Solve Homework: Find Acceleration & Reaction Force of Wedge

AI Thread Summary
The discussion focuses on solving a physics problem involving a wedge and a mass, with a specific emphasis on calculating the acceleration of the wedge and the reaction force between the wedge and the mass. The wedge has a mass of 4 kg and is inclined at 30 degrees, while a 1 kg mass rests on it, with a horizontal force of 10 N applied. Participants derive equations related to the forces acting on both the wedge and the mass, leading to the expression for acceleration. There is confusion regarding the variable 'A', which represents the acceleration of the wedge, and the need for additional equations to solve the problem effectively. The conversation highlights the importance of correctly identifying all forces and equations to arrive at the solution.
Bucky
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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.
 

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it seems you missed a equation.

you can achieve a=2g-root(3)R from
asin30 = g - Rcos30 (3)

then subtitute it to
5R = 4root(3)a + 20
it can give you R without a

besides, i really cann't understand then method you used in this problem and so as what the A stand for.i suggest you consider more ways.
 
enricfemi said:
it seems you missed a equation.

you can achieve a=2g-root(3)R from
asin30 = g - Rcos30 (3)

then subtitute it to
5R = 4root(3)a + 20
it can give you R without a

besides, i really cann't understand then method you used in this problem and so as what the A stand for.i suggest you consider more ways.

where does the 'missing equation' come from?
I can't actually recall why A is there, though I'm sure i need it :/ could it be the force of the mass pushing the wedge back?
 
Bucky said:
where does the 'missing equation' come from?
I can't actually recall why A is there, though I'm sure i need it :/ could it be the force of the mass pushing the wedge back?

:-p

you missed equation (3)

just tell me what the A stand for in the picture in order to let me understand it in your way.
 
sorry for the delay in replying.

A represents the acceleration of the wedge.
 
really sorry for so many days' delay.

in my peasonal sense,non-inertial frame will greatly predigest the problem.just use the wedge as frame.
in addition,you missed a force on wedge in the maps you gave.

happy new day!
 
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