In the figure, block m1 is loaded on block m2. If there is no relative sliding between the blocks and the inclined plane is smooth, the frictional force acting between the blocks is:
A
μm2sinθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
μm1cosθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
μm1sinθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Zero
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D Zero Assuming no friction between the surfaces, acceleration of
block 1 is
Since, the blocks can only move along the inclined plane, so all forces ⊥ to inclined plane will be balanced for respective blocks.
Acceleration of both blocks down the inclined plane is:
a1=Net Forcemass=m1gsinθm1=gsinθ
Acceleration of block 2
a2=Net Forcemass=m2gsinθm2=gsinθ
⇒ relative acceleration between blocks 1 and 2 is: (a1−a2)=(gsinθ−gsinθ)=0
Hence, we can say that there is no tendency of slipping between blocks, as both will move with same acceleration down the incline. ∴frictional force f=0 between block1 and block2.