understand why the object will continue moving up and with an acceleration of -10m/s^2.
https://www.dropbox.com/s/uhj7pbr2lse7env/Photo%2026-7-13%205%2047%2020%20PM.jpg

johnlittlephysics said

at 6:05 pm on Jul 26, 2013

There are three parts to the question:
1. After 3s, what is the velocity of the box?
Use Kinematics v-t graph or equation of motion method (shown here):
Taking upwards as +ve direction
v = u + at = 30 + (- 4)(3) = 18 m/s upwards. This actually leads to answer D immediately!
2. After 3s, what is the acceleration of the box?
Consider the free-body diagram of the box. When the rope snaps, the only force on the box is its weight, so acceleration is due to gravity at 10 m/s^2 downwards.
3. How would the box behave?
At 3 s, the box still have an upwards velocity of 18 m/s, so it would continue to move upwards. But it would slow down due to acceleration of gravity. It comes to rest eventually, before falling down again.
Refer to similar situation in assignment AS 2.3 question 3 (c) and (d).

Elizabeth Tong said

at 7:23 pm on Jul 26, 2013

Dear Mr Ang,

Whenever an object is traveling upwards, why would its acceleration always be -10 m/s^2 (taking downwards as positive)? If the object is thrown faster, would not the acceleration increase? In the case of my previous question, why is the object not traveling upwards at an acceleration of -4.0m/s^2 instead after the rope snapped? Thank you.

johnlittlephysics said

at 9:14 pm on Jul 26, 2013

1. If sign convention downwards is positive, acceleration due to gravity (downwards, due to its weight) would be positive: 10 m/s^2.
2. If you are referring to the acceleration due to the force of throwing, this acceleration only occurs when object is being thrown (e.g. hand still holding the object, contact force due to hand is acting). Once the object leaves the hand, the force due to the hand stops acting on the object, since there is no more contact (force).
In this case, only weight acts, and acceleration is due to gravity (10 m/s^2 downwards). (Fnet = ma, W = mg)
3. In previous question, as I mentioned, once rope snaps, no more tension, only weight acts on object,
Fnet = W, so a = g, acceleration due to gravity, in downwards direction, by Newton's 2nd law. Can u show me how you arrive at -4.0 m/s^2?

Elizabeth Tong said

at 9:31 pm on Jul 26, 2013

The box was traveling upwards at an acceleration of -4.0m/s^2, so I thought that the acceleration will remain even though the string snapped- but I understand the correct answer now.

johnlittlephysics said

at 9:41 pm on Jul 26, 2013

Ok. Note that acceleration is not caused by the velocity.
Acceleration of an object is caused by the net force (or vector sum of all the forces acting on that object).
This acceleration would then cause the velocity to change over time.

johnlittlephysics said

at 10:06 pm on Jul 26, 2013

Q24: the question mentions vertically upwards, so weight mg is involved.
F - mg = ma, so F = mg + ma
2F - mg = ma', then ma' = 2(mg + ma) - mg
a' = 2a + g > 2a.

johnlittlephysics said

at 10:18 pm on Jul 26, 2013

Q20: This is a challenging question. Let g = 10 m/s^2.
Forces on the system are the weight due to 20 kg at A and 10 kg at B and the friction f on table top.
Since system does not move initially, no net force, so forces towards left = forces to the right
20g = f + 10g, so f = 10g which would always oppose the motion of the 10 kg on the table top.
A: 0 + 20g = 10g + 10g (since weights are balanced at A and B, f does not act)
B: 5g + 20g + f = 25g + 10g (f acts to the left since side B has greater weight)
C: 10g + 20 g + f < 35g + 10g, so there is net force and system moves (accelerates)
D: 50g + 20g = f + 50g + 10g (f acts to the right since side A has greater weight)

Ng J-Cyn said

at 2:46 pm on Jul 30, 2013

Hello Mr Ang, but for Question 24 isn't the rocket thrusted into space and doesn't space have no gravity?

johnlittlephysics said

at 5:50 pm on Jul 30, 2013

In the above Q24, it states "initial thrust" and "vertically upwards", these suggest that the rocket is just fired from the Earth's surface. Otherwise it would have other indicators e.g, "in outer space".

