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Reply 180
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.
Original post by precious maro
thanks for this, I better go learn how to tackle the 7 marker.

Any time !
Reply 182
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
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?
Reply 184
Someone help me with january 2013 question 3c, confuses me.
Reply 185
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.
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?
Guys there are some really challenging papers 2008 and bellow have a go !!! http://fizx.wikidot.com/g481
Reply 188
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.
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.
Reply 190
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.
Reply 191
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
Reply 192
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.
Reply 193
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 :smile:
Reply 194
Can someone tell me the main points for trilateration? I get confused about whether signals are sent by the satellite or GPS :s-smilie:
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 :smile:


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 :s-smilie:


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.)
Reply 196
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!! :smile:
Reply 197
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,
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?
Reply 199
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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