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# Official OCR A G481 mechanics 20th May 2013 watch

1. (Original post by sohailkm96)
Can anyone define the strength of a material, for example on a stress/strain graph or force/extension graph. What would show that the material is strong?
If the material is strong then the graph would just have a straight line showing the material obeys Hooke's law and possible not showing it undergoing any plastic deformation, i doubt the graph would show its breaking point also if the question was to ask something about the strength...
If you were to work out the strength though (the stiffness) you'd work out the constant k, which would be the gradient in a f/x graph.
2. (Original post by precious maro)
thanks for this, I better go learn how to tackle the 7 marker.
Any time !
3. (Original post by blackstarz)
Yeah but i was doing some older papers where they didn't tell you the pivot where they gave you two forces and the weight and you had to determine where to take moments about to find one of the forces. So was curious as to whether we might have to do the same in these papers
Sometimes they tell you, sometimes they don't.. But if you take moments about a point that has an unknown force A acting on it you automatically eliminate A leaving you hopefully with only one unknown force, B
4. (Original post by RafJav)
If the material is strong then the graph would just have a straight line showing the material obeys Hooke's law and possible not showing it undergoing any plastic deformation, i doubt the graph would show its breaking point also if the question was to ask something about the strength...
If you were to work out the strength though (the stiffness) you'd work out the constant k, which would be the gradient in a f/x graph.

Ahh okay, so a strong material will have a really high ultimate tensile strength?
5. Someone help me with january 2013 question 3c, confuses me.
6. (Original post by blackstarz)
Yeah but i was doing some older papers where they didn't tell you the pivot where they gave you two forces and the weight and you had to determine where to take moments about to find one of the forces. So was curious as to whether we might have to do the same in these papers
In all of the papers I've done, from 2009 onwards, the moment questions do tell you which point you need to take moments about so i wouldn't worry about that...they might decide to throw one in just to put people off but if they do, the thing you're taking moment from would most likely be in equilibrium so it won't be too hard to work out.
7. (Original post by _hail)
Sometimes they tell you, sometimes they don't.. But if you take moments about a point that has an unknown force A acting on it you automatically eliminate A leaving you hopefully with only one unknown force, B
Ahh i see, so if there are 2 unknown forces and they ask you to determine one of the forces and you know the weight you take moments about the other force that has not been asked to find out?
8. Guys there are some really challenging papers 2008 and bellow have a go !!! http://fizx.wikidot.com/g481
9. (Original post by blackstarz)
Ahh i see, so if there are 2 unknown forces and they ask you to determine one of the forces and you know the weight you take moments about the other force that has not been asked to find out?
Exactly! And then you don't even have to think about the other force. And then sometimes for part b) of the question it's like: "Now find the other force." And provided it's in equilibrium, you don't have to bother with moments again- you can just resolve vertically so if the weight W is pointing down and you have your two unknowns A and B pointing up and you've figured out A in part a) you can do W=A+B so B=W-A.
10. (Original post by Joseph-)
Someone help me with january 2013 question 3c, confuses me.
The weight of the car acts directly down, so the horizontal component of the weight acts down the slope, so there is a larger force in the motion's direction (<---) Therefore as there is a greater force down the slope, the resultant force is increased, as mass is constant, the acceleration increases, increased acceleration means it has a higher velocity in the same time interval. Therefore it travels a greater distance.
11. (Original post by Joseph-)
Someone help me with january 2013 question 3c, confuses me.
Because the car is on an inclined slope going downwards, if you picture it physically, there'll be a component of the car's weight acting down the slope against the resistive force, so because of this the resultant braking force would be smaller and so the distance travelled would be greater...hopefully that makes sense.
12. (Original post by sohailkm96)
The weight of the car acts directly down, so the horizontal component of the weight acts down the slope, so there is a larger force in the motion's direction (<---) Therefore as there is a greater force down the slope, the resultant force is increased, as mass is constant, the acceleration increases, increased acceleration means it has a higher velocity in the same time interval. Therefore it travels a greater distance.
ty
13. (Original post by blackstarz)
Yeah but i was doing some older papers where they didn't tell you the pivot where they gave you two forces and the weight and you had to determine where to take moments about to find one of the forces. So was curious as to whether we might have to do the same in these papers
You can take the pivot about any point, it doesnt really matter. I had the same questions in Mechanics maths but my teacher says it doesnt matter as long as you get the clockwise and anticlockwise forces correct.
14. (Original post by sohailkm96)
Ahh okay, so a strong material will have a really high ultimate tensile strength?
Yup because if K is big, then F is big (as f=kx), and so if F is big, the stress will be big as F/A=Stress, and a large value for stress means a high ultimate tensile strength
15. Can someone tell me the main points for trilateration? I get confused about whether signals are sent by the satellite or GPS
16. (Original post by RafJav)
Yup because if K is big, then F is big (as f=kx), and so if F is big, the stress will be big as F/A=Stress, and a large value for stress means a high ultimate tensile strength
Ahh okay thanks a lot!

