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Two masses `A` and `B` of `10 kg` and `5 kg`, respectively , are connected with a string passing over a frictionless pulley fixed at the corner of a table as shown. The coefficient of static friction between `A` and the table is `0.2`. The minimum mass `C` that should be placed on `A` to prevent it from moving is equal to A. 15 kgB. 10 kgC. 5 kgD. 20 kg |
Answer» Correct Answer - A (a) ` mu (m_(A)+m_(C))g=m_(B)g` `therefore" " m_(C)=(m_(B))/(mu)=m_(A)=(5)/(0.2)-10=15 kg` |
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