Grade 12Mechanics
Two masses A and B of 10 kg and 5 kg resp are connected as shown.The coeff of friction of A with table is 0.2. Find the min mass of C that may be placed on A to prevent it from moving.
3 Answers
Pratham Ashish
hi,
let the boxes are not moving ,
in this case, for block b you can write,
5g - T = 0
let the mass of block c be x
if the boxes are not moving
friction force on A = 0.2 * ( c + 10) g
for block A ,
T - friction Force = 0
T - 0.2 * ( c + 10) g = 0
from eq .(1)..
T = 5g
Eq. 2 becomes,
5g - 0.2 * ( c + 10) g = 0
5g - 0.2 c g - 2 g = 0
3g = 0.2 c g
c = 15 kg
Mihna
For limiting friction, myu= mb/ma+mc
0.2=5/10+ mc
mc=15 Kg
So the mass of c is 15kg
That is clear from the solution and think it would help you
Gud luck
Diksha
For limiting condition , u = mb/ma+mc
0.2 =5/10+mc
2+0.2mc=5
mc =15kg
So the minimum mass of C that may be placed on A to prevent it from moving is 15kg
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