Yin Xihui said

at 6:06 pm on Jul 30, 2013

Mr Ang, I don't get how friction works on the wheels of a moving vehicle. E.g. Something like one wheel doesn't move or something.
In addition, is there any difference between the way friction acts on a bicycle (which is not motor-operated) and a motorcycle (which is moto-operated)? Thanks!

johnlittlephysics said

at 6:46 pm on Jul 30, 2013

1. You need to revise WS 2.1 Ex B Question 5 on friction on a wheel.
View the video on that question at http://nyghsec3physics.pbworks.com/w/page/54997383/self_learning
2. Distinguish between frictional force on which type of wheel, NOT whether it is motor-operated or not.
3. To summarize:
a. free rolling wheel (can roll on its own if body is pushed/pulled), e.g. of a cart, front wheel of normal bicycle or motor cycle.
- the direction of frictional force on a free rolling wheel is opposite to direction of motion of the vehicle.
b. driven or powered wheel (rotates because it is connected to car engine, motor bike engine, or by a chain to the pedals of bicycle - bicycle rear wheel).
- the direction of frictional force on a powered wheel is in same direction of motion as the vehicle.

Tan Chek Lin said

at 6:30 pm on Jul 30, 2013

Hi Mr Ang, must we know the equations of motions for this block test?

johnlittlephysics said

at 6:47 pm on Jul 30, 2013

Equations of motion is under Kinematics.

johnlittlephysics said

at 8:01 am on Jul 31, 2013

312 Yu Kee:
1. For free body diagrams, when we're asked to label forces acting on for eg. the plank, do we have to label the weight of the plank? Also, if there is a load being hung on the plank by the means of a string, is the tension acting on the plank as well?
2. I am quite confused by the term constant (net) force. Previously I've always assumed that as long as the word 'constant' appears in the question there would be no resultant force, but now constant net force seems to imply that there is constantly more resultant force acting on it, and hence instead of constant velocity, it's constant acceleration?

johnlittlephysics said

at 8:02 am on Jul 31, 2013

1. "weight" of plank is a force on the plank, so it must be included.
- this is due to contact by the string, we label it as "tension" acting on the plank.
2. net force or resultant force is the sum of forces acting on a body.
- the word "constant" is not meaningful unless it is used together with a physical quantity!
E.g. constant velocity means zero acceleration,
constant acceleration means velocity is changing at a constant rate,
constant net force F = ma means acceleration a is also constant.

johnlittlephysics said

at 3:57 pm on Sep 15, 2013

302 Ying Ting: why is tension the same for both sides of the pulley even though different masses are hung? In that case, is the magnitude of tension affected by the magnitude of masses?

johnlittlephysics said

at 4:13 pm on Sep 15, 2013

1. For a string (or rope) over pulley, there is usually an underlying assumption that its mass is negligible, and frictional force (contact between string & pulley) is negligible.
2. For a string over a pulley, we can imagine it is equivalent to pulling both ends of string with forces (due to the weight of masses m1 and m2 hung at each end). The tension becomes a force in the string that m1 pulls on m2 and m2 pulls on m1. Like an action-reaction pair, tension on both sides of pulley are the same in magnitude.
3. If pulley system is at rest, tension on each side is equal to the weight on each side.
4. If pulley is accelerating, Newton's 2nd law is applied to each mass (m1 or m2) (free body diagram), each weight will also affect the magnitude of the tension.

Goh Ying Ting said

at 6:27 pm on Sep 15, 2013

Ok I see thanks!:)

Tam Li June said

at 10:45 am on Oct 10, 2013

Hi Mr Ang, may you please explain question 5) of AS 2.3 as I still not very sure about it?
Thanks.

johnlittlephysics said

at 5:59 pm on Oct 10, 2013

Since answer is provided, pls ask more specifically what u don't understand.

Tam Li June said

at 10:56 am on Oct 10, 2013

For question 5), why do we take the mass as 60 and not 860? Thanks.

johnlittlephysics said

at 6:01 pm on Oct 10, 2013

The forces shown are for the freebody of the man alone (forces on the man). The lift mass is not relevant in this analysis.

Koh Xin Qi said

at 5:35 pm on May 4, 2014

Dear Mr Ang,

If the question is phrased this way 'the car accelerates at a constant velocity' does that mean that resultant force is zero and Fnet=0?

johnlittlephysics said

at 6:25 pm on May 4, 2014

This is an incorrect phrasing as the words are in contradiction!
"Accelerates" means velocity is changing and cannot be constant.
It can be rephrased as "the car accelerates uniformly" or "the car moves with uniform or constant acceleration" which means constant a and constant net force.
Where is this question taken from?

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