Solve Homework: Find Acceleration & Reaction Force of Wedge

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Homework Help Overview

The problem involves a wedge of mass 4 kg with an angle of 30 degrees, resting on a smooth horizontal table. A 1 kg mass is placed on the inclined face of the wedge, and a horizontal force of 10 N is applied towards the vertical face of the wedge. The objective is to find the acceleration of the wedge and the reaction force between the wedge and the mass.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss equations related to the forces acting on the wedge and the mass, including horizontal and vertical components. There are attempts to express the reaction force in terms of acceleration and to substitute variables in the equations. Questions arise about the meaning of certain variables, such as 'A', and the existence of a 'missing equation' that could clarify the problem.

Discussion Status

The discussion is ongoing, with participants providing insights and questioning each other's reasoning. Some guidance has been offered regarding the relationships between the variables, but there is no explicit consensus on the next steps or the correct interpretation of the equations involved.

Contextual Notes

Participants note potential confusion regarding the definitions of variables and the completeness of the equations used. There is mention of a non-inertial frame approach, suggesting different perspectives on how to analyze the problem.

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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