# There are two separate fixed frictionless inclined planes making angle 60^o and 30^o with the horizontal. Two blocks A and B are kept on both of them. Find the relative acceleration of A with respect to B?

Sep 23, 2017

The magnitude of the difference between their accelerations is $3.59 \frac{m}{s} ^ 2$. The direction of A's acceleration is ${30}^{\circ}$ below the direction of B's acceleration.

#### Explanation:

Assume block A is on the ${60}^{o}$ incline and B is on the ${30}^{\circ}$ incline. Let the forces that are the downslope components of the weight of the 2 blocks be called ${F}_{A \mathrm{ds}} \mathmr{and} {F}_{B \mathrm{ds}}$. (By downslope component, I mean the component of the weight that is parallel with, and down, the slope.)

Those components have the values
${F}_{A \mathrm{ds}} = m \cdot g \cdot \sin {60}^{\circ}$
${F}_{B \mathrm{ds}} = m \cdot g \cdot \sin {30}^{\circ}$

Since the inclines are frictionless, the Newton's 2nd Law formulas for them are
Block A: $m \cdot g \cdot \sin {60}^{\circ} = m \cdot {a}_{A}$
Block B: $m \cdot g \cdot \sin {30}^{\circ} = m \cdot {a}_{B}$

Canceling the mass terms and solving both for their acceleration yields
${a}_{A} = g \cdot \sin {60}^{\circ} = 0.866 \cdot g$
${a}_{B} = g \cdot \sin {30}^{\circ} = 0.5 \cdot g$

The magnitude of the difference between their accelerations is
$0.366 \cdot g = 0.366 \cdot 9.8 \frac{m}{s} ^ 2 = 3.59 \frac{m}{s} ^ 2$

Since both accelerate down their inclines, the direction of A's acceleration is ${30}^{\circ}$ below the slope of B's acceleration.

I hope this helps,
Steve