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

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Homework Help Overview

The problem involves a solid sphere placed in a frictionless wedge, where participants are tasked with calculating the force exerted by the wedge on the sphere. The scenario includes specific parameters such as the sphere's radius, mass, and the angle of the wedge.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss drawing a triangle of forces to analyze the system, with attempts to resolve forces using trigonometric relationships. There are questions about the accuracy of their calculations and the interpretation of angles in the context of equilibrium.

Discussion Status

Several participants are actively engaging with the problem, sharing their attempts and results. There is a recognition of the challenges faced in reaching the correct answer, and some guidance has been offered regarding the setup of the force triangle. However, no consensus has been reached on the correct approach or solution.

Contextual Notes

Participants express frustration with their results, indicating that intuitive approaches have not yielded the expected outcomes. There is a focus on ensuring the triangle of forces closes for equilibrium, and the discussion includes various interpretations of the angles involved.

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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?
 
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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.
 

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