27 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.mere ik weird logic se 250 aa ra
but tht seems wrong
@Percy i might used a method with no sense but i am gettig 240
using the normal force c puts on b
which comes out to be (10+10)10 = 200
and adding our fric force
which between b and c is
0.2 20 *10 = 40
adding both is 240
what?
i mean i found out how to do it but am getting stuck in the calc
normal frce rxn to gravi
on b
wait i got 240 i am pretty sure this is the right method
lemme send
ahoww
its kinda hard to explain i will click a photo wait
ok
@Sephrina @Percy
good questions , challenges the concepts
i was stuck for a good 10 minutes
i didnt understood what u did
how the eqn came
wait
first of all
if they almost sliding
The top most block must be moving
because its friction cant handle much
so we use kinetic coeff
then if they are abt to slide b and c will apply equal and opp friction on each other
also the lower most block will apply friction on C
the friction due to 2nd and 3rd block cancel out
only The top most and bottom most remain
so those friction add up?
and the force must be equal to them?
not exactly
kinda of u can say that
ok
I am sorry but I still don't get it
for 2 block problems we take the free body diagrams of each individual block and then solve the question
if the blocks are sliding then we assume different acceleration for each block
why can't we use the same approach here
u can do that always no matter how many blocks
yes u can , but assuming accelaration will only add more equations
Could you please walk me through the force balance or equation setup step by step? I think I’m missing how the net forces are being combined.
first we apply force on B block
sliding will occur between A and B block , as friction is not sufficient to stop it
there will be kinetic friction
uk = 0.4
F = 100x0.4 = umg = 40
so 40 force in backward direction
now when Sliding is almost about to happen B will apply friction on C and C will also apply on B both equal and opp in magnitude
wait
nah i did a mistake mb
lemme solve again
i am not getting answer
ohh okay...lets see if anyone else responds
btw this is the solution given by PW....if you are able to understand it pls explain it to me
the wording in question makes it seem like they want condition so just that slipping starts
but in solution they assumed just after slipping
i thought about this too but it doesnt make sense to me phyiscally , first they said C will Slip over D
Taking the above 4 boxes as a system
the kinetic fricn btw C and D = 40
and force ahead = F
F-40 = 40a
now taking A and B as a system there acc should be same if slipping is just about to start
F-f = 30a
there will be static fricn btw AB and C , = 0.3
300 x 0.3 = 90
F - 90 = 30a
now u have two eqn and 2 variables
yeah thats exactly what I thought
only for C and D
ig in a real world scenario C and D will slip over first before sliding starts btw B and C
but i dont think there is a way to know that while solving question
yeah exactly
+solved @Gamertug @Sephrina
Post locked and archived successfully!
Archived by
<@1035556259417571408> (1035556259417571408)
Time
<t:1748144464:R>
Solved by
<@700658749416669194> (700658749416669194), <@888280831863451688> (888280831863451688)