Normal Force Problem: Find Contact Force b/w Sphere, Wedge A & Wall B

AI Thread Summary
The discussion centers on calculating the normal contact force between a sphere, wedge A, and wall B in a frictionless scenario. The initial equations suggest that the normal force N equals mgcosθ, but this is challenged as incorrect. It is clarified that R, the force on wall B, has a component along N, which must be considered in the force balance. The correct relationship is established as R = Nsinθ, emphasizing the need for a free body diagram (FBD) to visualize the forces acting on the sphere. Ultimately, the resolution of forces reveals that N does not equal mgcosθ due to the influence of R's components.
rajumahtora
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Homework Statement



Find Normal Contact force between the sphere and wedge A and wall B.
All surfaces are frictionless and in rest.
Diagram :-
attachment.php?attachmentid=64277&stc=1&d=1385374828.jpg


2. The attempt at a solution

Let normal force on A be N and on B be R.
My Diagram :-
attachment.php?attachmentid=64278&stc=1&d=1385374984.jpg


By The Diagram,
N = mgcosθ ...eq(1)
R = Nsinθ ...eq(2)
= mgcosθsinθ

But the answer if :- N = (mg)/cosθ
it is not matching with eq(1).
How is this possible?
Please help me out and explain why it is wrong
 

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N is not equal to mgcosθ. There is a component of R along N. Draw a separate FBD for the sphere showing the three forces.
 
Pranav-Arora said:
N is not equal to mgcosθ. There is a component of R along N. Draw a separate FBD for the sphere showing the three forces.
But how N is not equal to mgcosθ and is R = Nsinθ correct component
 
rajumahtora said:
But how N is not equal to mgcosθ and is R = Nsinθ correct component

R=Nsinθ is correct. As I said before, there is a component of R along N so you will have to take that into account while balancing the forces in the direction normal to wedge. Do you see now?
 
Pranav-Arora said:
N is not equal to mgcosθ. There is a component of R along N. Draw a separate FBD for the sphere showing the three forces.
But how N is not equal to mgcosθ?
 
rajumahtora said:
But how N is not equal to mgcosθ?

R also has a component parallel to N.


Resolve N into horizontal and vertical components. The sum of all forces is zero, and that is valid to the horizontal and vertical components, separately.

ehild
 
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