Question

# The coefficient of static friction between the block of 2 kg and the table shown in figure is μs = 0.2. what should be the maximum value of m so that the blocks do not move? Take g = 10 m/s2. The string and the pulley are light and smooth.

A

2 kg

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

1 kg

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

0.4 kg

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

0.5 kg

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is C 0.4 kg Consider the equilibrium of the block of mass m. The forces on this block are (a) mg downward by the earth and (b) T upward by the string. Hence, T-mg=0 or, T=mg. ...............(i) Now consider the equillibrium of the 2kg block.This forces on this block are (a) T towards right by the string. (b) f towards left (friction) by the table, (c) 20 N downward (weight) by the earth and (d) N upward (normal force) by the table, For vertical equilibrium of this block, N = 20 N. . . . . . . (II) As m is the largest mass which can be used without moving the system, the friction is limiting. Thus, f = μs N. . . . . . . (III) For horizontal equilibrium of the 2 kg block, f = T . . . . . . (IV) Using equations (I), (III) and (IV) μs N = mg Or, 0.2 × 20 N = mg Or, m=0.2×2010kg=0.4kg.

Suggest Corrections
0
Join BYJU'S Learning Program
Select...
Related Videos
Rubbing It In: The Basics of Friction
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program
Select...