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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!

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!

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

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

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

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