# M1 question from examination style paper

A sledge S of mass 20kg lies on an icy straight track inclined at 25degrees to the horizontal. The coefficient of friction between S and the track is 0.2. The position of S is maintained by a horizontal force of P newtons which acts in the vertical plane containing the track. The sledge S is in limiting equilibrium and on the point of moving down the track. Modelling the sledge as a particle obtain to 3 s.f. the value of P.

It's question 7 of the examination style paper in the back of the Edexcel M1 book and annoyingly I can't seem to get the right answer. Any help greatly appreciated!
Mussycat
A sledge S of mass 20kg lies on an icy straight track inclined at 25degrees to the horizontal. The coefficient of friction between S and the track is 0.2. The position of S is maintained by a horizontal force of P newtons which acts in the vertical plane containing the track. The sledge S is in limiting equilibrium and on the point of moving down the track. Modelling the sledge as a particle obtain to 3 s.f. the value of P.

It's question 7 of the examination style paper in the back of the Edexcel M1 book and annoyingly I can't seem to get the right answer. Any help greatly appreciated!

Friction up slope F=(N/5)

The 2 equations I get are:

Horiz: P= Nsin25 -(N/5).cos25 (1)
vert: 20g = N.cos25 + (N/5).sin25. (2)

Get expression for N from(2)...

N = 20g/(cos25 + (1/5).sin25)

... and put into (1)

Get P = 47.7

Aitch
Thank you! That's the right answer but what is erm N/5 (sorry if I'm being thick )
Aitch
Friction up slope F=(N/5)

The 2 equations I get are:

Horiz: P= Nsin25 -(N/5).cos25 (1)
vert: 20g = N.cos25 + (N/5).sin25. (2)

Get expression for N from(2)...

N = 20g/(cos25 + (1/5).sin25)

... and put into (1)

Get P = 47.7

Aitch

For part 2, resolve perpendicular to slope and along slope: (because you want the resultant along the slope)

Perp: N= 20g.cos25 (1)
Along slope: Resultant =20g.sin25 - F (2)
F = (N/5) up slope

Get N/5 from (1) and put into (2) as F

Get Resultant down slope = 47.306

F = ma => 47.306 = 20a
a = 2.4

Aitch
Mussycat
Thank you! That's the right answer but what is erm N/5 (sorry if I'm being thick )

The frictional force up the slope is mu times the normal reaction:

N * 0.2 = F

Aitch
Oh right, we always use R for the normal reaction for some reason Cheers!