Newton's laws homework: inclined plane with pulley

[akapm90]

Homework Statement

A wedge-shaped block of mass M is placed on a smooth plane inclined at an angle α to the horizontal. The block M has a vertical rectangular slot. Inside this slot, a smaller block of mass m is hanging connected to light, inextensible string, and passes over an ideal pulley fixed to block M, and fixed to the right-hand wall. Assume all surfaces are frictionless and the pulley and string are massless.

Calculate:
1- The magnitude of the the acceleration of block M
along the inclined plane.
2- The tension T in the string.

Relevant Equations

$$\sum F_{mx} = -F_i = ma_x$$
$$\sum F_{my} = T - mg = ma_y$$
$$\sum F_M = - Mg \times \sin \alpha + F_i \times \cos \alpha + T(\sin \alpha - \cos \alpha) = Ma_M$$
$$ a \times \cos \alpha = a_{mx}$$

I have this homework, I spend 3 days trying to solve it but I still don't get it.
The forces acting over M are the weight, the normal, the tension (up and down), and some internal force with mass m.
In m the weight, tension and some internal force with mass M.
Since there is a tension acting horizontally (at first) in M, it contributes to its acceleration. My question is; as M goes down, this tension won't be parallel to the ground anymore. Does this mean that acceleration is variable? Still if we assume that we want the acceleration at that instant, we still have the acceleration of m in the y-axis, and I can't relate this to the acceleration of M. With the acceleration of m in the x-axis we have that it's the same from M.

 

[TSny]

I'm wondering if the question is asking for the acceleration $a$ and tension $T$ as functions of time, or if it is asking only for the values of $a$ and $T$ immediately after the system is released (while the upper portion of the string is horizontal)?

 

[akapm90]

It doesn't specify after the system is released, I assume refers to that. But not in function of time, it asks for a specific formula of $$a_M$$ using $$M, m, \alpha$$

 

[TSny]

It doesn't specify after the system is released, I assume refers to that. But not in function of time, it asks for a specific formula of $$a_M$$ using $$M, m, \alpha$$

OK. I see now that you already brought up this ambiguity in your first post. Thanks.

 

[TSny]

Still if we assume that we want the acceleration at that instant, we still have the acceleration of m in the y-axis, and I can't relate this to the acceleration of M. With the acceleration of m in the x-axis we have that it's the same from M.

The pulley moves with the block $M$. If the block moves a distance $S$ up the slope, then the pulley moves parallel to the slope the same distance $S$. Let $l_{1,0}$ represent the initial length of the portion of the string from the pulley $P$ to the fixed point $A$. Let $l_{2,0}$ represent the initial length of the string from $P$ to the mass $m$. After the block moves up the slope a distance $S$, these lengths become $l_1$ and $l_2$. See diagram below. Try to find expressions for the $x$ and $y$ coordinates of $m$ in terms of $\alpha$, $S$, $l_{1,0},$ and the total length of the string $L$. Hint: Use a triangle to relate $l_1$ to $l_{1,0}$, $S$, and $\alpha$.

 

[Steve4Physics]

Still if we assume that we want the acceleration at that instant, we still have the acceleration of m in the y-axis, and I can't relate this to the acceleration of M. With the acceleration of m in the x-axis we have that it's the same from M.

In addition to what @TSny has suggested, you could try this...

Let $A_x$ and $A_y$ be the horizontal and vertical components of M’s acceleration. ($\frac {A_y}{A_x} = \tan \alpha$).

Let $a_x$ and $a_y$ be the horizontal and vertical components of m’s acceleration.

$A_x =a_x$ (because M and m move together in the x-direction due to the vertical slot in M)

At the instant when the upper section of the string is horizontal:
$a_y = -A_x$ (because one section of the fixed-length string is horizontal and the other section is vertical)
Edit: That's wrong because it doesn't consider the vertical acceleration of the pulley. See Posts #10 and beyond for correction.

The upshot is that we have four values, $A_x, A_y, a_x$ and $a_y$. Given any one of them, we can find the other three using the above relationships.

 

[kuruman]

If the block moves a distance $S$ up the slope, . . .

Why up the slope? There is no mention that the block is given an initial velocity, therefore we must assume that it starts from rest on a frictionless incline.

 

[TSny]

Why up the slope? There is no mention that the block is given an initial velocity, therefore we must assume that it starts from rest on a frictionless incline.

Even though it starts at rest, we can still consider the configuration after the block has moved some distance. Of course, the block might move down the slope instead of up the slope, but the sign of $S$ handles both directions. (The block might even remain at rest for certain values of the parameters $m$, $M$, and $\alpha$.)

 

[TSny]

At the instant when the upper section of the string is horizontal:
$a_y = -A_x$ (because one section of the fixed-length string is horizontal and the other section is vertical)

I'm not seeing this. If $\alpha = 0$, so the upper section of the string remains horizontal, then $a_y = -A_x$. But this is not true for general $\alpha$.

 

[TSny]

@Steve4Physics: Your post made me think of an easier way to see how $a_y$ is related to $A_x$ and $A_y$. Imagine that the block is lifted vertically upward so that it only has a vertical acceleration $A_y$. Consider how $a_y$ is related to $A_y$ when the string is still horizontal.

If the block is imagined to have only a horizontal acceleration $A_x$, then we have your relation between $a_y$ and $A_x$.

For the actual motion of the block along the slope, we have a superposition of these two cases.

This might be kind of hand-wavy, but it yields the same result as what I get by going through the analysis associated with post #5.

 

[Steve4Physics]

@Steve4Physics: Your post made me think of an easier way to see how $a_y$ is related to $A_x$ and $A_y$. Imagine that the block is lifted vertically upward so that it only has a vertical acceleration $A_y$. Consider how $a_y$ is related to $A_y$ when the string is still horizontal.

If the block is imagined to have only a horizontal acceleration $A_x$, then we have your relation between $a_y$ and $A_x$.

For the actual motion of the block along the slope, we have a superposition of these two cases.

This might be kind of hand-wavy, but it yields the same result as what I get by going through the analysis associated with post #5.

Yes, Thanks. I see my mistake in Post #6. I overlooked the (obvious!) fact that the pulley itself has a vertical acceleration of $A_y$.

I don't think your explanation is too hand-wavy!

m’s vertical acceleration is the vector sum of:
- the vertical acceleration of the pulley (A_y) and
- the accleration caused by the changing length of the 'upper section of the string' (USS)

With the USS horizontal, this gives $a_y = A_y - A_x$.

On a slightly different point, I believe that the author of the question was asking only about the case where the USS is horizontal. Analysis with a non-horizontal USS would require the distance between the wall and pulley to be defined – which is not part of the question. (A good approximation for a horizontal USS would be a setup with a very large distance between the pulley and the wall.)