Centre of gravity (CG) of an object is an imaginary point where the entire weight if the object seems to act.
Category Archives: 05 Turning Effect of Forces
Balancing Toys
There are two main types of balancing toys.
Type 01: CG below the pivot point
Type 02: CG above the pivot
Trash bag – energy transfer and inertia
Turning Effect of Forces – Moment – Basic Concepts
Once you know the basic concepts, try applying the POM in other application questions. Refer to the two videos below.
Will it topple?
To indicate the weight, you have to draw the arrow from the CG vertically down.
Basic Concepts on Turning Effect of Forces (Moment)
Understanding Hydraulic System Overview
Understanding Hydraulic System 01
Understanding Hydraulic System + Lever 02
The Uncle (aka Physics Guru) in coffeeshop
How did he do that?!
Another type of gadget to hang your bags by the table is this. As long as the centre of gravity (CG) is directly below the support area (area of the hanger in contact with the table), the bag will not fall.
The Standing Broom Challenge
Real-life ‘roly-poly toy’ – China
Centre of Gravity
Double Pivots Question
A man is standing on a uniform plank of mass 20 kg and 6 m long. He is standing 1 m from the string at A.
a) Find the unknown forces FA and FB.
The man now walks to the left of A.
b) Determine the maximum distance in which the man can move before the plank topples.
Solutions:
a) Fa = 550 N and FB = 250 N
b) d = 0.67 m
Refer to the video tutorial for the explanation.
Click on this post for the comics on moments concepts
What is moment?
Minimum force F to push barrel up a step
Solutions: Option D
You can view the video tutorial here or the written solutions below.
(You can click here to view another post on similar question where angle is given instead)
Free-body Diagram
N2010 P2Q12 Fuel gauge using variable resistor
Solutions:
View the video on how the gauge works.
As the fuel level drops, the float which stays on the fuel surface will descend. The rod which is attached to the float will turn clockwise about the pivot X. As the rod turns, the resistance on the variable resistor increases. This increases the resistance of the circuit. Hence the current flowing through the circuit will decreases, causing the needle to deflect more to the left, indicating towards E (empty). Thus the reading on the fuel gauge decreases.
Solutions:
(i) Both the fixed resistor and sensor are in series.
Total effective resistance Re = 5000 + 1000 = 6000 ohms
V = IR
12 = I x 6000
I = 0.0020 A
Hence potential across Y, V = IR (where I is constant in a series circuit)
= 0.0020 x 1000
= 2.0 V
(ii) When the temperature increases and the resistance of sensor Y decreases, the total effective resistance of the circuit decreases. Current flowing through the circuit increases.
Since V = IR, where the R of the fixed resistor is a constant 5000 ohms, as current I increases, the potential difference across the 5000 ohms resistor will increase.
Moments – Where is the new CG?
Important Concepts on Moments
When a system is in equilibrium, 2 conditions are met.
1) Principle of Moment (POM):
Sum of anticlockwise moment = Sum of clockwise moment
2) Net force = 0N; all forces are balanced:
Total downward forces = Total upward forces
Whenever a force passed through the pivot point, that force will not create any moment as there is no perpendicular distance from the force to the pivot. Hence this force will not be included in the calculation of moments.
2013 Nov Sci Phy P2 Q9 – Pressure and Moment
Solutions:
(a) (i) A bigger force than F can be obtained due to the lever system and hydraulic system.
Lever system: Applying principle of moments, the anticlockwise moment by the F is equal to the clockwise moment by the force on piston A (note that the handle is pushing the piston down, but the piston A is pushing on the handle upwards – action = reaction). As the perpendicular distance from F to the pivot is greater than the perpendicular distance of the force by piston to the pivot, the force on the piston A is greater than F at handle.
Hydraulic system: As the pressure transmitted in the liquid is the same, pressure at piston A = pressure at piston B. As P = F/A and area of piston A is smaller than area of piston B, a larger force is obtained in piston B.
Hence these two systems allow the force on piston B to be greater than F at the handle.
(a) (ii) Both liquid and gas molecules are in a continuously random motion. But in liquid, the molecules are closely packed together and able to slide around one another. There is very little empty space between the molecules hence liquid is not compressible. Gas molecules are far apart from one another, hence gas can be easily compressed.
(b) P = F/A = 12000 / 0.060 = 200 000 Pa
(c) Velocity is the vector quantity while speed is a scalar. As the car goes round the bend, the direction of the car changes. Hence velocity is changing even though speed is constant.
Likewise, as the velocity is changing, the car is considered to have an acceleration (not in the sense of increasing speed though).
Power of Centre of Gravity (CG)
Centre of gravity (CG) is the point where the whole weight of the object seems to act.
Weight acts from the CG vertically down.
If the object is pivoted at the cg, the weight of the object, which acts vertically downwards from the cg, does not create any moment about the pivot point. As M = F x d, there is no perpendicular distance from the weight to the pivot.
Knowing the position of Centre of Gravity (CG)
Refer to this post for more information on balancing toys. https://evantoh23.wordpress.com/2013/10/04/balancing-toy-where-cg-is-above-pivot/
Balancing Toy : CG is above pivot point
The video below shows a type of balancing toy, which has the CG above the pivot point.
Due to the round bottom, when the toy is tilted, the position of the pivot changes. When tilted to the left as shown on the left diagram, the weight which acts from the centre of gravity (CG) is to the right of the pivot. Hence the weight about the pivot creates a clockwise moment that causes the toy to turn clockwise.
