Mechanics M1 - Resolving Forces Q

Can anyone help me work through this M1 mechanics question?

A skier of mass 78kg is pulled at constant speed up a slope, of inclination 12 degrees by a force of magnitude 210N acting upwards at an angle of 20 degrees to the slope. Find the magnitudes of the frictional force and the normal contact force acting on the skier.

I've managed to get myself stupidly confused about it lol, so any help appreciated :biggrin:

Reply 1

magiccarpet

hey man im in my 1st yr of law at uni so not done physics for nearly 6 months but i did get an A.

im guessing the frictional force is the opposite to the tangential cotnact force.

downwards u have 9.81*78kg=765.18N
thecomponent of this acting opposite the 12 degrees force pulling him upwards is (90-12=) 78 degrees away from the downwards line.

to find this do: cos 78 * 765.18N
this is the frictional force
im just getting a calculator...

Reply 2

magiccarpet

think friction is 656.37N -210=446N

Reply 3

magiccarpet

hey again.
for the normal contact force, this is the force at right angles to the surface.to me this is 90-12=78 degrees from the horizontal normal. i think we need to find all the forces in that direction in order for the forces to balance (he is moving at steady speed so must be in equilibrium: newtons 1st law). im just gonna do that now.

Reply 4

magiccarpet

hey tsr's getting busy so check my profile for my email add, is u send me urs, i can email u the rest of the answer if i cant get onto tsr. here's a crap diagram for part 1

Reply 5

C4>O7

I don't think thats right. I get friction = 35N and N.C.F.=820N

Reply 6

James.R

Hi thanks a lot for all your help - it's becoming clearer :biggrin:

My email is in my profile too :biggrin:

Reply 7

magiccarpet

ok for the next bit

resolve all forces horizontally and vertically remembering its at steady speed so all forces as for equilibrium:

the 210N at 12 degrees has a vertical componenet of:
210 x cos78= -180.13865N

and a horizontal component of
210 x cos 78=177.209331N

plus dont forget the 765.18N downwards of gravity x mass, this has to be added to the vertical component of the contact force to balance out the man's weight
so the normal contact force is vertically:
765.18N+ (-180.13N)= 585.04N

and horizontally:
210 x cos 78=177.209331N

so the force at right angles to the surface is :
F cosH=177.209=F cos 78
rearranging
FcosH/cos H=F as the cos H's cancel

177.209/78 cos= -206N

vertically:
585.04/cos12=693.24

so the normal tangential contactforce is -206N+693.24N
=487N

(hopefully!)

Reply 8

magiccarpet

C4>O7

I don't think thats right. I get friction = 35N and N.C.F.=820N

oh yea, sorry, need to subtract 210 from my friction value

Reply 9

James.R

Thanks a lot for all your help :biggrin:

(rep on its way)

Reply 10

magiccarpet

C4>O7

I don't think thats right. I get friction = 35N and N.C.F.=820N

you might be right i can t really remember what im doing. however i think my working is sound and the friction must be greater than the pulling force of 210N, as if that force wasnt there, he'd slide back down.

Reply 11

C4>O7

Look:
Resolving // to slope:
210cos20-78gsin12-f=0
f=38.4N

Resolving _|_ to slope:
N.C.F. +210sin20-78gcos12=0
N.C.F.=676N

EDIT: I must have been tired last night!..

sihijd.bmp102.1KB

Reply 12

magiccarpet

hey james, what was the right answer?

Reply 13

Newton

Resolving vertically:

N+210sin20=78gcos12

=>N=78gcos12-210sin20=747.7-71.8=675.9N

Resolving perpendicular to the plane:

210cos20=F+78gsin20=>F=210cos20-78gsin12=197.34-158.93=38.41N

Newton.

Reply 14

magiccarpet

what did i do wrong?