Statics, equilibrium sphere in a groove on an incline plane

[J-dizzal]

Homework Statement

The smooth homogeneous sphere rests in the 132° groove and bears against the end plate, which is normal to the direction of the groove. Determine the angle θ, measured from the horizontal, for which the reaction on each side of the groove equals the force supported by the end plate.

Homework Equations

ΣF=0

The Attempt at a Solution

edit: in the second equation F₃sin(24) is supposed to be positive.

 

[haruspex]

Your hand-drawn diagram is wrong, and your equations are wrong for both diagrams.

This is very much a 3D problem. The printed diagram shows two different views of the same thing.
The black and white view is looking down the slope, showing what the groove looks like in cross section.
In the coloured view, the groove is seen side on, running up the slope from the end plate. The deepest line of the groove is shown by the lowest of the three sloping dashed lines.

There are four forces on the ball, the three reaction forces and gravity.

 

[J-dizzal]

Ok, i don't see how 45 degrees wouldn't work then. because the sum of the two forces of the wedge = the force on the vertical plate.
I will try writing out the sum of forces equations again.

 

[haruspex]

Ok, i don't see how 45 degrees wouldn't work then. because the sum of the two forces of the wedge = the force on the vertical plate.
I will try writing out the sum of forces equations again.

It doesn't say the sum of forces from the groove equals the force from the end plate. It says the force on each side of the groove. I take this to mean three forces of equal magnitude.

 

[J-dizzal]

It doesn't say the sum of forces from the groove equals the force from the end plate. It says the force on each side of the groove. I take this to mean three forces of equal magnitude.

yea but i tried 22.5 deg

 

[haruspex]

yea but i tried 22.5 deg

I don't see how my interpretation of the set-up leads to that angle.

 

[J-dizzal]

I don't see how my interpretation of the set-up leads to that angle.

F_c = F_a+F_b at 45 degrees
F_c=F_a=F_b at 45/2 degrees
this is just my intuition though.

 

[haruspex]

F_c = F_a+F_b at 45 degrees
F_c=F_a=F_b at 45/2 degrees
this is just my intuition though.

You can't add forces like that. Forces are vectors. These three forces act in three different directions.

 

[J-dizzal]

You can't add forces like that. Forces are vectors. These three forces act in three different directions.

I edited my sketch above. I still have too many variables.

 

[haruspex]

I edited my sketch above. I still have too many variables.

The very first equation there is wrong. You seem to be inconsistent over which way x and y point. The F1 term has x horizontal, while the W term (and the zero F2, F3 terms) has x parallel to the slope.
Similarly in the second equation, y is vertical for the F1 and W terms but normal to the slope for the F2 and F3 terms. You also have a sign error there.
You don't need any more equations. The actual values of the forces don't matter, and cannot be determined. All that matters is the ratios between them.

 

[J-dizzal]

im going to try changing my axis so that they are at the same angle as θ.

edit: cancel that
edit': ill change the side view to the same angle as theta and the front view will remain the same.

 

[J-dizzal]

The very first equation there is wrong. You seem to be inconsistent over which way x and y point. The F1 term has x horizontal, while the W term (and the zero F2, F3 terms) has x parallel to the slope.
Similarly in the second equation, y is vertical for the F1 and W terms but normal to the slope for the F2 and F3 terms. You also have a sign error there.
You don't need any more equations. The actual values of the forces don't matter, and cannot be determined. All that matters is the ratios between them.

Edited my sketch again.
Would it be beneficial to solving the problem if i wrote out each force in unit vector notation?

 

[haruspex]

Edited my sketch again.
Would it be beneficial to solving the problem if i wrote out each force in unit vector notation?

The equations look good now. Don't forget, you also know F1=F2=F3. They're quite easy to solve - no need to go into vector notation.

 

[J-dizzal]

i don't see how this is easy to solve.

 

[haruspex]

i don't see how this is easy to solve.

Look at your x and y equations. You know the three reaction forces are to have equal magnitudes, so just replace them all by F.
That leaves you with W as the unknown you need to eliminate, so use the equations to do that. What one equation do you now have?

 

[J-dizzal]

Look at your x and y equations. You know the three reaction forces are to have equal magnitudes, so just replace them all by F.
That leaves you with W as the unknown you need to eliminate, so use the equations to do that. What one equation do you now have?

w= (F+Fsin(24)+Fsin(24))/(sinθ + cosθ)

 

[haruspex]

w= (F+Fsin(24)+Fsin(24))/(sinθ + cosθ)

No, you still have W in there.
You have two equations involving F, W and theta. Combine them into one equation in a way which eliminates W.

