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

1 Answer
Sep 23, 2017

The magnitude of the difference between their accelerations is 3.59ms2. The direction of A's acceleration is 30 below the direction of B's acceleration.

Explanation:

Assume block A is on the 60o 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 FAdsandFBds. (By downslope component, I mean the component of the weight that is parallel with, and down, the slope.)

Those components have the values
FAds=mgsin60
FBds=mgsin30

Since the inclines are frictionless, the Newton's 2nd Law formulas for them are
Block A: mgsin60=maA
Block B: mgsin30=maB

Canceling the mass terms and solving both for their acceleration yields
aA=gsin60=0.866g
aB=gsin30=0.5g

The magnitude of the difference between their accelerations is
0.366g=0.3669.8ms2=3.59ms2

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