Newtons laws of motion ...PLEASE HELP?
A particle of mass 0.5 kg lies on a rough plane inclined at an angle (alpha) to the horizontal, where sin(alpha) = (5/13).
The particle is acted on by a horizontal force of 8 N and is about to move up a line of greatest slope.
(a) Show that the value of the coefficient of friction between the particle and the plane is 0.71.
ANS: This part is alright. I got [(71/13)/(100/13)] = 0.71.
(b) Determine, with working, whether or not the particle will move when the force of 8 N is removed?????
ANS: I used F=ma. But then, can I allow the acceleration to be ZERO. Only then, I obtained Max Frictional Force = 5(sin alpha) = 5(5/13) = 25/13.
THEN, I see that Max Frictional Force < 0.71(Normal Contact Force = 60/13)
So, the particle does not move!
CAN I ALLOW THE ACCELERATION TO BE ZERO?????????????????????????
THAN KS
The particle is acted on by a horizontal force of 8 N and is about to move up a line of greatest slope.
(a) Show that the value of the coefficient of friction between the particle and the plane is 0.71.
ANS: This part is alright. I got [(71/13)/(100/13)] = 0.71.
(b) Determine, with working, whether or not the particle will move when the force of 8 N is removed?????
ANS: I used F=ma. But then, can I allow the acceleration to be ZERO. Only then, I obtained Max Frictional Force = 5(sin alpha) = 5(5/13) = 25/13.
THEN, I see that Max Frictional Force < 0.71(Normal Contact Force = 60/13)
So, the particle does not move!
CAN I ALLOW THE ACCELERATION TO BE ZERO?????????????????????????
THAN KS
I'll just focus on part (b).
The force down the incline is given by mgsin(alpha) = mg*5/13 = 0.385mg
The force of friction will oppose this force, and is given by uN = 0.73*mgcos(alpha)
It is simple to show that if sin(alpha) = 5/13, then cos(alpha) = 12/13
So therefore Net Force of friction opposing the downward force is 0.73*mg*12/13 = 0.674mg.
Therefore the friction force (0.674mg) is greater than the gravitational force (mg*5/13), and therefore the block will not move.
It is incorrect to simply set a = 0 initially in the formula F = ma, because it is unknown whether a is 0 or not. The above calculations have to be done first, and now we know that a = 0.
The force down the incline is given by mgsin(alpha) = mg*5/13 = 0.385mg
The force of friction will oppose this force, and is given by uN = 0.73*mgcos(alpha)
It is simple to show that if sin(alpha) = 5/13, then cos(alpha) = 12/13
So therefore Net Force of friction opposing the downward force is 0.73*mg*12/13 = 0.674mg.
Therefore the friction force (0.674mg) is greater than the gravitational force (mg*5/13), and therefore the block will not move.
It is incorrect to simply set a = 0 initially in the formula F = ma, because it is unknown whether a is 0 or not. The above calculations have to be done first, and now we know that a = 0.
My teacher said that it is perfectly acceptable to start by assuming something and if it proves to be inconsistent then the assumption is proved to be wrong. I do this frequently. But you should clearly say what you're trying to show.
And if a mass is on a inclined plane with no other external forces save reaction, weight and friction then to judge whether it would slide or not you could use evaluate tan(alpha) and the coefficient of friction and compare their values. If tan(alpha)>coefficient then it will slide. Otherwise it wouldn't. It's a great way to check.
And if a mass is on a inclined plane with no other external forces save reaction, weight and friction then to judge whether it would slide or not you could use evaluate tan(alpha) and the coefficient of friction and compare their values. If tan(alpha)>coefficient then it will slide. Otherwise it wouldn't. It's a great way to check.
Original post by boykaA
A particle of mass 0.5 kg lies on a rough plane inclined at an angle (alpha) to the horizontal, where sin(alpha) = (5/13).
The particle is acted on by a horizontal force of 8 N and is about to move up a line of greatest slope.
(a) Show that the value of the coefficient of friction between the particle and the plane is 0.71.
ANS: This part is alright. I got [(71/13)/(100/13)] = 0.71.
(b) Determine, with working, whether or not the particle will move when the force of 8 N is removed?????
ANS: I used F=ma. But then, can I allow the acceleration to be ZERO. Only then, I obtained Max Frictional Force = 5(sin alpha) = 5(5/13) = 25/13.
THEN, I see that Max Frictional Force < 0.71(Normal Contact Force = 60/13)
So, the particle does not move!
CAN I ALLOW THE ACCELERATION TO BE ZERO?????????????????????????
THAN KS
The particle is acted on by a horizontal force of 8 N and is about to move up a line of greatest slope.
(a) Show that the value of the coefficient of friction between the particle and the plane is 0.71.
ANS: This part is alright. I got [(71/13)/(100/13)] = 0.71.
(b) Determine, with working, whether or not the particle will move when the force of 8 N is removed?????
ANS: I used F=ma. But then, can I allow the acceleration to be ZERO. Only then, I obtained Max Frictional Force = 5(sin alpha) = 5(5/13) = 25/13.
THEN, I see that Max Frictional Force < 0.71(Normal Contact Force = 60/13)
So, the particle does not move!
CAN I ALLOW THE ACCELERATION TO BE ZERO?????????????????????????
THAN KS
The logic is there but you can't 'assume' the acceleration is 0. If you're lucky you might get a mark for it but most the time you can't assume something like that. Because the question is basically asking you if it is still in equilibrium, you can't 'assume' that it's acceleration is 0 because you are assuming what the question wants you to show.
Original post by k4l397
The logic is there but you can't 'assume' the acceleration is 0. If you're lucky you might get a mark for it but most the time you can't assume something like that. Because the question is basically asking you if it is still in equilibrium, you can't 'assume' that it's acceleration is 0 because you are assuming what the question wants you to show.
nope, you can.
You assume something. (use of IF)
It proves inconsistent/consistent
Thus the assumption is wrong/valid
(edited 10 years ago)
Original post by RoyalBlue7
nope, you can.
You assume something. (use of IF)
It proves inconsistent/consistent
Thus the assumption is wrong/valid
You assume something. (use of IF)
It proves inconsistent/consistent
Thus the assumption is wrong/valid
This is only partly correct. It is ok to assume something, and show that such an assumption leads to a contradiction. That would be a valid proof by contradiction that the initial assumption was wrong.
However if you assume something, but do not end up showing a contradiction then, this by no means indicates the initial assumption was true.
In this case, the acceleration is indeed 0, and therefore assuming it is 0 from the start is not a mathematically correct approach.
Original post by Doctor_Einstein
This is only partly correct. It is ok to assume something, and show that such an assumption leads to a contradiction. That would be a valid proof by contradiction that the initial assumption was wrong.
However if you assume something, but do not end up showing a contradiction then, this by no means indicates the initial assumption was true.
In this case, the acceleration is indeed 0, and therefore assuming it is 0 from the start is not a mathematically correct approach.
However if you assume something, but do not end up showing a contradiction then, this by no means indicates the initial assumption was true.
In this case, the acceleration is indeed 0, and therefore assuming it is 0 from the start is not a mathematically correct approach.
Yes I get it - aiming for a contradiction is only helpful.
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