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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
2 Answers
- Wayne DeguManLv 77 years ago
Please refer to the sketch below.
(↖) R = 8sinα + 0.5gcosα
=> R = 8(5/13) + 0.5g(12/13) => 100/13 Newtons
(↗) Friction + 0.5gsinα = 8cosα....(1)
=> Friction = 8cosα - 0.5gsinα
i.e. Friction = 8(12/13) - 0.5g(5/13) => 71/13 Newtons
Also, Friction = µR
so, 71/13 = (100/13)µ
=> µ = (71/13)(13/100) = 71/100 = 0.71
When the 8 Newton force is removed we only have friction and the downward force due to gravity.
Downward force => 0.5gsinα = 5(5/13) = 1.92 Newtons
Frictional force => µ(0.5gcosα) = 0.71(5)(12/13) = 3.28 Newtons
The force due to gravity must exceed the frictional force in order for the particle to move.
as downward force < friction the particle will not move
Note: as motion is not occurring we take acceleration as zero and balance the resulting forces
i.e. (1) could be written as 8cosα - Friction - 0.5gsinα = 0.5(0)
=> 8cosα - Friction - 0.5gsinα = 0
so, Friction = 8cosα - 0.5gsinα...as before
:)>
- EliotLv 57 years ago
I'm not sure what your question means. Basically, you are asked to determine whether the total acceleration is zero or not. Since that is what you are asked to find, then you cannot just assume it!
