M1 friction/resolving forces help!

superstarrr

I'm really struggling with this question. Any help would be appreciated.

A parcel of mass 5kg, rests on a rough plane inclined at 45deg to the horizontal. The parcel is held in equilibrium by a force of magnitude 20N inclined at 30deg to a line of greatest slope of the plane, as shown in the diagram. Draw a diagram showing all the forces acting on the parcel, and show that the normal contact force between the parcel and the plane has magnitude 24.6N, correct to 3sf.
Given that the equilibrium is limiting, with the parcel on the point of moving down the plane, find the coefficent of friction between the parcel and the plane.

diagram

I don't think your diagram makes sense. The parcel would slide down because of gravity, it is held in position by a force going up the plane.

Picture 30.png18.8KB

Reply 2

OP

copied the diagram out the text book!

Reply 3

superstarrr

copied the diagram out the text book!

OK but there should be an arrow(s)

Reply 4

superstarrr

copied the diagram out the text book!

Here is the first part.

superstarrr

Here is second part. Mu appears rather large. Do you have the solution in your textbook?

Picture 32.png43.4KB

Reply 6

OP

Yes that's the answer as it's a 'particularly rough slope'. Thanks so much for your time!

Reply 7

OP

superstarrr

Yes that's the answer as it's a 'particularly rough slope'. Thanks so much for your time!

Sorry just checked and the actual answer and it's 0.703. So I'm still confused !

Reply 8

Square

13

Working on the answer for you now, will edit this post with the solution.

Okay I find the easiest way to go about this is to consider ALL the forces acting on the point:

So if we take i and j to be our vector directions parallel and perpendicular to the slope all the forces are: weight, reaction, friction and the horizontal force: T
Splitting these into components we get:

Fi + Rj - mgcos45j - mgsin45i + Tsin30j + Tcos30i = 0 (no resultant force as parcel is stationary on slope)

Now if we resolve parallel we just consider the i terms, this gives:

F - mgsin45 + Tcos30 = 0 [call this equation 1]

And similarly for j

R - mgcos 45 + Tsin30 = 0 [call this equation 2]

We are asked to work out the coefficent of friction which I will call M (its actually mu but i dont know how to do that on a keyboard!).

F = MR putting this into equation one gives:

MR - mgsin45 + Tcos30 = 0

We are asked to work out M so thats okay being in the equation and now the only other unkown we have is R, so we need to find a way to get a value for R which we get from equation 2:

R = mgcos45 - Tsin30

So if we sub this back into equation 1 we get:

M(mgcos45 - Tsin30) - mgsin45 + Tcos30 = 0

Therefore M = (mgsin45 - Tcos30)/(mgcos45 - Tsin30)

= 0.703 the right answer.

Hope this helps if you dont understand anything then let me know.

My way is similar to the way that steve2005 did it but I tend to thing of things in terms of vectors rather than resolving its easier in my opinion! Also I have left the all the calculations until the last stage which rules out any rounding or evaluation errors.

Reply 9

OP

Thanks a lot :smile:

Reply 10