mechanics!!!!!

[burnt kayak]

help plz

[wooden orchid]

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[burnt kayak]

part 2

so b

[proven bison]

So am assuming u found tension and friction force?

[burnt kayak]

yes

[proven bison]

Ok u drew the FBD of 13 kg block?

[burnt kayak]

uhhh

dont think so

[proven bison]

Hmm draw all types of force action on 13kg block , what two forces u think would act in it

[proven bison]

also just ask if u don't understood anything

[burnt kayak]

sorry was playing roblox

hmm

the one going up

R

and weight

[proven bison]

Mhm true

What about sideways , any force in sideways?

[burnt kayak]

well if its on the floor i guess not

oh maybe friction idk

[proven bison]

Mhm ur close

It's on the floor that's true but there's a rope connected sideways , so u get tension on ryt side

and uk friction force always opposes the given force , so here in case friction force would act left side

[burnt kayak]

SORRY GOT DISTRACTED AGAIN

[burnt kayak]

but on b doesnt it reach the floor

[proven bison]

💀

[proven bison]

Ye so the ball reaches the floor , now the tension get eliminated but friction force still stays

[burnt kayak]

yes

[burnt kayak]

busy life ya know

my feet pics keep getting leaked

[proven bison]

Bru..

[proven bison]

Ok so u can get acceleration from friction force ryt?

[burnt kayak]

yes

[proven bison]

Ok then u can find distance with
two formula
V= u+at and s= ut+at²/2

Put v = 0 , u = 6 ,a= whatever u get

[burnt kayak]

oo got it

thx

can u help me with another question too

[proven bison]

Sure

[burnt kayak]

not very good at mechanics 😣

[proven bison]

Hmm I see , so it's a similar kind of question , just that circular mass is at inclined plane

Now acc to u , keeping that circular mass in inclined, what different u think would be in this question

Cuz only one thing has been changed , except everything is same

[proven bison]

Btw would appreciate if u reply bit quick

[burnt kayak]

sorrie

[burnt kayak]

the angle thing on p

[proven bison]

Hmm yea , so u gotta take component of mg parallel to inclined

So this case ur tension equal mgsin30 = mg/2

So yea this was the only thing different in this question, had to take component of weight since it's inclined ,
Rest of stuff is similar to previous question so u can do it

[burnt kayak]

okie dokie