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13_02dynamics

Page history last edited by johnlittlephysics 9 years, 2 months ago

02 Dynamics



Post your questions/comments below. 

Comments (22)

johnlittlephysics said

at 10:48 pm on Feb 26, 2013

Push a object and observe when it moves. Observe the forces, net force and motion of the object.
http://phet.colorado.edu/en/simulation/forces-1d
1. download the simulation onto a PC or laptop
2. will need Java software installation to run.

johnlittlephysics said

at 12:24 am on Apr 20, 2013

312 Tian Tian: I have problem with question 3(d) in As 2.3. I am not very sure what the question is asking about and why the answers are supposed to be 10s, 25m.

johnlittlephysics said

at 12:32 am on Apr 20, 2013

Draw a sketch diagram to show motion of the object, position over time and other quantities.
From 3(b), at the end of 10 s, the box is already 12.5m above ground & speed is 2.5 m/s.

Part (c): it then experienced an acceleration a=-0.25 m/s^2 causing it to slow down.
From this moment till highest point, u = 2.5 m/s, at max height, v=0
Applying v = u + at, time t = 10 s (another 10 s longer to reach highest point).
Applying v^2 = u^2 + 2as, so s = 12.5 m.

Total height reached above ground =12.5+12.5= 25 m (parts b and d).

johnlittlephysics said

at 1:44 pm on Apr 20, 2013

302 Denise: I would like to clarify, does mass of an object affect terminal velocity? For example, 2 objects with the same size and shape but one has a larger mass than the other, will both of them eventually reach the same terminal velocity when dropped from rest?

johnlittlephysics said

at 2:21 pm on Apr 20, 2013

2 objects of same size and shape would experience the same air resistance R when moving at the same speed v.
The following discussion is similar to that for AS2.2 Q4 drag force and acceleration of a moving car.

Forces acting on a falling object: weight W downwards and air resistance R upwards (if not neglected).
net force F = W - R, so ma = W - R, so acceleration a = (W - R)/m.
When terminal velocity is reached, a = 0, so R=W = mg.
If mass m is different, R is different so v is different.
Conclusion: 2 identical objects of different mass would reach a different terminal velocity.

johnlittlephysics said

at 9:08 pm on May 6, 2013

302 Agalia: I don't understand how to do the challenge yourself question in WS 2.3.

johnlittlephysics said

at 9:13 pm on May 6, 2013

From Archimedes principle, upthrust exerted on the block = weight of fluid displaced.
Consider a freebody of the block: 2 forces are upthrust and its weight.
Hence, weight of block = upthrust on it = weight of fluid displaced.
mass of block m x g = mass of fluid displaced x g
Hence, m = volume of fluid displaced x density of fluid = 1/4 x volume of block x fluid density
m = ¼ x (2 x 4 x 5) x 2000 = 20 000 kg.

Hua Yiting said

at 10:54 pm on May 9, 2013

Hi Mr Ang, I don't understand the pop quiz question 6, why was it initially at rest? The three forces cannot balance. And why the answer is B? Thank you!

johnlittlephysics said

at 4:44 pm on May 11, 2013

Dear Yiting,

The answer should be D (I gave wrong answer B), thanks for alerting me.
The question states the body was initially at rest (before the effect of the forces happen). The 3 forces acting on it are not balanced. Using vector resolution method, each 40 N has a horizontal component acting to the right of 40 cos 60 = 20 N. The net force initially is 50 - (2 x 20) = 10 N to the left (in direction XG). So body accelerates in that direction.
When a 10 N force is also added to the right (in direction GX). The net force becomes zero, but body continues moving in same direction (XG) at constant speed.

johnlittlephysics said

at 2:26 pm on Jul 27, 2013

302 Li June: May I check if it two similar objects but with different masses will fall at same velocity but different acceleration?

johnlittlephysics said

at 3:08 pm on Jul 27, 2013

Similar would mean objects with same size and shape, except different mass here.
Two general cases: without and with significant air resistance:
Case 1: Assume no air resistance (negligible).
Only force on both objects is weight W = mg, bigger mass, larger weight, same acceleration g due to gravity. Can be measured through experiment.
Conclusion: both have same constant acceleration g, same speed at any instant, and travel same distance in same time.

