The Student Room Group

Mechanics Problem

Please i need a little help on this

A small parcel of mass 2Kg moves on a rough plane inclined at an angle of 30 degrees to the horizontal. the parcel is pulled up a line of greatest slope of the plane by means of a light rope which is attached to it. The rope makes an angle of 30 degrees the plane.The voefficient of friction between the parcel and the plane is 0.4

Given that the tension in the rope is 24N
a) find to 2 significant figures,the acceleration of the parcel.
b) Show that ,when the parcel comes to this position of rest,it immmediately starts to move down the plane again.

yes i know its mouth full but your help wil be much appreciated
first draw a diagram n show all th forces actin on th parcel ,resolved forces also

use F = coef.of friction *R ,where resolvin perpendicular to plane R is 2(9.8)cos30

so, F=2(9.8)cos 30 * 0.4

T along direction of motion is 24cos30

usin F=ma up th plane
T - F - component of weight down th plane = ma
24cos30 - 2(9.8)cos30(0.4) - 2(9.8)sin30 = 2a

solve to find a, a=2.10ms^-2

i hav no idea wats goin on in part (b) though..what position of rest does it myn in th question??

neway hope this helps :smile:
Reply 2
A small parcel of mass 2Kg moves on a rough plane inclined at an angle of 30 degrees to the horizontal. the parcel is pulled up a line of greatest slope of the plane by means of a light rope which is attached to it. The rope makes an angle of 30 degrees the plane.The voefficient of friction between the parcel and the plane is 0.4.

first analyse what we have:
this force is moving it up: 24 * cos 30
the forces moving it down are:
1. friction which is calculated like this
(mu)R = f(max)
r + 24 * sin 30 - 2 * 9.8 * cos 30 = 0
r = 19.6 * cos 30 - 24 * sin 30
r = 4.9740
f(max) = 0.4(4.9740) = 1.9896
2. the parcel's weight
weight = 2 * 9.8 * sin 30 = 19.6 * sin30
so this is the equation

24 * cos 30 - 1.9896 - 19.6 * sin30 = 2a
a = 4.497 = 4.5 (2 s.f)

the second part iam not sure what he means by "this place of rest) , care to explain.
Reply 3
LostLilAngel
first draw a diagram n show all th forces actin on th parcel ,resolved forces also

use F = coef.of friction *R ,where resolvin perpendicular to plane R is 2(9.8)cos30

so, F=2(9.8)cos 30 * 0.4

T along direction of motion is 24cos30

usin F=ma up th plane
T - F - component of weight down th plane = ma
24cos30 - 2(9.8)cos30(0.4) - 2(9.8)sin30 = 2a

solve to find a, a=2.10ms^-2

i hav no idea wats goin on in part (b) though..what position of rest does it myn in th question??

neway hope this helps :smile:

the rope makes an angle with the plane , so u need to resolve it's forces.
hey thanx 4 pointin tht out codefusion i kinda 4got abt resolivin it perpendicular t plain :p:
Reply 5
u r welcome
Reply 6
could you pls tell me in which year did this question come?
Original post by ZestDamsel
Please i need a little help on this

A small parcel of mass 2Kg moves on a rough plane inclined at an angle of 30 degrees to the horizontal. the parcel is pulled up a line of greatest slope of the plane by means of a light rope which is attached to it. The rope makes an angle of 30 degrees the plane.The voefficient of friction between the parcel and the plane is 0.4

Given that the tension in the rope is 24N
a) find to 2 significant figures,the acceleration of the parcel.
b) Show that ,when the parcel comes to this position of rest,it immmediately starts to move down the plane again.

yes i know its mouth full but your help wil be much appreciated


I imagine you meant THE position of rest, i.e. when the velocity first becomes zero. In which case (b) is simply a matter of showing that the friction force that will then be acting up the plane is insufficient to prevent motion.
Original post by LostLilAngel
first draw a diagram n show all th forces actin on th parcel ,resolved forces also

use F = coef.of friction *R ,where resolvin perpendicular to plane R is 2(9.8)cos30

so, F=2(9.8)cos 30 * 0.4

T along direction of motion is 24cos30

usin F=ma up th plane
T - F - component of weight down th plane = ma
24cos30 - 2(9.8)cos30(0.4) - 2(9.8)sin30 = 2a

solve to find a, a=2.10ms^-2

i hav no idea wats goin on in part (b) though..what position of rest does it myn in th question??

neway hope this helps :smile:


Why don't you preview your post before submitting it and then correct all then typos
Reply 9
Original post by brianeverit
Why don't you preview your post before submitting it and then correct all then typos


Have you looked at the dates on these posts? :biggrin:
Original post by davros
Have you looked at the dates on these posts? :biggrin:


What have the dates got to do with it?
Reply 11
Original post by brianeverit
What have the dates got to do with it?


You appear to be "helping" people who last posted in 2006 :smile:
Original post by davros
You appear to be "helping" people who last posted in 2006 :smile:


(I apologise. I did not notice the date. But my comment still holds for some posts.