Question

Neglecting friction every where, find the acceleration of M. Assume m > m'.

Answer 1

A = \frac{g (m - m')}{2M + m + m'}

Answer 2

The problem asks for the acceleration of the large block M, neglecting friction everywhere and assuming m > m'.

Let A be the magnitude of the acceleration of block M. Since m > m', the system on the right (involving mass m) will exert a larger pulling force than the system on the left (involving mass m'). Therefore, block M will accelerate to the left. We will assume A is positive for acceleration to the left.

Let's analyze the forces and kinematics for each part of the system using Newton's second law in the ground frame.

1. Right System (Mass m):

Let T_1 be the tension in the string.
Let a_1_rel be the acceleration of the block m (on the table) relative to block M, directed to the right. This also means the hanging mass m accelerates downwards with a_1_rel relative to M.

• For the hanging mass m:
  The acceleration of the hanging mass m is purely vertical. Its vertical motion is coupled to the horizontal motion of the block m (on table) relative to the pulley (which is on M). Thus, the absolute vertical acceleration of the hanging mass m is a_1_rel.
  Applying Newton's second law:
  mg - T_1 = m a_1_rel (Equation 1)
• For the block m (on table):
  The block m is accelerating to the right relative to M with a_1_rel. Since M is accelerating to the left with A, the absolute acceleration of the block m (on table) is a_1_rel - A to the right.
  Applying Newton's second law:
  T_1 = m (a_1_rel - A) (Equation 2)
  Adding Equation 1 and Equation 2:
  mg = m a_1_rel + m (a_1_rel - A)
mg = m (2 a_1_rel - A)
g = 2 a_1_rel - A
  From this, we find a_1_rel:
  a_1_rel = (g + A) / 2
  Substitute a_1_rel back into Equation 2 to find T_1:
  T_1 = m ((g + A) / 2 - A)
T_1 = m (g + A - 2A) / 2
T_1 = m (g - A) / 2 (Equation 3)

2. Left System (Mass m'):

Let T_2 be the tension in the string.
Let a_2_rel be the acceleration of the block m' (on the table) relative to block M, directed to the left. This also means the hanging mass m' accelerates downwards with a_2_rel relative to M.

• For the hanging mass m':
  The absolute vertical acceleration of the hanging mass m' is a_2_rel.
  Applying Newton's second law:
  m'g - T_2 = m' a_2_rel (Equation 4)
• For the block m' (on table):
  The block m' is accelerating to the left relative to M with a_2_rel. Since M is accelerating to the left with A, the absolute acceleration of the block m' (on table) is a_2_rel + A to the left.
  Applying Newton's second law:
  T_2 = m' (a_2_rel + A) (Equation 5)
  Adding Equation 4 and Equation 5:
  m'g = m' a_2_rel + m' (a_2_rel + A)
m'g = m' (2 a_2_rel + A)
g = 2 a_2_rel + A
  From this, we find a_2_rel:
  a_2_rel = (g - A) / 2
  Substitute a_2_rel back into Equation 5 to find T_2:
  T_2 = m' ((g - A) / 2 + A)
T_2 = m' (g - A + 2A) / 2
T_2 = m' (g + A) / 2 (Equation 6)

3. Block M:

The horizontal forces acting on block M are due to the tensions in the strings pulling the pulleys attached to M.
The tension T_1 pulls the right pulley to the left.
The tension T_2 pulls the left pulley to the right.
Since we assumed M accelerates to the left with A:
Applying Newton's second law for M:
T_1 - T_2 = M A (Equation 7)

Now, substitute the expressions for T_1 (Equation 3) and T_2 (Equation 6) into Equation 7:
m (g - A) / 2 - m' (g + A) / 2 = M A

Multiply the entire equation by 2 to clear the denominators:
m (g - A) - m' (g + A) = 2 M A
mg - mA - m'g - m'A = 2 M A

Rearrange the terms to solve for A:
mg - m'g = 2 M A + mA + m'A
g (m - m') = A (2M + m + m')

Finally, solve for A:
A = \frac{g (m - m')}{2M + m + m'}

Since m > m', the numerator (m - m') is positive, and the denominator (2M + m + m') is also positive. Thus, A is positive, confirming our initial assumption that block M accelerates to the left.