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Tag Archives: Topic Dynamics
Finding Forces on Multiple Bodies
BMW Backed Impeller Thrusted Electric Wingsuit Can Fly at 185 MPH
Gravity Flying Jet Suit
Jet Suit Flight and Talk by Richard Browning
Setting Record
Real Ironman
Another kind of jet suit by another company
During impact of a free falling ball, the force on ground is greater than the weight of ball
In this post, it shows a free-falling ball from a height of 1.0 m. During the impact, the direction of the force on the ground is downwards and the force on the ground by the ball is greater then the weight.
As the ball is free-falling, the only force acting is its weight downwards. Hence a common misconception is to think that the force on the ground during impact is equal to the weight. This is wrong.
The normal force (force on the ball by the ground = stopping force on the ball by the ground) is greater than the weight.
The force on the ball by the ground is equal and opposite to the force on the ground by the ball. Hence the magnitude of the force on the ground is greater than the weight.
Similar concept can be applied if a man jumps off from a height. But in this case, the man’s leg will exert a stopping force over a short distance. That stopping force, once again, is greater than the weight of the man.
Man Jumps Vertically Upwards, Pressure On Ground Is Greater During The Jump
This concept is similar to a 2016 O-Level Pure Physics Question P2 Q2, on why the pressure acting on the ground is greater during the jump, compared to when he is standing stationary on the ground.
During the jump, his leg will exert an upward force. This upward force (equivalent to normal force or force on the man by the ground) is greater than the weight of the man. Hence there is a net (resultant force) upwards, causing him to accelerate upwards.
That force on the man by the ground is equal and opposite to the force on the ground by the man. This is an action-reaction pair. Since the force exerted on the ground by the man is greater (greater than weight), the pressure exerted on the floor is greater.
(NOTE: Normal force and Weight is not an action-reaction pair)
Heavy and light shuttlecocks – which will reach the floor first when released?
Refer to this post on falling of basketball and bowling ball
Refer to this post on a video of falling feather and bowling ball
Basketball vs bowling ball – which will reach the floor first when dropped?
Refer to this post on terminal velocity of shuttlecocks
Refer to this post on a video of falling feather and bowling ball
PP N2010 P1 Q4 Two objects same size and shape but one is heavier. Motion when both released
Solutions: Option C
When a body is released from rest, the only force acting on the body is its weight due to gravity. Both bodies experience acceleration due to gravity. Hence for both bodies (regardless of mass), will have the same initial acceleration of 10 m/s2.
As the two bodies are of the same size and shape, they will experience the same air resistance for any particular speed. As speed increases, air resistance increases.
For terminal velocity to be reached, air resistance has to be equal to the weight. Since weight is greater for the ball with larger mass, the air resistance has to be bigger. Thus, the ball has to accelerate more (air resistance increases with speed) for the larger air resistance to be equal to the weight. Hence the ball with larger mass will have larger terminal velocity.
[NOTE]
Do not confuse ‘speed of the body is independent of the mass’ as learned in Work Done, Energy and Power. This concept is based on the assumption that there is no air resistance. So not applicable in this question as for terminal velocity to occur, air resistance must be present.
Static Electricity – Oil Drop
Skydiving
Newton’s Law of Motion
Newton’s First Law of Motion
3E1 2011 Project Gp2 – Newton’s Laws of Motion
Another great video done! Well done!
Daniel Lim (leader)
Lim Shi Shuan
Yong Jun Xuan
Jonathan Ho
Low Kai Kang
Description of video:
Our video is an re-enactment of how the the apple dropped on Sir Issac Newton’s head, resulting in the three laws of motion.
We also make use of our demonstration to show the three laws
N2005/P2/A1 – Tension in Spring
The figure below shows a 0.40 kg mass hanging at rest from a spring.
(a) State what is meant by the mass of an object.
Mass is the amount of matter in the body. SI unit is kg.
(b) (i) On the figure above, draw an arrow showing the ling of action and the direction for each of the two forces that act on the mass. Write the name of the force next to each arrow.
(ii) The gravitational field strength is 10 N/kg. Calculate the values of the two forces you have drawn in (i).
W = mg = 0.40 x 10 = 4 N
Weight = Tension = 4 N (Newton’s 1st law, net force = 0N)
(c) The mass is pulled downwards and then released. Explain, in terms of any changes in the forces acting on the mass, why the mass accelerates upwards.
As the spring is pulled downwards, there is a gain in elastic potential energy. The tension in the spring increases compared to previously.
Since Tension > Weight, there is an resultant force acting upwards, hence there is an acceleration upwards.
N2008/P1/Q5 – Closed Loop Triangle for System in Equilibrium
A heavy sack is suspended from a ceiling by a rope.
The sack is then pulled to one side and held stationary.
Which diagram represents the forces acting on the sack in its new position?
Solutions: Ans: D
Initial Acceleration of a body (releases from rest or moves off from rest)
When a ball is released from rest or a car moves off from rest, the initial velocity for both cases are 0 m/s. But what is the initial acceleration? 0 m/s2 ? IT IS NOT!
