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## Homework Statement

A wedge of mass 4kg and angle 30 degrees is at rest on a smooth horizontal table. a mass of 1kg is placed on the smooth inclined face of the wedge and a horizontal force of 10N is applied to towards the vertical face of the wedge. Show that the acceleration of the wedge is

(40 - root(3g) ) / 17 ~= 1.35m/s.

Also find the reaction force between the wedge and the mass

## Homework Equations

## The Attempt at a Solution

(from diagrams)

Equation of wedge horizontally

4A = 10 - Rsin30 (1)

Equation of mass horizontally

A + acos30 = Rsin30

A = Rsin30 - acos30 (2)

Equation of mass vertically

asin30 = g - Rcos30 (3)

Substitute (2) into (1)

4(Rsin30 - acos30) = 10 - Rsin30

5Rsin30 - 4acos30 = 10

5Rsin30 = 10 + 4acos30

5R* 1/2 = 4a * root(3)/2 + 10

5R/2 = 2 root(3)a + 10

5R = 4root(3)a + 20

R = ((4root(3)a) / 5) + 4

Now I have 'R' in terms of 'a', but I'm not sure where to go. My notes on this trail off...I seem to substitute R back into equation (1), along with equation (2) but that will just give me R in terms of a again.