How Is the Force Exerted by the Wedge on the Sphere Calculated?

  • Thread starter Thread starter Dtails
  • Start date Start date
  • Tags Tags
    Ball Wedge
AI Thread Summary
The discussion revolves around calculating the force exerted by a wedge on a solid sphere placed within it, focusing on a scenario with frictionless surfaces and specific angles. The participants explore the application of force equations and trigonometric relationships to determine the correct values for the forces acting on the sphere. Despite attempts to use vector resolution and triangle of forces, the calculations yield incorrect results, leading to frustration. The importance of correctly applying sine and cosine functions in relation to the angles involved is emphasized. Ultimately, the conversation highlights the challenges of solving equilibrium problems in physics, particularly when dealing with angles and forces.
Dtails
Messages
9
Reaction score
0

Homework Statement

A solid sphere of radius 1 m and mass 5 kg
is placed in a wedge with θ = 29 ◦
The inner surfaces of the wedge
are frictionless.
The acceleration of gravity is 9.8 m/s2 .
Determine the force exerted by the wedge
on the sphere at point B. Answer in units
Figure, ASCII style!
| /
|O/ <-b
v

point a is on the flat side towards the ball.

Edit: I realize that the ascii diagram may suck, but.

Homework Equations


Fnet=0

The Attempt at a Solution



Alright, so I tried this.
Fx=a+bcos(119)
Fy=bsin(119)-5*9.8.This shouldn't be as annoying as it is. I mean, by my method, B is easily given, but the answer i get is wrong. Which means I'm not seeing something. Ideas?
 
Physics news on Phys.org
If you draw the triangle of forces acting on the ball, there the the horizontal reaction A (towards the right), joined to the reaction B upwards and to the left, and the vertical force representing the weight of 5 kg. (5*9.8)

|<\
|**\B
|***\
v---->
...A

The * are just fillers to make a nicer triangle, ignore them.
The angle between A and B is 29 degrees.
So if you try to solve this right triangle, you will get the values of A and B.
 
Just tried the triangle, just gave me the same answer as the vector resolution. :/

5*9.8/sin61 = 56.024349 Which apparently is not the answer.

This is why it's annoying! The intuitive answers are always wrong!
 
Sin(\theta) = opposite / hypotenuse
Cos(\theta) = adjacent / hypotenuse
 
Well I hope I know that xP But yeah. From your triangle, the 29 degree shoots up to the top of the triangle, and the 61 becomes the most useful for a sin substitution. hence 5*9.8/b=sin61, so 5*9.8/sin61=b. :/ Not sure if I'm screwing up somewhere.
 
The (lousy) triangle I drew is a triangle of forces which has to close for equilibrium.
Force A is the normal (90degrees) to the vertical side, so it is horizontal.
Force B is the normal to the slanting side, so it makes 29 degrees with the horizontal.
Well, the vertical component is mg=5*9.8, which I am sure you know.
Want to give it another try, if you are solving for B?
 
Hah, I fail epiclly at Trig. xD Sin29 worked like a charm, on the last try, too. Thanks, man.
 
Back
Top