Case 2: with air resistance (in real life, if height of falling is large)
When falling, net force on each object Fnet = W - R, where R is air resistance which increase with speed.
Initially W is much larger than R, so greater mass means greater W, greater Fnet and higher acceleration.
Conclusion: object with greater mass will fall faster and hit ground earlier.

Also refer to a my discussion of a related question from Denise above.

johnlittlephysics said

at 8:31 am on Jul 28, 2013

312 Yu Kee: For worksheet 2.6, vector resolution of forces qn 2, after drawing out the vector triangles I still don't see a difference between the pushing and pulling of the roller, won't the force still be the same?

johnlittlephysics said

at 8:36 am on Jul 28, 2013

This question involves vector resolution. In both cases,
1. Draw freebody diagram to show all the forces on the roller
2. Vertical force on the ground by roller = normal contact force by the ground on roller (action-reaction pair)
These are N1 and N2.
3. Resolve F to get the vertical component Fy = F sin 45 in both cases.
See details at https://www.dropbox.com/s/1udwswzy8zoa5yx/Photo%2028-7-13%208%2006%2042%20AM.jpg

johnlittlephysics said

at 8:08 pm on Jul 28, 2013

302 Carmen: May I ask how can I approach a vector resolution question as I'm still not very sure of this method.

johnlittlephysics said

at 8:25 pm on Jul 28, 2013

At this level, vector resolution method would be an additional method available (other than usual vector addition method) to solve problems involving vectors (e.g. forces).
Some general guidelines:
1. Drawing a clearly labelled freebody diagram of a body of interest is still important.
2. Most of the problems involve forces acting vertically (e.g. weight) and horizontally (e.g. friction on the floor). Any force F that acts at an angle can then be resolved into vertical Fy and horizontal components Fx by drawing a separate triangle and using sine or cosine functions. Use these components (instead of original F) in subsequent analysis.
3. For each direction, an equation can be formed either based on Newton's 1st or 2nd law of motion.
E.g. In the vertical direction: A body does not move vertically, net force is zero (Newton's 1st law),
Total upward forces = total downward forces
E.g. In the horizontal direction: A body accelerates, net force (horizontal forces) = ma (Newton's 2nd law)
4. Special cases: e.g. Body moving along a slope, consider and resolve all forces along the slope and perpendicular to the slope. Subsequent analysis uses these 2 perpendicular directions.

johnlittlephysics said

at 9:17 pm on Jul 29, 2013

302 Li June: May I check with you how do we approach bonus question in quiz 6, Dynamics 1?

johnlittlephysics said

at 9:21 pm on Jul 29, 2013

Quiz 6B, Q2 bonus.
As the strings change their positions, the direction of the tensions change correspondingly. Draw these on the vector triangle so that any change in angle and length (magnitude of tension) can be seen clearly.
See https://www.dropbox.com/s/y0m6gs4e7uezdul/Photo%2029-7-13%209%2013%2008%20PM.jpg

johnlittlephysics said

at 6:01 pm on Jul 30, 2013

302 Li June: May I check how do we approach question 5 of AS 2.3 as I am unclear of what the answer key is trying to say?

johnlittlephysics said

at 6:26 pm on Jul 30, 2013

Analyze the forces on the man and draw and label a freebody diagram. Besides his weight, the only contact force is upward force N on him by the weighing scale. The reaction force to this force is also N on the scale by him, causing the reading on the weighing scale. See https://www.dropbox.com/s/slvu5akunp4s5px/Photo%2030-7-13%206%2015%2034%20PM.jpg
Note the weight of the lift is not relevant here.

johnlittlephysics said

at 9:36 pm on Sep 30, 2013

302 Li June: Just to check, how do we determine the position of the angle provided in the question in the vector diagram we draw, in the case of vector resolution? Like how do we know where it should be located at when not stated in the question? Do we follow the property of like similar triangles?
One such question is the question 3 in AS 2.5.

johnlittlephysics said

at 9:40 pm on Sep 30, 2013

Draw a sketch diagram of the system if a diagram is not provided.
An angle is usually given with reference to a given line or side or vector.
E.g. angle wrt the vertical or horizontal or clockwise from a given force.
In AS 2.5 Q3, it is given "ramp that is inclined at 11 degrees above horizontal floor", so angle is measured from horizontal line. Similar triangle is not needed here.

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