Revisiting the Feynman Ball on an Inclined Plane Problem

[John Helly]

Homework Statement

I was looking at the first problem in the Feynman Tips on Physics suggested problems. It appears to me that the solution offered is backwards in that the force on the wall Fw = tan(alpha) when it seems it should be Fw=sec(alpha). Can anyone suggest what I'm misunderstanding?

Relevant Equations

Fw = tan(alpha) * kg-wt
Fp = (1/cos(alpha)) * kg-wt

Fw = sec(alpha) * kg-wt
Fp = tan(alpha) * kg-wt

 

[haruspex]

Homework Statement: I was looking at the first problem in the Feynman Tips on Physics suggested problems. It appears to me that the solution offered is backwards in that the force on the wall Fw = tan(alpha) when it seems it should be Fw=sec(alpha). Can anyone suggest what I'm misunderstanding?
Relevant Equations: Fw = tan(alpha) * kg-wt
Fp = (1/cos(alpha)) * kg-wt

Fw = sec(alpha) * kg-wt
Fp = tan(alpha) * kg-wt

We don’t all have a copy of that work, useful as it may be. Please state the question.

 

[John Helly]

Right. Sorry. Here it is.

1.1 A ball of radius 3.0 cm and weight 1.00 kg rests on a plane tilted at an angle alpha with the horizontal and also touches a vertical wall. Both surfaces have negligible friction. Find the force with which the ball presses on each plane.

Fw is the force on the wall. Fp is the force on the plane.

 

[kuruman]

Why did you write
Fw = tan(alpha) * kg-wt ?
Fw is a horizontal force because the wall is vertical and can only exert a force perpendicular to its surface.

Please post the free body diagram that you used to write your equations.

 

[John Helly]

That's what is given as an answer in the solutions. I don't think it's correct. I think the answer is backwards.

I think the horizontal force on the vertical wall should be Fw=(1/cos(alpha)) * kg-wt. Not Fw=tan(alpha)*kg-wt.

It sounds like you agree. I'll try to draw a decent diagram and post it if this is not clear.

 

[TSny]

I think the horizontal force on the vertical wall should be Fw=(1/cos(alpha)) * kg-wt.

What does this expression predict for F_w when $\alpha = 0$? Does that seem right?

 

[kuruman]

What does this expression predict for F_w when $\alpha = 0$? Does that seem right?

I just realized that the "-" in Fw=(1/cos(alpha)) * kg-wt. is not a "minus" sign but a "times" sign. Nevertheless, the expression is still wrong for the reason you mentioned in #6.

 

[John Helly]

Good point. So it looks like the original published answer is correct.

 

[TSny]

Good point. So it looks like the original published answer is correct.

Yes, the published answer is correct. If you would like to discover why you weren't getting the right answer, post your attempt along with a good free-body diagram for the ball.

 

[John Helly]

Will do.

 

[TSny]

I just realized that the "-" in Fw=(1/cos(alpha)) * kg-wt. is not a "minus" sign but a "times" sign.

The kg-wt ("kilogram-weight") is a unit of force defined as the force that will cause a 1 kg mass to accelerate with the acceleration due to gravity at the surface of the earth. This unit is also called the kilogram-force.

 

[kuruman]

Yes, of course. How silly of me not to think of the unit context in an algebraic expression without numerical values.

 

[John Helly]

Here's what I came up with. I can see it when it rotates down but still seems kind of odd.

 

[kuruman]

This diagram is confusing. I suggest that you redraw it showing the three forces that act on the ball only and find expressions for them. Then use Newton's 3rd law to find the force that the ball exats on the wall and on the plane.

 

[John Helly]

Ok.

 

[John Helly]

Better?

 

[John Helly]

This seems to make sense.

 

[PeroK]

View attachment 328971
Better?

Two of those forces are forces exerted by the ball on the surfaces. The forces on the ball are in the opposite direction. As the ball is static, the total force on the ball must be zero - which is the key point.

 

[John Helly]

Mahalo. I was trying to interpret it geometrically and could't get the directions right.

 

[haruspex]

Mahalo. I was trying to interpret it geometrically and could't get the directions right.

Is the wall pulling the ball towards itself or pushing back on it?
Same question for the ramp.
What are the two vertically upward arrows?

 

[kuruman]

There are three entities exerting forces on the sphere

1. Earth
2. Wall
3. Inclined plane

Look at the figure below which is the beginning of a free body diagram for the sphere as your system. You are asked to finish drawing it. The wall and the inclined plane are not part of the system so that forces that the sphere exerts on them are not to be included.

The first step is to draw one and only one arrow representing the force that each entity exerts on the sphere. There should be 3 arrows with their tails at the point where the force is exerted. You may assume that the "Earth" arrow has its tail at the center of the sphere. Go for it.

 

[John Helly]

Ok. Correct?

 

[kuruman]

That's the diagram but not the equation. Why should the weight be equal to the sum of the other forces?

Newton's second law says that the net force is equal to mass times acceleration. Here the acceleration is zero. The net force is the vector sum of the three forces. You have to find that and set its horizontal and vertical components equal to zero.

 

[John Helly]

Right. More head scratching.

 

[kuruman]

Once you're done scratching, you may consider figuring out the horizontal and vertical components of each vector in order to find algebraic expressions for the horizontal and vertical components of the net force.

 

[John Helly]

Will do.

 

[PeroK]

Will do.

Newton's second law can be written in vector form:$$\vec F = m\vec a$$This is actually shorthand for three equations, one for each spatial dimension:$$F_x = ma_x, \ F_y = ma_y, \ F_z = ma_z$$In this (two dimensional) problem, this reduces to two equations that must be satisfied. It's common to take $x$ as the horizontal direction and $y$ as the vertical. So, you have two equations to balance.

The next thing I would do is show the horizontal and vertical components of the force you've labelled $F_p$.

 

[John Helly]

Mahalo.

 

[John Helly]

Mahalo.

Well, this is what I came up with and it's obviously wrong but I don't know why.

 

[TSny]

You're pretty close. Your forces in the diagram look good. But you need to be careful with the signs of the x and y components of the forces. Usually, we take the positive x-direction to be to the right and the positive-y direction to be upward. If so, then what should be the sign of the x-component of $\vec {F}_p$? What should be the sign of the y-component of $\vec {F}_g$?

 

[kuruman]

I concur with @TSny. For future reference, when the sum of any number of vectors is zero and you add them using the graphical "tip-to-tail" graphical construction, you end with a closed polygon. The number of sides is equal to the number of forces that are added.

Here you have a right triangle because there are 3 forces and $F_w$ and $F_g$ are perpendicular (see below). All you have to do is figure out which of the interior acute angles should be labeled $\alpha$ and then write down the usual trig relations.