A system consists of Block A and B of masses 5kg 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=10m/s2).
A
60N
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
30N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
120N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
90N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A60N 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=50N The static friction between block A and B is f2s=μ2s(N1)=0.6(50)=30N
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=60N Hence, if the force F is above 60N, they will move.