Due to momentum it will go over the vertical position and tilts right as shown in the right diagram. Likewise, as weight is to the left of the pivot, and the weight creates a anti clockwise moment.
Hence this toy will always have a restoring moment which will return the toy to its vertical position where the weight is directly above the pivot. In this position, the weight which passes though the pivot does not create any moment (no perpendicular distance). Hence will be at rest.
Refer to this following video. Another type of balancing toy in which the CG is below the pivot point. Likewise, there is a restoring moment which brings the toy back to its original position. When in rest, the CG is directly below the pivot point.
Balancing Toys
Balancing toys always return to its original position as their centre of gravity (CG) are below the pivot (balancing) point. This kind of system will create a restoring moment which helps to return the system to its original position when displaced slightly.
Such systems are in stable equilibrium. Characteristic of such system is that when displaced, the CG rises.
3E1 2011 Project Gp4: Moments
Another great effort from Ryan’s team!
Done By: Ryan, Davis, Yuan Jie, Yu Xiang and Zhi Hao
3E1 2011 Project Gp 1 – Moments
Hi 3E1,
This is the first video project submitted by Bernice Goh, Tang Yun Hui, Rachel Lee, Kho Yun Suen, Jezebel Tay! Great job!
I like the video! It’s professionally done up =)
Comment of the video and answer their question!
Physics of Construction Lifting Truck
1) Why are the 4 extended legs necessary?
This is to increase the area of base of the truck to increase the stability. From our theory, as long as the weight acting vertically downward from the centre of gravity is within the area of base, there will be a restoring moment to bring the truck back to its original position, hence increases its stability.
2) In what situation will the truck topple? (note the the centre of gravity (CG) of the truck is in general very low due to mass concentration at the base)
Basically, there are 2 situdation:
– if the load that is lifted is too heavy, the overall new CG of the truck and load might shift outside the area of base, hence causing the truck to topple. i.e. the clockwise moment created by the load is created than the anticlockwise moment created the weight of truck.
– If the angle of tilt is too much which increases the perpendicular distance from load to pivot. Likewise, the overall CG might be outside the area of base and there is a net clockwise moment.
3) What is the purpose of the metal plate underneath the legs?
The purpose is to increase the area of base. The whole weight of the truck is spread over the 4 legs. Since P = F/A, with a bigger base area, the pressure acting on the ground will be reduced. This is to minimise any damages done to the ground.
Restoring moment
This Lego structure which Daniel and Danry made is ‘Stable equilibrium’
Characteristics of stable equilibrium:
– when displaced, centre of gravity (cg) rises.
– the weight of the basket will create a restoring moment which returns the basket to its original position.
– when at rest, the weight of CG is directly below the pivot.
Below is a brief explanation
Sent from my iPad
Minimum force to lift a drum over a step
Given that the mass of the drum is 20 kg. What is the minimum force, F, required to just movee th drum off the ground? (Assume uniform mass of drum and g = 10 N/kg)
Knowing the position of the pivot and identifying the perpendicular distance is important.
Consider the 3 variations of the same question as shown below.
Type 1:Â Very straight forward as the pivot is same level as the CG of the drum. Hence p
endicular distance is easy to identify.
Type 2:Â Using pythagoras theorem to find the necessary perpendicular distance.
Type 3:Â Using Toa, Cah and Soh to find the necessary perpendicular distance.
Click here to view another similar question
Moments Question using Simultaneous Equations
A non-uniform plank weighing 300 N is set up as shown. The spring balance reads 160 N when a small boy stands at A and 760 N when he stands at B. (Note: Non-uniform plank means the CG of the plank is not at its centre)
a) Find the weight of the boy.
b) How far is the centre of gravity of the plank from point A?
Solutions:
a) 300 N and b) 6.67 m
Click on the video tutorial if you can’t derive the 2 equations.
Something about Moments
Turning Effect of Force
The force diagrams shows all the forces acting on a beam of length 3x. Which force system causes only rotational motion of the beam without any linear movement?
Solutions: Option B
One has to note that the beam does not have a fixed pivot which it can turn about.
Hence beam is free to move when the 4 forces are acting on it. To create only rotational motion (it will rotate about a fixed position (centre of beam)) only, there will not be any linear movement (the centre of beam moves away from its original centre regardless if it is rotating).
To create only rotational motion, the only possible way is when the beam rotates about its centre point. Hence resultant moment created by the forces is either clockwise or anticlockwise moment.
Taking moment about centre, you have to do calculation on Clockwise Moment and Anticlockwise Moment.
For B:Â Taking moment about centre and consider length x as 1 m, left hand side of pivot, moment = 4 x 1.5 + 2 x 0.5 = 7Nm anticlockwise moment right hand side of pivot, Anticlockwise moment = 3 x 1.5 + 5 x 0.5 = 7Nm anticlockwise moment
Hence only system B is able to have rotational motion anticlockwise moment.
For e.g C: LHS of pivot, moment = 2 x 1.5 – 4 x 0.5 = 1Nm anticlockwise moment RHS of pivot, moment = 3 x 1.5 + 5 x 0.5 = 2Nm anticlockwise moment
Though system will rotate anticlockwise, but the anticlockwise moment on both sides are not equal , thus will cause linear movement.
Upward forces = downward forces does not mean body will be at rest. Rather only if the body is at rest / moving at constant velocity, then the forces are balanced and resultant is zero. (Newton’s 1st law)
Evan's Space 