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# 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.

Pratham Ashish
17 Points
11 years ago
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
15 Points
2 years ago
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
15 Points
2 years ago
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