Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Find the value of pushing force P just to move the block. If the coeffecient of friction between block and the floor is given as \mu = 0.6.

Answers (1)

Minimum Force to avoid sliding of body down on an inclined plane -

   

 


 

P=W \left[\frac{sin(\lambda-\theta)}{cos(\theta+\lambda)} \right ]

P = Pulling force 

\lambda= Angle of Inclination

\theta= Angle of friction

- wherein

For equilibrium 

R+P sin\alpha=W cos\lambda

P cos \alpha+F=W sin \lambda

Use F=\mu R

 

3 + mg sin \theta = P + \mu mg cos \theta

3 + \frac{100}{\sqrt{2}} = P + 0.6 \times \frac{100}{\sqrt{2}}

P = 32 N

Posted by

Satyajeet Kumar

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Trending Articles/News

anam-khan

JEE Main Result 2024 Live: JEE Mains session 2 results at jeemain.nta.ac.in soon; rank predictor
anam-khan

JEE Main 2024 session 2 final answer key out at jeemain.nta.ac.in; steps to check probable scores
anam-khan

JEE Main Result 2024 Live: Download session 2 final answer key at jeemain.nta.ac.in; rank predictor, cutoff

Latest Question