# Two blocks A and B of equal masses are attached to a string passes over a smooth pulley fixed to a wedge as shown find the magnitude of acceleration of the center of mass of the two blocks when they are released from rest . neglect friction ?

## Jul 27, 2018

#### Answer:

$a = 3.6 \frac{m}{s} ^ 2$

#### Explanation:

I am going to add 3 words to the question:
"magnitude of the acceleration of the center of mass".

Assume that mass A is on the left and mass B is on the right. Define right to be positive direction (for force, and acceleration).

The component of each block's weight parallel with the respective incline:

${W}_{A} = - m \cdot g \cdot \sin {30}^{\circ}$

${W}_{B} = m \cdot g \cdot \sin {60}^{\circ}$

The weights are translated to upward pull on the string. Therefore, the net force acting on the each mass along the respective incline:
Mass A

${F}_{\text{netA}} = - {W}_{A} + {W}_{B} = - m \cdot g \cdot \sin {30}^{\circ} + m \cdot g \cdot \sin {60}^{\circ}$
${F}_{\text{netA}} = m \cdot g \cdot \left(- \sin {30}^{\circ} + \sin {60}^{\circ}\right)$
${F}_{\text{netA}} = m \cdot g \cdot \left(- 0.5 + 0.866\right) = 0.366 \cdot m \cdot g$

Mass B

${F}_{\text{netB}} = - {W}_{A} + {W}_{B} = - m \cdot g \cdot \sin {30}^{\circ} + m \cdot g \cdot \sin {60}^{\circ}$
${F}_{\text{netB}} = m \cdot g \cdot \left(- \sin {30}^{\circ} + \sin {60}^{\circ}\right)$
${F}_{\text{netB}} = m \cdot g \cdot \left(- 0.5 + 0.866\right) = 0.366 \cdot m \cdot g$

Acceleration of A:

${F}_{\text{netA}} = m \cdot {a}_{A}$
${a}_{A} = {F}_{\text{netA}} / m = \frac{0.366 \cdot \cancel{m} \cdot g}{\cancel{m}}$

Notice that if you go thru the same calculation for B, you get the same result. Therefore the center of mass will accelerate at

$a = 0.366 \cdot 9.8 \frac{m}{s} ^ 2 = 3.6 \frac{m}{s} ^ 2$ (2 sig figs)

I hope this helps,
Steve