A sphere held steady on a slope by a rope

[Theexploer]

Homework Statement

- Calculate the intensities of the forces applied to the sphere. Choose the cOy coordinate system shown in the diagram.
- Calculate the radius of the sphere.

friction is non-existent.

Relevant Equations

A sphere of mass m = 150g of radius r = O'C = O'A is held in equilibrium on a perfectly smooth inclined plane by a wire AB of length l = 0.25m.
The angle between the inclined plane and the horizontal is a = 10*.
The angle between the inclined plane and the wire AB is Beta = 20*.

The sum of the forces should be 0.
Sin A'C'B' = px/b
px = mg . sin alpha
P should be px = - m.g. sin alpha and py = m.g.cos alpha
Finally i fund as result F = -0.8 and R = -1.23
but for the second question i didn't fund the radius of the circle.

 

[kuruman]

I think you need a better drawing. The one you show is deceptive in that it shows that the string is horizontal which would make $\beta=\alpha$ whilst it is given that $\beta=2\alpha.$ A drawing (more or less) to scale would be easier to analyze.

 

[Theexploer]

I think you need a better drawing. The one you show is deceptive in that it shows that the string is horizontal which would make $\beta=\alpha$ whilst it is given that $\beta=2\alpha.$ A drawing (more or less) to scale would be easier to analyze.

Unfortunately that's the only drawing I got on the book.

 

[Halc]

Drawing, however poor, is probably enough.
Getting R requires a demonstration that O' - A - B lie on a straight line (as drawn, but how do you prove that?), and then it's just algebra with quantities that are known.

The forces involved are irrelevant to the computation of radius.

 

[Theexploer]

Drawing, however poor, is probably enough.
Getting R requires a demonstration that O' - A - B lie on a straight line (as drawn, but how do you prove that?), and then it's just algebra with quantities that are known.

The forces involved are irrelevant to the computation of radius.

But the question asks for the intensities of the forces applied to the sphere which i'm not sure about my results, and to calculate by the given written informations find the radius of the cirlcle that's what i'm looking for.

 

[BvU]

Homework Statement: - Calculate the intensities of the forces applied to the sphere. Choose the cOy coordinate system shown in the diagram.
- Calculate the radius of the sphere.

friction is non-existent.
Relevant Equations: A sphere of mass m = 150g of radius r = O'C = O'A is held in equilibrium on a perfectly smooth inclined plane by a wire AB of length l = 0.25m.
The angle between the inclined plane and the horizontal is a = 10*.
The angle between the inclined plane and the wire AB is Beta = 20*.

The sum of the forces should be 0.
Sin A'C'B' = px/b
px = mg . sin alpha
P should be px = - m.g. sin alpha and py = m.g.cos alpha
Finally i fund as result F = -0.8 and R = -1.23
but for the second question i didn't fund the radius of the circle.

Hello again ,

Like in your other thread I am misssing things. What do the various symbols represent ? $px, \ py,\ P, \ b, \ ?$ (I suspect $P = p = m\vec g \$ (correct ?)

No idea if, similarly, $A = A'$ etc. Why confuse everybody ?

Sin A'C'B' = px/b

Do you mean $\sin(\angle \text {ABC})$ ?

With $\vec p_x = mg\sin\alpha$ you have the x component (along the incline) of the weight . It is compensated by the x-component of the pulling force $\vec T$

It took me a while to come up with this drawing.... (read $mg\cos\alpha$ instead of $mg\operatorname{scos}\alpha$ )

Easy to see $\sin \beta = R/(AB + R)$ -- no physics for part 1.

For part 2 you have $mg\sin\alpha = T \cos\beta$ with $T$ the tension from the rope.
Three forces act on the sphere: $mg, T$ and the normal force $N$ . Done .

$\$

 

[BvU]

Unfortunately that's the only drawing I got on the book.

That's not an argument. If I can draw this, so can you !

$\$

 

[Theexploer]

Hello again ,

Like in your other thread I am misssing things. What do the various symbols represent ? $px, \ py,\ P, \ b, \ ?$ (I suspect $P = p = m\vec g \$ (correct ?)

No idea if, similarly, $A = A'$ etc. Why confuse everybody ?


Do you mean $\sin(\angle \text {ABC})$ ?

With $\vec p_x = mg\sin\alpha$ you have the x component (along the incline) of the weight . It is compensated by the x-component of the pulling force $\vec T$

View attachment 340375

It took me a while to come up with this drawing.... (read $mg\cos\alpha$ instead of $mg\operatorname{scos}\alpha$ )

Easy to see $\sin \beta = R/(AB + R)$ -- no physics for part 1.

For part 2 you have $mg\sin\alpha = T \cos\beta$ with $T$ the tension from the rope.
Three forces act on the sphere: $mg, T$ and the normal force $N$ . Done .

$\$

Thank you very much for this detailed explanation