Take note of this misconception! Whenever there is a resultant or net force, there is an acceleration.
Basic mistakes in your Dynamics Test! =(
Some of you are driving up my blood pressure!!! =( sob sob sob!
Vector Diagrams
Vector diagram is commonly referred to parallelogram rule.
Vector diagrams are generally divded into 3 categories:
1) 2 forces acting on a body, there is a resultant force. (Newton’s 2nd Law)
2) 3 forces acting on a body, the body is at rest / in equilibrium or moving at constant speed. (Newton’s 1st Law, Net force = 0N)
   – The resultant force due to any 2 forces is equal and opposite to the 3rd unknown force.
3) 3 forces acting on a body, the body is at rest / in equilibrium or moving at constant speed (Newton’s 1st Law, Net force = 0N)
   – Only 1 known force, angles given, using closed loop triangle to find the 2 unknown forces.
Let’s look at some examples for each category.
1) 2 forces acting on a body, there is a resultant force. (Newton’s 2nd Law)
2) 3 forces acting on a body, the body is at rest / in equilibrium or moving at constant speed. (Newton’s 1st Law, Net force = 0N)
   – The resultant force due to any 2 forces is equal and opposite to the 3rd unknown force.
   – You can either use parallelogram rule or closed looped triangle.
Example 1: Body at rest – Using Parallelogram Rule
Example 1: Body at rest – Using Closed-Looped Triangle
Example 2: Body at constant speed – Using Parallelogram Rule
Example 2: Body at constant speed – Using Closed-Looped Triangle
3) 3 forces acting on a body, the body is at rest / in equilibrium or moving at constant speed (Newton’s 1st Law, Net force = 0N)
   – Only 1 known force, angles given, using closed loop triangle to find the 2 unknown forces.
Click here to view the construction of vector diagram using Parallelogram Method.
Important Concepts on Newton’s 1st and 2nd Law
Basic important concepts on Newton’s Laws which you have to understand.
a) On a rough floor, when the applied force on a moving object is removed, the resultant force acting on object is backwards (frictional force). This resultant force causes the object to decelerate. (Newton’s 2nd Law)
A good example is that of a wakeboarder. When the tension in cable is equal to the opposing force, the wakeboarder moves at constant speed, balanced forces, net force is 0N. But the moment he releases the cable, the only force acting is the opposing force. The resultant force backwards (opposing force) causes him to slow down (decelerate). Hence resultant force backwards does not mean the wakeboarder is moving backwards.
b) On a frictionless floor, when the applied force on a moving object is removed, the resultant force acting on object is 0N, and the object will move with a constant speed in a straight line. (Newton’s 1st law)
View the video tutorial to understand the 2 concepts.
Basic concepts on Netwon’s 1st and 2nd Law of Motion
Whenever an object is at rest or moving at constant speed in a straight line (constant velocity), Newton’s 1st Law of motion must come to your mind.
Concepts on Newton’s 1st Law are: Net force = resultant force = 0N, a = 0m/s2 , all forces are balanced.
If the object is accelerating or decelerating, that means the forces are unbalanced. Newton’s 2nd Law of Motion must come to your mind.
Concepts on Newton’s 2nd Law are: There is net force or resultant force, unbalanced forces, there is acceleration or deceleration.
This video tutorial gives you an overview basic concepts of Newton’s Laws and understanding of these are vital.
Resultant of 2 forces – Common mistake
A force of 3.0 N and a force of 4.0 N act on an object at the same time. Which of the following forces cannot be the resultant force on the object?
A) 0.5 NÂ Â Â Â B) 3.5 NÂ Â Â Â C) 4.0 NÂ Â Â Â D) 5.0 N
Solutions: A) 0.5 N
To achieve maximum resultant force from 3.0 N and 4.0 N, both forces have to be parallel and in the same direction. Hence the maximum resultant force will be 3 N + 4 N = 7 N
To achieve minimum resultant force from 3.0 N and 4.0 N, both forces have to be parallel and in the opposite direction. Hence the minimum resultant force will be 4 N – 3 N = 1 N
Hence any magnitude of resultant force between 1 N and 7 N is possible as the 2 forces can be positioned at an angle to each other.
Therefore, only option (A) cannot be the resultant force on the object as it is less than the minimum resultant force of 1 N.
Vertical Resultant Force of ball rolling off platform
A ball of mass 5 kg rolls along a smooth horizontal surface until it falls off the edge at time = 3s and touches the lower surface at t = 5s. Sketch the graph which represents how the resultant vertical force F acting on the ball varies with time as the ball moves from A to B.
The question only ask for vertical resultant force, hence there is no need to consider the horizontal forces. Anyway the ball is moving horizontally at constant speed, hence no acceleration, forces are balanced. It will move at constant speed (Newton’s 1st law) When the ball is rolling on the upper/lower platform, the weight is equal to the normal force acting on the ball by the platform, hence net force is 0N.
