Get Answers to all your Questions

header-bg qa

A block of mass \sqrt{3} kg is kept on a rough surface having coefficient of static friction \frac{1}{}{2\sqrt{3}}. What is the minimum value of F (in N) such that the block shown in arrangement just start moving?

Option: 1

20


Option: 2

10


Option: 3

12


Option: 4

15


Answers (1)

best_answer

Given,

mass of the block, m=\sqrt{3}kg

Coefficient of static friction, \mu=\frac{1}{2\sqrt3}

F.B.D of the block-

\\ N=mg+Fsin60^0

limiting friction-

\\ f=\mu N\\ \Rightarrow f=\mu(mg+\frac{\sqrt{3}F}{2})\ ...(1)

For the minimum value of F, the block will just begin to move unaccelerated.

\\ f=Fcos60^0\ ...(2)

Comparing equation (1) and equation (2)-

\\ \frac{F}{2}=\mu(mg+\frac{\sqrt{3}F}{2})\\\\ \Rightarrow F=\frac{2\mu mg}{1-\sqrt{3}\mu}\\

Substituting the values-

F=20N

Posted by

Kuldeep Maurya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE