wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A system consists of Block A and B of masses 5 kg each and connected to a pulley as shown in figure. Find the minimum value of force F at which they will start moving? (Assume g=10 m/s2).


A
60 N
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
30 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
120 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
90 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 60 N
Let us find the forces acting on each block, so the FBDs of the blocks are as shown



From the FBDs we have
N1=5g=50 N
The static friction between block A and B is
f2s=μ2s(N1)=0.6(50)=30 N

Now, as per the question, the blocks just start moving, means, the friction will be limiting friction.

So, for block B:
T=f2s
For block A:
T+f2s=F
F=2f2s=2×30=60 N
Hence, if the force F is above 60 N, they will move.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Warming Up: Playing with Friction
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon