A 3kg block is pulled by a force which is inclined at 37∘ to the horizontal table. The friction coefficient between the table and block is 1/3. For what minimum value of this force, will the block start sliding?
A
5N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
10N
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
20N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
25N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B10N Free body diagram of the given block can be given as
If we have calculate the minimum force to just slide Fmincos37∘=fs
where fs is the static friction ⇒Fmincos37∘=μN ....(i)
Now balancing the vertical force we get N+Fminsin37∘=mg⇒N+35Fmin=30⇒N=30−35Fmin
Now putting this in eq(i) 45Fmin=13(30−35Fmin
On solving this we get Fmin=10N