(Original post by Majjie)
Can someone tell me the main points for trilateration? I get confused about whether signals are sent by the satellite or GPS
Radio/microwaves are sent from the SATELLITE, they are received by the GPS. The time taken for it to reach the GPS is determined, the speed of the waves are known ( speed of light, c, 3x10^8 ms^-1) so we can use time = distance/speed. To determine the distance that the car is away from the satellite. This is done by three or more satellites, as we know it is a particular distance from the satellite, but not exactly where it is from each satellite, we draw a circle, of radius of that distance. Drawing circles from each satellite, shows the possible places where the GPS/car is, where these circles intersect shows where the car MUST be. (This is the principle of trilateration.)
17. (Original post by sohailkm96)
Ahh okay thanks a lot!

Radio/microwaves are sent from the SATELLITE, they are received by the GPS. The time taken for it to reach the GPS is determined, the speed of the waves are known ( speed of light, c, 3x10^8 ms^-1) so we can use time = distance/speed. To determine the distance that the car is away from the satellite. This is done by three or more satellites, as we know it is a particular distance from the satellite, but not exactly where it is from each satellite, we draw a circle, of radius of that distance. Drawing circles from each satellite, shows the possible places where the GPS/car is, where these circles intersect shows where the car MUST be. (This is the principle of trilateration.)
Thanks!!
18. Can anyone go through how airbags/ seatbelts reduce impact on passengers?
This is what Ive got so far, Is there anything eles I need to say?
Seatbelts- prevents driver from moving through the windsheild due to its mass
Stretches a little which increases the distance the force can act it

Airbags: Works with seatbelts as it needs this time to inflate.
Sudden deceleration triggers reaction to inflate the bag.
Increases distance and time that force acts in.
reduces the deceleration on the driver.

THUS REDUCING THE AVERAGE FORCE IMPACTED ON PASSENGERS,
19. (Original post by Majjie)
Can anyone go through how airbags/ seatbelts reduce impact on passengers?
This is what Ive got so far, Is there anything eles I need to say?
Seatbelts- prevents driver from moving through the windsheild due to its mass
Stretches a little which increases the distance the force can act it

Airbags: Works with seatbelts as it needs this time to inflate.
Sudden deceleration triggers reaction to inflate the bag.
Increases distance and time that force acts in.
reduces the deceleration on the driver.

THUS REDUCING THE AVERAGE FORCE IMPACTED ON PASSENGERS,

For airbags I got down about accelerometer and how they work.
Also for Seatbelts (may not be right) but I put also about increases time which decreases deacceleration thus decreasing force due to F=ma

Anyone else got anymore to add on?
20. (Original post by sohailkm96)
The weight of the car acts directly down, so the horizontal component of the weight acts down the slope, so there is a larger force in the motion's direction (<---) Therefore as there is a greater force down the slope, the resultant force is increased, as mass is constant, the acceleration increases, increased acceleration means it has a higher velocity in the same time interval. Therefore it travels a greater distance.
can you say the resultant force is backwards but it is less since this horizontal component of the weight cancels some of it out, this gives a lower deceleration, and a lower deceleration means the car at the inclined plane moves faster and thus get to travel further.

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