| Two masses A and B of 10 kg and 5 kg are connected with a string passing over a frictionless pulley fixed at the corner of a table (as shown in figure). The coefficient of friction between the table and block is 0.2. The minimum mass of C that maybe placed on A to prevent it from moving is equal to |
 |
A) 15 kg
B) 10 kg
C) 5 kg
D) zero
Correct Answer: A
Solution :
| Let T be the tension in the string, |
| f = frictional force between block A and table, |
| m = minimum mass of C. |
| For the just motion of block A on table |
| \[T=\,f=\mu R=\,\mu (m+m')\,g\] |
| = 0.2(10+m')g ... (i) |
| For the just motion of block B |
| T = 5g ... (ii) |
| From Eqs.(i) and (ii), we get |
| \[5g=0.2\text{ }\left( 10+m' \right)\text{ }g\] |
| \[M'=\left( 52 \right)/0.2\text{ }=15\text{ }kg.\] |
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