Velocity-time and Displacement-time graph for Ball dropped and rebounces back
There is a new updated post on this concepts. Refer to the link below
https://evantoh23.wordpress.com/2020/04/13/displacement-time-and-velocity-time-graph-of-ball-thrown-up-and-comes-down/
When a ball is released from a height, it will accelerate on the way down due to the resultant force (weight) acting downwards. Just before the ball touches the ground, the velocity of the ball is the maximum. When it hits the ground, the speed decreases to 0 m/s instantly. When it rebounces back in the opposite direction, the initial velocity is the maximum. Assume ideal situation (no air resistance, no energy lost to sound or heat). The ball will rebounce back to its original height.
In reality, there is work done against air resistance, energy converted to heat and sound when ball hits the ground, hence the ball will never reach its original height.
Note that in both situations, conservation of energy always applied. All energy is conserved, just that energy of the ball is converted to other forms like heat and sound.
Click here to view a comic on forces explanations of ball thrown up
Click here to view the post on velocity-time and displacement-time of ball thrown up
Velocity-time and Displacement-time graph for Ball thrown up and comes down
There is a new updated post on this concept. Refer to the link below
https://evantoh23.wordpress.com/2020/04/13/displacement-time-and-velocity-time-graph-of-ball-thrown-up-and-comes-down/
A ball is thrown vertically upwards from the hand and lands back onto the hand. It is important to note that once the ball leaves the hand, the  resultant force acting on the ball is only its weight! And it is acting downwards throughout the motion.Â
Velocity-Time Graph
Displacement-Time Graph
Key points to note when sketching the v-t graph:
1) Fixing the direction up to be positive (can be down as positive if you want)
2) At t = 0s, the initial velocity of the ball is maximum.
3) As it goes up vertically, due to the weight acting downwards, the ball decreases in speed (decelerates)
4) At the highest point, the ball is momentarily at rest (v =0m/s)
5) On the way down, due to the weight of the ball acting downward, the ball accelerates downwards.
Ball thrown vertically up from building and falls to ground
A rock is thrown vertically upwards with a velocity of 29.4 m/s from the top of a building 78.4 m high. After how long will the rock reach the ground below? (Gravity =Â 10 m/s2)
Many are familiar with calculation when ball is released from rest. This question involves ball being thrown upwards, and then it falls to the ground. To solve this question, just treat the motion as 2 separate parts.
1) Ball going up to max height,
2) Ball going down (similar to ball released from rest from highest point)
N2004P1Q4 – Dynamics + Kinematics
Trolley X and trolley Y are joined by a stretched spring. Trolley X has twice the mass of trolley Y.
When the trolleys are released, the acceleration of X is  to the right. What is the initial acceleration of trolley Y to the left?Solutions:Key concept here is Newton’s 3rd Law. Action equals reaction. Both trolleys are connected by the spring, hence the tension force experienced by each trolley is equal and opposite.
Fx   =   Fy
ma      =   ma
2m x 2Â Â Â =Â Â Â m x a
a       =  4 m/s2
Another important concept to know. In Fig A, a box of m is pulled with a force of 40 N. Given that the frictional force is 10 N. The resultant force is 30N. Box is accelerating.
In Fig B, if another identical box m is added on top of the existing box (no change to the floor), the total mass is 2m, the frictional force now will double = 20 N. Hence resultant force now will be 20N. The acceleration of the box will be lower than previously.
Dynamics and kinematics: Force-time graph, find frictional force
A 10 kg box is initially at rest on a rough horizontal surface and a horizontal force is then applied to it. The force varies as shown in the diagram.
a) If the box moves with an acceleration of 2m/s2 in the first 15 s, what is the frictional force between the surfaces?
Resultant force on box = ma
Applied force – frictional force = ma
40 – frictional force = 10 x 2
frictional force = 20 N
b) How does the velocity change during the next 5 s?
Next 5 s after time 15 s, there is no applied force. Hence the resultant force acting on box isonly frictional force 20N (in direction opposite to motion of box).
F = ma
-20 = 10 x a
a = – 2 m/s2
The velocity of box at time 15s is required.
Hence when t = 0s, box accelerates at 2m/s2 from rest,
a = (v – u) / t
2 = (v – 0) /15
v = 30 m/s
(Hence the final velocity of the box at time 15s is 30 m/s, just before applied force of 40 N is released)
Find velocity of box at end of next 5s (time = 20s):
a = (v – u) /t
– 2 = (v – 30) / 5
v = 20 m/s.
Therefore, velocity decreases at a constant rate of -2 m/s2 from 30 m/s to 20 m/s in next 5s.
c) Plot a velocity-time graph and calculate the distance travelled by the box in the 30 s.
Resultant force and Vector Diagram
A trolley is acted on by three forces, P, Q and R in magnitude and direction as shown in the figure below. A single force F could balance all three forces. What is the magnitude of F?
Solutions:There are 3 different forces, P, Q and R acting on the trolley.Consider the forces P and R only, as they are acting perpendicular to each other, the resultant force falls exactly on Q. (using the vector diagram method/parallelogram method).The resultant of P and R is Q, plus the existing Q, will give a final resultant force of 2Q acting on the trolley.
Evan's Space 