 

[J-dizzal]

No, you still have W in there.
You have two equations involving F, W and theta. Combine them into one equation in a way which eliminates W.

I don't see how, because w is multiplied by two different trig functions?

 

[J-dizzal]

I don't see how, because w is multiplied by two different trig functions?

only thing i can come up with to eliminate w is to call is mg , but it makes m a variable

 

[haruspex]

I don't see how, because w is multiplied by two different trig functions?

It's just standard handling of simultaneous equations. Get one equation into the form W = (some function of the other variables), then use that to substitute for W in the other equation.

 

[J-dizzal]

It's just standard handling of simultaneous equations. Get one equation into the form W = (some function of the other variables), then use that to substitute for W in the other equation.

(F-1.8135F sinθ)/(sinθ+cosθ)=0 is that looking better?

edited

 

[J-dizzal]

It's just standard handling of simultaneous equations. Get one equation into the form W = (some function of the other variables), then use that to substitute for W in the other equation.

am i close?

 

[haruspex]

(F-1.8135F sinθ)/(sinθ+cosθ)=0 is that looking better?

edited

Doesn't seem quite right. Please don't plug in numbers like 1.8135, keep it all symbolic, or it makes it hard for to me to follow your working.

 

[J-dizzal]

i never had a cos(24), i had a sin(24) and i turned that into a decimal and mulitplied by 2, because there are 2 sin(24) terms.
Ill try again without decimal numbers

edit: w=F+2Fsin(24)/(-sinθ+cosθ)

 

[haruspex]

i never had a cos(24), i had a sin(24)

Yes, sorry, my mistake. But your equation at post #16 is wrong.
In the latest version of the image in the OP, you have, effectively:
$W \sin(\theta)=F$, $W\cos(\theta)=2F\sin(\alpha)$, where alpha = 24 degrees.

Start again from there, and please post working as typed text, not images.

 

[J-dizzal]

Yes, sorry, my mistake. But your equation at post #16 is wrong.
In the latest version of the image in the OP, you have, effectively:
$W \sin(\theta)=F$, $W\cos(\theta)=2F\sin(\alpha)$, where alpha = 24 degrees.

Start again from there, and please post working as typed text, not images.

θ=arcsin[-F((-sinθ+cosθ)/(F+2Fsinα))]
I need to get rid of F now.

edit: could i just set F=1?

 

[haruspex]

θ=arcsin[-F((-sinθ+cosθ)/(F+2Fsinα))]
I need to get rid of F now.

edit: could i just set F=1?

F just cancels out, doesn't matter what you set it to.
I think there's something wrong with your equation, though. You've made this unnecessarily complicated. Why won't you just do as I suggested: $W = \frac{2F\sin(\alpha)}{\cos(\theta)}$, then use that to replace W in the other equation.

 

[J-dizzal]

F just cancels out, doesn't matter what you set it to.
I think there's something wrong with your equation, though. You've made this unnecessarily complicated. Why won't you just do as I suggested: $W = \frac{2F\sin(\alpha)}{\cos(\theta)}$, then use that to replace W in the other equation.

Well i didnt see that until now.
im still getting an F in there though ; θ=arccos(Fsin(α))

edit I am plugging W into F-Wsin(θ)=0

 

[haruspex]

Well i didnt see that until now.
im still getting an F in there though ; θ=arccos(Fsin(α))

edit I am plugging W into F-Wsin(θ)=0

I can't tell where you are going wrong unless you post your working.

 

[J-dizzal]

I can't tell where you are going wrong unless you post your working.

plugging w into the first eq i get

F-(2Fsin(α))/cosθ)=0
-2Fsin(α)/cosθ=-F
2Fsin(α)=Fcosθ
θ=arccos(Fcosθ)​

edit

F-(2Fsin(α))/cos(θ))sin(θ)=0
(2Fsin(α)/cos(θ))sin(θ)=0
2Fsin(α)/cos(θ) = F/sin(θ)
2Fsin(α)=Fcos(θ)/sin(θ)
Fsin(α)=cos(θ)/sin(θ)
θ=arctan(1/Fsin(θ))​

Where does F cancel out?

edit. i tried doing another substitution for F, but its getting to complicated. i think I am missing something obvious but i don't see it.
Or i end up with cosθ=cosθ but doesn't get me anywhere toward solving for θ