Pulley system on rough surface.

[Buffu]

Homework Statement

[/B]
Question :- Find the acceleration of block of mass $M$. The coefficient of friction between blocks is $\mu_1$ and between block and ground is $\mu_2$.

free body diagram at the end.

Variables :-
$f_1$ - friction between blocks.
$f_2$ - friction between block and ground.
$N_2$ - Normal reaction between blocks.
$N_1$ - Normal reaction between block and ground.
$T$ - Tension in the string.
$a$ - acceleration of the blocks since string is in-extendable.
$M < m$

Homework Equations

Since small block will move down, and big one right.
$$ma = mg - (f_1 + T) \qquad (1)$$
$$T - (N_2 + f_2) = Ma \qquad (2)$$
$$N = Mg + f_1 \qquad (3)$$

If big block moves right then, pulleys will move right and small block will move right. Thus big block is non inertial frame. So we add the adequate pseudoforces to balance out unbalanced forces.

$N_2 = ma \qquad (4)$

The Attempt at a Solution

$(1) + (2)$

$$a = mg - f_1 - N_2 - f_2 $$

$$a = mg - N_2 - ( f_2 + f_1 )$$

$$a = mg - N_2 - ( \mu_1 N_2 + N_1 \mu_2 )$$

substituting $(3)$ and $(4)$

$$a(m + M) = mg - ma-(ma\mu_1 + \mu_2(Mg + f_1))$$
$$a(2m + M + \mu_1) = mg - \mu_2(Mg + ma\mu_1)$$
$$a(m(2 + \mu_1 + \mu_1 \mu_2) + M) = g(m - \mu_2 M)$$
$$a = {g(m - \mu_2 M)\over (m(2 + \mu_1 + \mu_1 \mu_2) + M)}$$

But the given answer is $a = {[2m - \mu_2(M + m)]g\over M + m[5 + 2(\mu_1 + \mu_2)]}$.
I think the given solution is incorrect, because i can't find any error in my calculations but i am not sure.
Please check my free body diagrams in the picture below and my equations. I think any error in those would be sufficient for me to proceed ahead.

My kindest Thanks for your time and help.

 

[TSny]

Is it correct to assume that the vertical acceleration of m equals the horizontal acceleration of M?

 

[haruspex]

N2 - Normal reaction between blocks.
N1- Normal reaction between block and ground.

Did you mean those the other way around? It's a bit inconsistent with the other suffixes.

a - acceleration of the blocks

Their horizontal accelerations are the same, clearly, but what about the vertical acceleration of m?

TSny beat me to it.

 

[TSny]

I believe you are missing some of the tension forces T that act on the big block M.

 

[Buffu]

Is it correct to assume that the vertical acceleration of m equals the horizontal acceleration of M?

I think vertical acceleration will also be $a$.

Let total length of string be $x$ and displacement of smaller block be $u$ and bigger block be $p$ in some time $t$.
So,

$$u + p = x$$
$${(du)^2\over dt^2} + {(dp)^2\over dt^2} = {(dx)^2\over dt^2}$$
$$\text{vertical acceleration of m} + \text{horizontal acceleration of M} = 0$$
$$\text{vertical acceleration of m} = -a$$
if we take magnitude, then $\text{vertical acceleration of m} = a$

May you please point my mistake ?

Did you mean those the other way around? It's a bit inconsistent with the other suffixes.

Their horizontal accelerations are the same, clearly, but what about the vertical acceleration of m?

TSny beat me to it.

Sorry for inconsistency. I wrote what i mean, did not mean it other way.

 

[haruspex]

Let total length of string be xxx and displacement of smaller block be uuu and bigger block be p

There are two horizontal strings that change length, but only one vertical.

 

[TSny]

Let total length of string be $x$ and displacement of smaller block be $u$ and bigger block be $p$ in some time $t$.
So,

$$u + p = x$$
$${(du)^2\over dt^2} + {(dp)^2\over dt^2} = {(dx)^2\over dt^2}$$
$$\text{vertical acceleration of m} + \text{horizontal acceleration of M} = 0$$

I don't follow this. The string can be broken into four sections as shown

 

[Buffu]

I don't follow this. The string can be broken into four sections as shown
View attachment 107810

So should it be like this :-

$$L_1 + L_2 + L_3 + L_4 = x$$
$${d(L_1)\over dt^2} + {d(L_2)\over dt^2} + {d(L_3)\over dt^2} + {d(L_4)\over dt^2} = {d(x)\over dt^2} $$
$${d(L_1)\over dt^2} + 0 + 2{d(L_3)\over dt^2} = 0$$
$$\text{vertical acceleration of m} + 2\text{horizontal acceleration of M} = 0$$
$$\text{vertical acceleration of m} = |2\text{horizontal acceleration of M}| $$
$$\text{vertical acceleration of m} = |2a| $$

 

[TSny]

Yes, that looks good. (Your notation for second derivatives is a bit off.)