A-LEVEL MATHS HELP! Asap
Hello, Okay so I was doing this question and I did get the answer but I don't think I got there right if you know what I mean. But I was just wondering if at A on the ground if it has a horizontal and vertical component because say if the way I did it was right I did not take into account the horizontal and vertical components of A.
Also this is for part A.
Also this is for part A.
(edited 1 year ago)
Original post by skyeforster15
Hello, Okay so I was doing this question and I did get the answer but I don't think I got there right if you know what I mean. But I was just wondering if at A on the ground if it has a horizontal and vertical component because say if the way I did it was right I did not take into account the horizontal and vertical components of A.
Also this is for part A.
Also this is for part A.
I would have resolved horizontally as that means the weight and RA have no effect. So the resolved friction must balance the resolved RB and its a couple of lines.
Original post by mqb2766
I would have resolved horizontally as that means the weight and RA have no effect. So the resolved friction must balance the resolved RB and its a couple of lines.
That doesn't really make sense because there are no horizontal forces, there are only forces which act either paralell or perpendicualr to the plane or going vertical eg, weight.
Original post by skyeforster15
That doesn't really make sense because there are no horizontal forces, there are only forces which act either paralell or perpendicualr to the plane or going vertical eg, weight.
You can resolve forces in any two mutually perpendicular directions e.g. horizontal/vertical or along/perpendicular to. If an object is in motion it usually makes sense to take one of the directions as the direction of motion, but for an equilibrium situation you usually just choose whichever pair of directions is easiest, or reduces the number of forces under consideration.
Original post by davros
You can resolve forces in any two mutually perpendicular directions e.g. horizontal/vertical or along/perpendicular to. If an object is in motion it usually makes sense to take one of the directions as the direction of motion, but for an equilibrium situation you usually just choose whichever pair of directions is easiest, or reduces the number of forces under consideration.
Yes thank you, but I was just wondering if friction, the Ra and the Rb have vertical and horizontal componments.
Original post by mqb2766
I would have resolved horizontally as that means the weight and RA have no effect. So the resolved friction must balance the resolved RB and its a couple of lines.
Are these forces correct?
Original post by skyeforster15
Are these forces correct?
You realize if you resolve horizontally you get
mu Rp cos(theta) = Rp sin(theta)
so
tan(theta) = mu
But those equations look about right, so eliminate Ra and rearrange, ....
If you choose to resolve in the horizontal direction you eliminate the weight and Ra in the first place.
(edited 1 year ago)
Original post by mqb2766
You realize if you resolve horizontally you get
mu Rp cos(theta) = Rp sin(theta)
so
tan(theta) = mu
But those equations look about right, so eliminate Ra and rearrange, ....
If you choose to resolve in the horizontal direction you eliminate the weight and Ra in the first place.
mu Rp cos(theta) = Rp sin(theta)
so
tan(theta) = mu
But those equations look about right, so eliminate Ra and rearrange, ....
If you choose to resolve in the horizontal direction you eliminate the weight and Ra in the first place.
Yes, I see that now.
I think I have done that many moments questions that I have confused myself but now it makes sense again!
(edited 1 year ago)
Original post by skyeforster15
Yes, I see that now.
You should have come across the classic
mu = tan(theta)
relationship for limiting friction for a box on a plane inclined at an angle theta (its one way to measure mu). This is just a variation on that classic problem. The hint here is that there is no friction at A so there is no horizontal force applied at A and the weight is vertical so there is a direct relationship between the horizontally resolved friction and the reaction at P.
Original post by mqb2766
You should have come across the classic
mu = tan(theta)
relationship for limiting friction for a box on a plane inclined at an angle theta (its one way to measure mu). This is just a variation on that classic problem. The hint here is that there is no friction at A so there is no horizontal force applied at A and the weight is vertical so there is a direct relationship between the horizontally resolved friction and the reaction at P.
mu = tan(theta)
relationship for limiting friction for a box on a plane inclined at an angle theta (its one way to measure mu). This is just a variation on that classic problem. The hint here is that there is no friction at A so there is no horizontal force applied at A and the weight is vertical so there is a direct relationship between the horizontally resolved friction and the reaction at P.
No, I have actually never see this or came across this before.
Original post by skyeforster15
No, I have actually never see this or came across this before.
You should have. So something like
https://revisionmaths.com/advanced-level-maths-revision/mechanics/coefficient-friction
So if you put a block on a plane and slowly rotated the plane until the block just begins to slip, tan of the angle is the coefficient of friction. Its why you rarely see mu>1, as this corresponds to material which would only slip when the incline angle is > 45 degress. Rubber is an example where mu>1.
Its usually derived in the way you wanted to do (along the plane), but its easier if you just resolve horizontally and "ignore" the weight. The result is the same.
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