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

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 3K views
rajumahtora
Messages
18
Reaction score
0
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
 

Attachments

  • paint2.jpg
    paint2.jpg
    7.4 KB · Views: 532
  • paint1.jpg
    paint1.jpg
    7.9 KB · Views: 705
  • paint.jpg
    paint.jpg
    13.6 KB · Views: 540
Physics news on Phys.org
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.