Masses sliding on a smooth wedge

[Karol]

Homework Statement

[/B]Mass m lies on a Weighing scale which is on Wagon M. the inclined surface is smooth, between m and M there is enough friction to prevent m from moving.
1) What does the weigh show?
2) What is the minimum coefficient of friction between m and M to prevent slipping?
3) What is M's acceleration parallel to the slope if there is no friction between m and M? can it be bigger than $g\sin\alpha$?
4) What does the Weighing scale show in this case?

Homework Equations

Mass-acceleration: F=ma

The Attempt at a Solution

1) F=ma. $a_{\alpha}$ is parallel to the slope:
$$(M+m)g\sin\alpha=(M+m)a_{\alpha}~~\rightarrow~~a_\alpha=g\sin\alpha,~~a_y=a_\alpha\sin\alpha=g\sin^2\alpha$$
W is the weight, what the scale shows:
$$W=mg\sin\alpha$$

2)
$$a_x=a_\alpha\cos\alpha=g\sin\alpha\cos\alpha$$
$$f=ma_x:~mg\mu=mg\sin\alpha\cos\alpha~~\rightarrow~~\mu_{\rm min}=\sin\alpha\cos\alpha$$
3)
$$(M+m)g\sin\alpha=Ma_\alpha~~\rightarrow~~a_\alpha=\frac{M+m}{M}g\sin\alpha$$
Yes, it's bigger than $g\sin\alpha$
4)
$$a_y=a_\alpha\sin\alpha=\frac{M+m}{M}g\sin^2\alpha$$
$$W=ma_y=\frac{m}{M}(M+m)g\sin^2\alpha$$

 

[TSny]

One way to check an answer is to consider limiting cases. Does your answer for (1) give you a reasonable result for $\alpha$ going to $0$ or $\pi/2$?

 

[haruspex]

W is the weight, what the scale shows:

W=mgsinα​

Justify that claim. Consider the FBD of m.

 

[Karol]

One way to check an answer is to consider limiting cases. Does your answer for (1) give you a reasonable result for $\alpha$ going to $0$ or $\pi/2$?

$a_\alpha=g\sin\alpha$ seems reasonable since when α→0 a_α→0
The weight, as the scale shows:
$$mg-W=ma_y=mg\sin^2\alpha~~\rightarrow~~W=(1-\sin^2\alpha)mg$$

 

[haruspex]

$a_\alpha=g\sin\alpha$ seems reasonable since when α→0 a_α→0
The weight, as the scale shows:
$$mg-W=ma_y=mg\sin^2\alpha~~\rightarrow~~W=(1-\sin^2\alpha)mg$$

Right. Which simplifies to...?

 

[Karol]

1)
$$mg-W=ma_y=mg\sin^2\alpha~~\rightarrow~~W=(1-\sin^2\alpha)mg=mg\cos^2\alpha$$
4)
$$mg-W=ma_y=m\frac{M+m}{M}g\sin^2\alpha~~\rightarrow~~W=\left[ 1-\frac{M+m}{M}g\sin^2\alpha \right]mg$$

 

[TSny]

1)
$$mg-W=ma_y=mg\sin^2\alpha~~\rightarrow~~W=(1-\sin^2\alpha)mg=mg\cos^2\alpha$$

OK

4)
$$mg-W=ma_y=m\frac{M+m}{M}g\sin^2\alpha~~\rightarrow~~W=\left[ 1-\frac{M+m}{M}g\sin^2\alpha \right]mg$$

You have $a_y=\frac{M+m}{M}g\sin^2\alpha$. Does this behave properly as $\alpha \rightarrow \pi/2$?

You need to go back and rework parts (2), (3), and (4). Use separate FBD's for m and M.

 

[Karol]

You need to go back and rework parts (2), (3), and (4). Use separate FBD's for m and M.

The first equation is for M and the second for m. W is m's weight as M sees it, i.e. the force applied on M by m:
$$\left\{\begin{array}{l} (W+mg)\sin\alpha-W\mu\cos\alpha=Ma_\alpha \\ ma_\alpha\cos\alpha =W\mu \end{array}\right. ~~\rightarrow~~W=\frac{mMg\sin\alpha\cos\alpha}{(\mu\cos^2\alpha-\sin\alpha\cos\alpha)m+M\mu}$$
W becomes 0 when α=0, wrong. i expect it to be mg, am i right?

