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Grade: 12
        

4403-2121_3899_laws1.bmp


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.

10 years ago

Answers : (3)

Pratham Ashish
17 Points
							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 








						
10 years ago
Mihna
15 Points
							
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
one year ago
Diksha
15 Points
							
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
one year ago
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