Mechanics M1 Help- Rep for those you help

I would greatly appreciate help with these questions....Solutions are greatly appreciated.

Q.1) A small parcel of mass W=5Kg is held in equilibrium on a rough plane by the action of a horizontal force of magnitude H=30N acting in a vertical plane through a line of greatest slope. The plane is inclined at an angle of 15 degrees to the horizontal. The parcel is modelled as a particle. The particle is on the point of moving up the slope.
a) Find the normal reaction, R, on the parcel
b) Find the coefficient of friction between the parcel and the plane.

Q.2) A ring of mass 0.6kg is threaded on a fixed, rough horizontal curtain pole. A light inextensible string is attached to the ring. The string and the pole lie in the same vertical plane. The ring is pulled downwards by the string which makes an angle (alpha) to the horizontal, there tan (alpha)=2/4

The tension, T, in the string is 2N. Find the coefficient of friction between the ring and the pole.

Q.3) A parcel of mass W=4kg lies on a rough plane inclined at an angle (alpha) degrees to the horizontal, where tan (alpha) is equal to 3/4. The parcel is held in equlibrium by means of horizontal force H= 20Ns. The force acts in a vertical plane through a line of greatest slope of plane. The parcel is on the point of sliding down the plane.

Find the coefficient of friction between the parcel and the plane.

Q.4) A box of mass 1.2kg is placed on a plane which is inclined at an angle of A=29 degrees to the horizontal. The coefficient of friction between the box and the plane is 1/6. The box is kept in equilibrium by a light string which lies in a vertical plane containing a line of greatest slope of plane. The string makes an angle of B= 20 degrees with the plane. The box is in limiting equilibrium and is about to move up the plane. The tension in the string is T newtons. The box is modelled as a particle. Find the value of T.

Thanks for the help in advance.......

Reply 1

bikerx23

do your own homework

Reply 2

Bowie88

i would do it if i was shown how to.........thanks for your concern anyways. So anyhelp??????????

Reply 3

mizfissy815

Are these from the M1 heinmann book? If so, what questions?

Reply 4

Bowie88

mizfissy815

Are these from the M1 heinmann book? If so, what questions?

Im not sure...........wait, its Heinemann Modular Mathematics for Edexcel AS and A-Level "Mechanics 1." But these questions are from a worksheet from somewhere else. Do they look familiar?

Reply 5

*soph*

I've had a look at the first two...

1) Draw the diagram so H pulls up the slope, F is the friction down the slope counteracting it, the weight (5 x g) pulls straight down from the particle and the reaction force is perpendicular to F.

By resolving forces -
R = 5g cos 15
= 47.33N

F = H - 5gsin15
= 17.32

F = co-efficient x R
Co-efficient = 0.366

2) The particle has a weight of 0.6g, the reaction force R is pulling the ring up, friction is perpendicular to R and the tension force is alpha degrees below the horizontal on the other side of the particle. If tan alpha = 2/4 then alpha = 26.565

Again, by resolving forces -
R = 0.6g + 2sin alpha
R = 6.774425593

F = 2cos alpha
F = 1.788855181

F <= co-efficient x R
co-efficient >= 0.264

Hope they're right... I have a tendency to make silly mistakes in mechanics sometimes but I've done my best! Be interested to know what other people come up with as well though.

Reply 6

mizfissy815

Ok...I can't really be bothered to answer the questions for you but I'll help...

Firstly, ALWAYS draw a diagram illustrating where all the forces are acting.

Main pionts...
-Resolve the parcel to its components (ie mgsinθ and mgcos&#952 :wink:

mgsinθ of the parcel always acts down the slope.
-R- The normal reaction force always acts perpendicular to the plane.
-Acting opposite to R is the mgcosθ of the parcel.
-The force holding anything up or pulling any object down has a tension.
-Friction=µR - and always acts opposite to the direction of motion or intended motion in the case of limiting equilibrium.
-In limiting equilibrium questions the resultant force is zero…so opposite forces are equal.
-Make equations and use simultaneous equations method to find 2 unknowns.

Did I miss anything?