Can You Solve This Newton's Law of Motion Problem?

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SUMMARY

This discussion focuses on solving a Newton's Law of Motion problem involving two blocks on a frictionless surface connected by a string over a massless pulley. The user initially attempted to derive the equations of motion using free body diagrams but encountered errors in their calculations. Key equations include force balance in both the X and Y axes for each block, specifically equations involving tension (T), mass (M and N), and gravitational acceleration (g = 9.8 m/s²). The discussion highlights the importance of incorporating the normal reaction force into the equations and suggests using coordinate systems aligned with the planes of motion for clarity.

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rajumahtora
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To Admin - This is not my homework, My teacher solved it but I am trying to solve this problem using other methods. If I posted this in wrong group, please message me and I will Post this in correct corner.
Problem - To find acceleration in both blocks sliding on smooth surface,string and pulley is frictionless and massless. Refer to diagram below.Use g = 9.8 m/s2
attachment.php?attachmentid=64116&stc=1&d=1384951351.jpg


My Solution -
Free Body Diagram Of Block of mass M kg -
attachment.php?attachmentid=64114&stc=1&d=1384951351.jpg


The Equation of forces in Y axis is
9.8M - Tsinα = Macosα ...eq(1)
The Equation of forces in X axis is
Tcosα = MAsinα ...eq(2)

Free Body Diagra of Block of Mass N kg -
attachment.php?attachmentid=64115&stc=1&d=1384951351.jpg


The Equation of forces in Y axis is
Tsinβ - 9.8N = Nasinβ ...eq(3)
The Equation of forces in X axis is
Tcosβ = Na.cosβ ...eq(4)

Now, after simultaneously solving eq(1) and eq(3), I am getting wrong answer
Now, eq(4) is tell that T = Na which is not possible.
I guess I missed Normal Reaction Force but how to use it in equation
 

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Yes, you definitely missed the Normal force. Add it to your diagram and equations like any other force.

Life would be much easier if you chose coordinates parallel to the planes rather than vertical and horizontal.
 
rajumahtora said:
The Equation of forces in Y axis is
9.8M - Tsinα = Macosα ...eq(1)
The Equation of forces in X axis is
Tcosα = MAsinα ...eq(2)

You have resolved the forces incorrectly.The normal force from the wedge will have a component in both x and y direction.
 

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