Questions

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

#Mechanics #JEEPhysics

Easy Math Editor

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Comments

6#8 6#8

&7# &7#

Kushal Patankar - 6 years, 1 month ago

thanks @Kushal Patankar @Nishant Rai for the help

Tanishq Varshney - 6 years, 1 month ago

Always ready to help you , friend.

Kushal Patankar - 6 years, 1 month ago

@Kushal Patankar – Can you guys help me here?

@Kushal Patankar @Nishant Rai @Tanishq Varshney

satvik pandey - 6 years, 1 month ago

Tanishq I need your help here

Kushal Patankar - 6 years, 1 month ago

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 - 6 years, 1 month ago

$b)$ $F - mg \sin \theta - \eta_s mg \cos \theta =0$

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

$d)$ $ma = F + mg \sin \theta - \eta_k mg \cos \theta \rightarrow a = \frac{g}{6}$

Nishant Rai - 6 years, 1 month ago

@Nishant Rai @Saurabh Patil @satvik pandey @Kushal Patankar plz help by posting solution

Tanishq Varshney - 6 years, 1 month ago

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

satvik pandey - 6 years, 1 month ago

no problem, the above problem was sorted out

Tanishq Varshney - 6 years, 1 month ago

@Tanishq Varshney – How is preparations going on, for ADVANCE?

satvik pandey - 6 years, 1 month ago

@Azhaghu Roopesh M @Raghav Vaidyanathan @Rohit Shah @Ronak Agarwal @Mvs Saketh

Tanishq Varshney - 6 years, 1 month ago

What is answer to F

Kushal Patankar - 6 years, 1 month ago

$\frac{mg}{6}$

Tanishq Varshney - 6 years, 1 month ago

answers to $b,c,d$ ?

Nishant Rai - 6 years, 1 month ago

@Nishant Rai – b) 7mg/6

c) 5mg/6

d) g/3

Tanishq Varshney - 6 years, 1 month ago

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 - 6 years, 1 month ago

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 - 6 years, 1 month ago

Do you think static friction will act at its maximum value.

Kushal Patankar - 6 years, 1 month ago

@Tanishq Varshney @Kushal Patankar

Nishant Rai - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$