 

[PeroK]

View attachment 105776

The first equation is for M and the second for m. W is m's weight as M sees it, i.e. the force applied on M by m:
$$\left\{\begin{array}{l} (W+mg)\sin\alpha-W\mu\cos\alpha=Ma_\alpha \\ ma_\alpha\cos\alpha =W\mu \end{array}\right. ~~\rightarrow~~W=\frac{mMg\sin\alpha\cos\alpha}{(\mu\cos^2\alpha-\sin\alpha\cos\alpha)m+M\mu}$$
W becomes 0 when α=0, wrong. i expect it to be mg, am i right?

Why not try part 2). Once you get that, move on to parts 3 & 4.

Also, for 3 & 4, the equations of motion for m and M are coupled, so you may not be able to solve them independently.

 

[Karol]

Why not try part 2). Once you get that, move on to parts 3 & 4.

Post #8 is for part (2), but the answer doesn't comply with the boundary condition $\alpha=0$. i expect, for $\alpha=0$ that W=mg, but if i substitute $\alpha=0$ in my W, i get W=0

 

[PeroK]

Post #8 is for part (2), but the answer doesn't comply with the boundary condition $\alpha=0$. i expect, for $\alpha=0$ that W=mg, but if i substitute $\alpha=0$ in my W, i get W=0

I don't really understand post #8. From part 1 you have $W_1 = mgcos^2(\alpha)$

So:

a) What is the maximum frictional force between $m$ and $M$?

b) What horizontal frictional force between $m$ and $M$ is needed to sustain the motion without slipping?

 

[Karol]

The maximum friction force between m and M is $f=mg\mu\cos^2\alpha$. the horizontal friction force needed to sustain the motion is what i wrote in the OP:
$$a_x=a_\alpha\cos\alpha=g\sin\alpha\cos\alpha$$
$$f=ma_x,~~W\mu=ma_x,~~mg\mu\sin^2\alpha=mg\sin\alpha\cos\alpha~~\rightarrow~~\mu_{\rm min}=\frac{1}{\tan\alpha}$$

 

[PeroK]

The maximum friction force between m and M is $f=mg\mu\cos^2\alpha$. the horizontal friction force needed to sustain the motion is what i wrote in the OP:
$$a_x=a_\alpha\cos\alpha=g\sin\alpha\cos\alpha$$
$$f=ma_x,~~W\mu=ma_x,~~mg\mu\sin^2\alpha=mg\sin\alpha\cos\alpha~~\rightarrow~~\mu_{\rm min}=\frac{1}{\tan\alpha}$$

You've got the $sin^2$ and $cos^2$ mixed up. It's $\mu \ge tan \alpha$.

Anyway, I think 3 & 4 are quite tricky. You can see by energy considerations that $M$ has more energy in case 3 after both masses have fallen a given vertical height, as $m$ has no horizontal velocity in case 3, unlike case 1. Or, to put it another way:

The larger $m$ becomes, the more force it exerts on $M$ and the greater the acceleration of $M$. But, the greater the vertical acceleration, the less proportion of the normal weight of $m$ is felt by $M$.

Can you see how to use this to set up some equations for $a$ and $W_3$?

 

[Karol]

I made in post #1 for case 3:
$$(M+m)g\sin\alpha=Ma_\alpha~~\rightarrow~~a_\alpha=\frac{M+m}{M}g\sin\alpha$$
$$a_y=a_\alpha\sin\alpha=\frac{M+m}{M}g\sin^2\alpha$$
$$mg-W_3=ma_y~~\rightarrow~~W=(g-a_y)m=\left[ 1-\frac{M+m}{M}\sin^2\alpha \right]mg$$

 

[TSny]

In post #8 you assume correctly that both blocks have the same component of acceleration, $a_{\alpha}$, along the incline . Is this true for parts 3 and 4?

For these parts, in what direction must the two blocks have the same component of acceleration?

EDIT: Backing up to your first post, you wrote for part 3 that $(M+m)g\sin\alpha=Ma_\alpha$. Can you explain how you got the left side?

 

[Karol]

For these parts, in what direction must the two blocks have the same component of acceleration?

$a_y$ is the same for M and m. did you see my new, completed, post #14?

 

[TSny]

I made in post #1 for case 3:
$$(M+m)g\sin\alpha=Ma_\alpha~~\rightarrow~~a_\alpha=\frac{M+m}{M}g\sin\alpha$$

I don't see how you got the left side of the first equation.

$a_y$ is the same for M and m.

Yes.

 

[PeroK]

I made in post #1 for case 3:
$$(M+m)g\sin\alpha=Ma_\alpha~~\rightarrow~~a_\alpha=\frac{M+m}{M}g\sin\alpha$$

I see how you got this, but it's not right. You imagine that $m$ is pushing with its full, normal weight down on $M$. But, because they are both accelerating downwards, it should be:

$(M + W_3)$ instead of $(M + m)$

 

[Karol]

3) The first is FBD for M and the second is for m:
$$\left\{ \begin{array}{l} [W+Mg]\sin\alpha=Ma_\alpha \\ mg-W=ma_y=ma_\alpha\sin\alpha \end{array}\right.~~\rightarrow~~a_\alpha=\frac{(M+m)\sin\alpha}{M+m\sin^2\alpha}$$

 

[PeroK]

3) The first is FBD for M and the second is for m:
$$\left\{ \begin{array}{l} [W+Mg]\sin\alpha=Ma_\alpha \\ mg-W=ma_y=ma_\alpha\sin\alpha \end{array}\right.~~\rightarrow~~a_\alpha=\frac{(M+m)\sin\alpha}{M+m\sin^2\alpha}$$

If you add "g" somewhere, you've got it!

 

[Karol]

$$\left\{ \begin{array}{l} [W+Mg]\sin\alpha=Ma_\alpha \\ mg-W=ma_y=ma_\alpha\sin\alpha \end{array}\right.~~\rightarrow~~a_\alpha=\frac{(M+m)\sin\alpha}{M+m\sin^2\alpha}g$$

 

[TSny]

$$\left\{ \begin{array}{l} [W+Mg]\sin\alpha=Ma_\alpha \\ mg-W=ma_y=ma_\alpha\sin\alpha \end{array}\right.~~\rightarrow~~a_\alpha=\frac{(M+m)\sin\alpha}{M+m\sin^2\alpha}g$$

Looks right. It's interesting to consider the limiting case where m >> M and α → 0.
EDIT: The behavior of $a_\alpha$ for $\alpha$→ 0 is very different for M = 0 compared to M ≠ 0 with M << m. Graphing $a_\alpha$ vs $\alpha$ for various M/m values is useful. I think this has something to do with childhood memories of squeezing a slippery watermelon seed between thumb and forefinger to launch the seed across the room (not at anybody of course ).

 

[Karol]

The behavior of $a_\alpha$ for $\alpha$ → 0 is very different for M = 0 compared to M ≠ 0 with M << m. Graphing $a_\alpha$ vs $\alpha$ for various M/m values is useful

I will try and post here the graphs

 

[Karol]

all the graph's results must be multiplied by g, the gravity constant.
I don't understand why, for M=0, a_α tends to infinity. it must be a mistake of mine in the graph.

 

[PeroK]

https://www.physicsforums.com/attachments/106086 all the graph's results must be multiplied by g, the gravity constant.
I don't understand why, for M=0, a_α tends to infinity. it must be a mistake of mine in the graph.

$M \rightarrow 0$ is equivalent to $m >> M$

 

[Karol]

$M\rightarrow 0$ is equivalent to m>>M

Why the acceleration, parallel to the slope, as $\alpha\rightarrow 0$, tends to infinity? no matter what the sizes of m and M are they must stay in place, $a_\alpha\rightarrow 0$, since it is almost a flat table that they lie upon

 

[PeroK]

Why the acceleration, parallel to the slope, as $\alpha\rightarrow 0$, tends to infinity? no matter what the sizes of m and M are they must stay in place, $a_\alpha\rightarrow 0$, since it is almost a flat table that they lie upon

In the case $M \rightarrow 0$, $m$ is much larger and squeezes $M$ and shoots it out like the watermelon seed in post #22. The limit of this as $\alpha \rightarrow 0$ is 0 for any finite $m$ and $M$.

Taking $M = 0$ and $\alpha \rightarrow 0$ breaks the maths, which can happen.

 

[Karol]

Taking M=0 and α→0 breaks the maths, which can happen.

I don't understand much in mathematics but is here a problem? if M is very tiny, then, for M=0 there is a jump in the graph, from 0 to infinity. it's the first time i hear about it. as i have said, not that i heard about too much about mathematics

 

[PeroK]

I don't understand much in mathematics but is here a problem? if M is very tiny, then, for M=0 there is a jump in the graph, from 0 to infinity. it's the first time i hear about it. as i have said, not that i heard about too much about mathematics

Look, $M=0$ is simply not valid in the first place. How can you calculate the acceleration of something with zero mass?

 

[Karol]

M=0 is simply not valid in the first place. How can you calculate the acceleration of something with zero mass?

Thank you TSny and PeroK