m1 Chap5 Eg.8 p98 question Watch

This discussion is closed.
Malik
Badges: 12
Rep:
?
#1
Report Thread starter 13 years ago
#1
Example 8 Chapter 5 Page 98 M1 Book

A parcel of mass 5kg is released from rest on a rough ramp of inclination Ø=arcsin 3/5 (that is sin Ø=3/5) and slides down the ramp. After 3 seconds the parcel has a speed of 4.9ms. Treating the parcel as a particle, find the coefficent of friction between the parcel and the ramp.

Please look at the attachment now.

Resolving Perpendicular to ramp:
R- 5g Cos Ø=0
R=5g x 4/5
R=4g

Resolving down the ramp:
5g Sin Ø - F = ma
5g x 3/5 - F= 5a
3g - F = 5a (1)

v=u+at t=3 u=0 v=4.9
a= 4.9/3

Putting a=4.9/3 into (1)
F= 3g -(5x4.9)/3
F= 21.23 R=4g

Finding coffiecient just plug and play into F=uR.
I Know this is right its in the book so dont bother working it out.
What I want help in now is ,can anyone tell me how to resolve this same question horizontally and vertically. By doing it this way is it even possible to get answer for the question?
This is my attempt at it:

Vertically:
R cosØ+ F SinØ = 5g
4/5R + 3/5F= 5g
4R+ 3F= 25g

Horizontally(this is what i cant figure out):
F cos Ø - R Sin Ø = 5a?
4F + 3R = 25a

I tried solving these two equations but cant get F=21.23 and R=4g.
All I want to know is can this question only be solved perpendicular and parallel? or am I just messing up the horizontal resolving.

lol longest post I've ever made be very grateful for anyones help.
Attached files
0
goku999
Badges: 3
Rep:
?
#2
Report 13 years ago
#2
im not sure but its unwise to solve it horizontally and vertically.
Only one of the forces is horizontal/vertical

But two forces are parallel and perpendicular to the plane.

It may be solvable but its prone to mistakes...


yes my 100th post....woohoo :eek:
0
Revelation
Badges: 1
Rep:
?
#3
Report 13 years ago
#3
hmm, I dunno looks too complicated atm (too much m1 revison today :P) but take a look at Chapter 4, Example 11. The example resolves the forces perpendicular/parallel to the plane and horizontally/vertically.
0
Revelation
Badges: 1
Rep:
?
#4
Report 13 years ago
#4
(Original post by mala2k)
Horizontally(this is what i cant figure out):
F cos Ø - R Sin Ø = 5a?
4F + 3R = 25a
Hmm, its not = the total F (5a) but isnt it a component of it?

F*cos Ø - R*Sin Ø = 5*Cos Ø*a ??
0
X
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of the Arts London
    MA Performance Design and Practice Open Day Postgraduate
    Thu, 24 Jan '19
  • Brunel University London
    Undergraduate Experience Days Undergraduate
    Sat, 26 Jan '19
  • SOAS University of London
    Postgraduate Open Day Postgraduate
    Sat, 26 Jan '19

Are you chained to your phone?

Yes (61)
19.93%
Yes, but I'm trying to cut back (121)
39.54%
Nope, not that interesting (124)
40.52%

Watched Threads

View All