Question

A block of mass 'm' is placed on a smooth wedge of inclination \(theta\). The whole system is accelerated horizontally so that the block doesn't slip on the wedge. What is the magnitude of force exerted by the wedge on the block?

Answer 1

mgtan theta? o.O

Answer 2

no;
it is \[\frac{mg}{\cos \theta}.\]
o.O

Answer 3

but how ?

Answer 4

OH :o gotit

Answer 5

see

Answer 6

nice:)

Answer 7

i m ready:)

Answer 8

wt happened?

Answer 9

wait :(

Answer 10

ok.

Answer 11

=>f=ma
=>N=mgcos theta + masin theta
=>mgcos^2 theta+ mg sin^2 theta(whole upon)/cos theta
=>mg/cos theta

did you get anything o.O

Answer 12

I SUCK at explaining nvm

Answer 13

mg sin theta*

Answer 14

i also did it first time
but wt is this in 3rd step
mgcos^2 theta+ mg sin^2 theta(whole upon)/cos theta
how did u gt this?

Answer 15

however i understood the rest part.

Answer 16

mgcos^2 theta+ mg sin^2 theta(whole upon)/cos theta
how did u gt this?

Answer 17

wait !

Answer 18

k

Answer 19

idk o.o

Answer 20

o.O

Answer 21

sorry :|

Answer 22

np:)

Answer 23

see you could have done this too by resolving force and considering concept of pseudo force but it is better not to go through that...here just imagine why a block won't sleep...let's say it exerts no weight (that is mg) then there will be no slipping so in the above figure we can see that weight should be equal to
m*g= N cos theta
\[m*g= N \cos \theta ..... N= \frac{ mg }{ \cos \theta }\] and it is the normal force that is exerted by wedge on block... hope this helps you :)