Question

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?

Answer 1

The magnitude of the difference between their accelerations is `3.59 m/s^2`. The direction of A's acceleration is `30^@` below the direction of B's acceleration.

Explanation:

Assume block A is on the `60^o` incline and B is on the `30^@` incline. Let the forces that are the downslope components of the weight of the 2 blocks be called `F_(Ads) and F_(Bds)`. (By downslope component, I mean the component of the weight that is parallel with, and down, the slope.)

Those components have the values
`F_(Ads) = m*g*sin60^@`
`F_(Bds) = m*g*sin30^@`

Since the inclines are frictionless, the Newton's 2nd Law formulas for them are
Block A: `m*g*sin60^@ = m*a_A`
Block B: `m*g*sin30^@ = m*a_B`

Canceling the mass terms and solving both for their acceleration yields
`a_A = g*sin60^@ = 0.866*g`
`a_B = g*sin30^@ = 0.5*g`

The magnitude of the difference between their accelerations is
`0.366*g = 0.366*9.8 m/s^2 = 3.59 m/s^2`

Since both accelerate down their inclines, the direction of A's acceleration is `30^@` below the slope of B's acceleration.

I hope this helps,
Steve