Questions

Q1) A man wants to slide down a block of mass mm which is kept on a fixed inclined plane of inclination 30o30^{o} as shown . Initially, the block is not sliding. to just start sliding, the man pushes the block down the incline with a force FF. 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 22.

a)what is the value of FF

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 FF, the acceleration would be

Note by Tanishq Varshney
4 years, 5 months ago

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6#8 6#8

&7# &7#

Kushal Patankar - 4 years, 5 months ago

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thanks @Kushal Patankar @Nishant Rai for the help

Tanishq Varshney - 4 years, 5 months ago

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Always ready to help you , friend.

Kushal Patankar - 4 years, 5 months ago

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@Kushal Patankar Can you guys help me here?

@Kushal Patankar @Nishant Rai @Tanishq Varshney

satvik pandey - 4 years, 5 months ago

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Tanishq I need your help here

Kushal Patankar - 4 years, 5 months ago

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Let coefficient of friction be ηs,ηk\eta_s , \eta_k then ηsNηkN=2 \frac{\eta_s N}{\eta_k N}=2 \rightarrow ηsηk=2\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+mgsinθ=ηsmgcosθ......(i) F+ mg \sin \theta = \eta_s mg \cos \theta ......(i) in case of just sliding downwards.

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

equating the first two equations, we get ηk=233\eta_k = \frac{2}{3\sqrt{3}}

On substituting the value of ηk\eta_k in eq (ii)(ii), we get mg6\frac{mg}{6}

Nishant Rai - 4 years, 5 months ago

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

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

d)d) ma=F+mgsinθηkmgcosθa=g6 ma = F + mg \sin \theta - \eta_k mg \cos \theta \rightarrow a = \frac{g}{6}

Nishant Rai - 4 years, 5 months ago

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

Tanishq Varshney - 4 years, 5 months ago

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

satvik pandey - 4 years, 5 months ago

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no problem, the above problem was sorted out

Tanishq Varshney - 4 years, 5 months ago

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

satvik pandey - 4 years, 5 months ago

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What is answer to F

Kushal Patankar - 4 years, 5 months ago

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mg6\frac{mg}{6}

Tanishq Varshney - 4 years, 5 months ago

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answers to b,c,db,c,d ?

Nishant Rai - 4 years, 5 months ago

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@Nishant Rai b) 7mg/6

c) 5mg/6

d) g/3

Tanishq Varshney - 4 years, 5 months ago

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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 - 4 years, 5 months ago

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i have a doubt. On solving the equations, we get ηs=433\eta_s = \frac{4}{3\sqrt{3}} which is greater than tanθ \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 tanθ<η \tan \theta < \eta .

Nishant Rai - 4 years, 5 months ago

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

Kushal Patankar - 4 years, 5 months ago

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