Q1) A man wants to slide down a block of mass \(m\) which is kept on a fixed inclined plane of inclination \(30^{o}\) as shown . Initially, the block is not sliding. to just start sliding, the man pushes the block down the incline with a force \(F\). Now the block starts accelerating. To move it downwards with constant speed, the man starts pulling the block with the same force. Surfaces are such that ratio of maximum static friction to kinetic friction is \(2\).

a)what is the value of \(F\)

b) If the man wants to move the block up the incline, what is minimum force required to just start the motion

c) What is the minimum force required to move it up the incline with constant speed

d) If the man continues pushing the block by force by force \(F\), the acceleration would be

## Comments

Sort by:

TopNewest6#8

&7#

Log in to reply

@Kushal Patankar @Nishant Rai for the help – Tanishq Varshney · 1 year, 8 months ago

thanksLog in to reply

– Kushal Patankar · 1 year, 8 months ago

Always ready to help you , friend.Log in to reply

here?

Can you guys help me@Kushal Patankar @Nishant Rai @Tanishq Varshney – Satvik Pandey · 1 year, 8 months ago

Log in to reply

here – Kushal Patankar · 1 year, 8 months ago

Tanishq I need your helpLog in to reply

Let coefficient of friction be \(\eta_s , \eta_k\) then \( \frac{\eta_s N}{\eta_k N}=2 \rightarrow \) \(\frac{\eta_s}{\eta_k} = 2 \)

for the block to just move downward, forces should balance each other along the incline.

i am just writing the equations, check if these are correct.

\( F+ mg \sin \theta = \eta_s mg \cos \theta ......(i) \) in case of just sliding downwards.

\( F= mg \sin \theta - \eta_k mg \cos \theta .....(ii) \) in case of pulling it upwards.

equating the first two equations, we get \(\eta_k = \frac{2}{3\sqrt{3}} \)

On substituting the value of \(\eta_k\) in eq \((ii)\), we get \(\frac{mg}{6}\) – Nishant Rai · 1 year, 8 months ago

Log in to reply

\(c)\) \( F - mg \sin \theta - \eta_k mg \cos \theta =0 \)

\(d)\) \( ma = F + mg \sin \theta - \eta_k mg \cos \theta \rightarrow a = \frac{g}{6} \) – Nishant Rai · 1 year, 8 months ago

Log in to reply

i have a doubt. On solving the equations, we get \(\eta_s = \frac{4}{3\sqrt{3}} \) which is greater than \( \tan \theta\) , which means the body will not move on its own down the inclined plane. Then how in the second case, the body starts accelerating downwards , and a pulling force is required to move the block with a constant speed? i mean the body cannot move on its own if \( \tan \theta < \eta \) . – Nishant Rai · 1 year, 8 months ago

Log in to reply

– Kushal Patankar · 1 year, 8 months ago

Do you think static friction will act at its maximum value.Log in to reply

@Tanishq Varshney @Kushal Patankar – Nishant Rai · 1 year, 8 months ago

Log in to reply

Tanishq Varshney Kushal Patankar Can You please provide the exact solutions for these problems ? Please Help.

example link 1

example link 2

example link 3 – Nishant Rai · 1 year, 8 months ago

Log in to reply

What is answer to

F– Kushal Patankar · 1 year, 8 months agoLog in to reply

– Tanishq Varshney · 1 year, 8 months ago

\(\frac{mg}{6}\)Log in to reply

– Nishant Rai · 1 year, 8 months ago

answers to \(b,c,d\) ?Log in to reply

c) 5mg/6

d) g/3 – Tanishq Varshney · 1 year, 8 months ago

Log in to reply

@Azhaghu Roopesh M @Raghav Vaidyanathan @Rohit Shah @Ronak Agarwal @Mvs Saketh – Tanishq Varshney · 1 year, 8 months ago

Log in to reply

@Nishant Rai @Saurabh Patil @satvik pandey @Kushal Patankar plz help by posting solution – Tanishq Varshney · 1 year, 8 months ago

Log in to reply

– Satvik Pandey · 1 year, 8 months ago

Looks I am late. :) There was some problem with internet connection that's why I was unable to help you.Log in to reply

– Tanishq Varshney · 1 year, 8 months ago

no problem, the above problem was sorted outLog in to reply

– Satvik Pandey · 1 year, 8 months ago

How is preparations going on, for ADVANCE?Log in to reply