Question

A block of mass m slides down on rough inclined plane of angle x with the horizontal with an acceleration "a" along the plane the value of reaction is?

Answer 1

@experimentX @ghazi @dumbcow @mukushla

Answer 2

well you have to consider reaction of that mass at an instant ...so it will be = \[m*g*\cos \theta\] where theta is the inclination ....

Answer 3

okay wait

Answer 4

didn't see acceleration

Answer 5

The answer should be

\[m \sqrt{g^2 + ^2 - 2 a g sinx}\]

Answer 6

i think your answer needs modification

Answer 7

\[m g cosx\]

Answer 8

i got this...

Answer 9

Wat can be Done here,,,lol

Answer 10

well mgcosx is the answer when block is stationary ..it's moving so you've to resolve resultant acceleration first

Answer 11

oh......)) Can u show..that

Answer 12

yea wait...

Answer 13

Ok....

Answer 14

find R first by parallelogram law of vector addition and then use cos component of that R to multiply with m*g to find the reaction force

Answer 15

hope that helps you

Answer 16

theta is angle from horizontal = x

Answer 17

mg sinx = ma

sinx = a/g

Answer 18

i think your answer will be = \[m*g* (\sqrt{a^2+g^2\cos^{2}x})* \cos x\]

Answer 19

hmm..that's good ....sorry sometimes complex thinking never helps... :) good job

Answer 20

thxxx

Answer 21

:)