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
Messages
18
Reaction score
0

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
 

Attachments

  • paint.jpg
    paint.jpg
    7.7 KB · Views: 864
  • paint1.jpg
    paint1.jpg
    14.2 KB · Views: 822
Physics news on Phys.org
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
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top