Get Answers to all your Questions

header-bg qa

A force of 100N is just sufficient to pull a block of mass 10\sqrt{3} kg on rough horizontal surface. What is angle friction? (g=10m/s2)

Option: 1

30^{\circ}


Option: 2

45^{\circ}


Option: 3

60^{\circ}


Option: 4

50^{\circ}


Answers (1)

best_answer

Given-

Applied force,F=100N

Mass of the block, m=100\sqrt3 kg

As the Force F is just sufficient to pull the block, it must be equal to the limithing friction force.

F.B.D of the block-

\\ N=mg\\ f_{l}=F\\

Angle of friction (\theta) is defined as-

\\ tan\theta=\frac{f_{l}}{N}\\ \theta=tan^{-1}\frac{F}{N}=\frac{100}{100\sqrt3}\\ \\ \Rightarrow \theta=30^0

Posted by

chirag

View full answer

JEE Main high-scoring chapters and topics

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