For 1c, how do you know they're asking for displacement and not distance? And how do you know if the distance is going to be the same as displacement? (that the particle hasn't changed direction?)
Because it says it in the question, find the position vector! The position vector by definition is the displacement from the origin and r = integration of v It’s not asking for distance travelled so don’t need to worry about change of direction Also it’s moving in 2D space and your working with vectors. If you want to calculate distance between two points just use vectors and distance formula. Also if you want to work out distance traveled between different time values then do definite integration on v between these t values i.e the distance travelled is the area under the velocity time graph Just integrate the i and j components of v separately and apply the boundary condition to find the c’s
The particle at point C is attached to the rod. It is different from a particle RESTING on the rod, in which case it would have a reaction force that makes a difference to the system. You can think of this attached particle as being "glued" onto the rod and the force that keeps it glued on overrides any other force that may act on it such as the reaction force (but it doesn't affect the system). It will only have its weight force acting on the system. I hope that helps.
Particle being attached means that it will not move from its position regardless of its weight/friction etc. you could say that it has a reaction force as it’s pushing down on the rod due to its weight however this reaction force is only felt by the particle C not by the rod ie will not effect the rod. That’s why it’s not needed when taking moments from point A. Only the weight of the particle effects the rod in pushing it down. Reaction at C is only needed to be considered if particle C is allowed to be free ie not attached and it follows the particle as it moves up or down the beam.
Friction works against the component of X that’s parallel to it (ie parallel to the inclined plane). 5gsin(alpha) also works against this same component of X. So Friction and 5gsin(alpha) are both directed down the plane in this question and only the parallel component of X acts up the plane
You could initially say that friction works in the same direction as the parallel component of X and against 5gsin(alpha) After doing the maths you will end up with a negative friction which would mean the direction of friction is meant to be the other way ie down the plane
For 1c, how do you know they're asking for displacement and not distance? And how do you know if the distance is going to be the same as displacement? (that the particle hasn't changed direction?)
Because it says it in the question, find the position vector! The position vector by definition is the displacement from the origin and r = integration of v
It’s not asking for distance travelled so don’t need to worry about change of direction
Also it’s moving in 2D space and your working with vectors. If you want to calculate distance between two points just use vectors and distance formula.
Also if you want to work out distance traveled between different time values then do definite integration on v between these t values
i.e the distance travelled is the area under the velocity time graph
Just integrate the i and j components of v separately and apply the boundary condition to find the c’s
When do you multiply a constant to a vector?
why no normal reaction for particle at point C (q4)... im confused by the idea of a particle, do particles have weights only but no normal reactions?
The particle at point C is attached to the rod. It is different from a particle RESTING on the rod, in which case it would have a reaction force that makes a difference to the system.
You can think of this attached particle as being "glued" onto the rod and the force that keeps it glued on overrides any other force that may act on it such as the reaction force (but it doesn't affect the system). It will only have its weight force acting on the system.
I hope that helps.
Particle being attached means that it will not move from its position regardless of its weight/friction etc.
you could say that it has a reaction force as it’s pushing down on the rod due to its weight however this reaction force is only felt by the particle C not by the rod ie will not effect the rod.
That’s why it’s not needed when taking moments from point A. Only the weight of the particle effects the rod in pushing it down.
Reaction at C is only needed to be considered if particle C is allowed to be free ie not attached and it follows the particle as it moves up or down the beam.
GradeAUnderA?
for question 2 the friction works against x but why cant it be 5gsin(alpha) working against x and the friction?
Friction works against the component of X that’s parallel to it (ie parallel to the inclined plane). 5gsin(alpha) also works against this same component of X.
So Friction and 5gsin(alpha) are both directed down the plane in this question and only the parallel component of X acts up the plane
You could initially say that friction works in the same direction as the parallel component of X and against 5gsin(alpha)
After doing the maths you will end up with a negative friction which would mean the direction of friction is meant to be the other way ie down the plane
Hope this made sense
How to get 90% in paper 3? What should I do?
Focus on bigger topics that keep coming up every time. Keep doing different past papers. Keep going.
@@MathsCoach what are the most common topics that youve seen been asked for stats+mech