Assume that friction between A and C is so high that relative motion will occur between them under no conditions. Now, draw the FBD for B ( a weight vector of magnitude 5g downwards and a tension vector of magnitude T upwards). Then draw the FBD for the combined mass A-C. This can be done explicitly by considering A and C separately and drawing FBD of each, assuming the frictional coefficient between them to be f. But for the sake of simplicity, since A and C have no relative motion, we consider them to be a single system with mass (10+m)kg where m kg, the mass of C, is to be found out. Now, draw the FBD for this single system, four vectors, viz. (10+m)g downwards, being balanced by reaction from table R upwards, frictional force 0.2R=0.2(10+m)g towards left and tension T towards right (since the rope is massless, tension is T everywhere). Clearly, for the limiting equilibrium of the system, we must have T=5g, and 0.2(10+m)g=T. Thus 0.2(10+m)g=5g, which means m=15. Thus mass of C has to be at least 15 kg.
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Kuldeep Guha Mazumder
·
1 year, 6 months ago

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Right .. I tried eliminating b by drawing a force vector of 50 N on the right. Then I tried Making equation but when i got to making an equation for C I couldn't find the friction between A and C
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T Sidharth
·
1 year, 7 months ago

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What have you tried? Where are you stuck in?
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Calvin Lin
Staff
·
1 year, 7 months ago

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TopNewestAssume that friction between A and C is so high that relative motion will occur between them under no conditions. Now, draw the FBD for B ( a weight vector of magnitude 5g downwards and a tension vector of magnitude T upwards). Then draw the FBD for the combined mass A-C. This can be done explicitly by considering A and C separately and drawing FBD of each, assuming the frictional coefficient between them to be f. But for the sake of simplicity, since A and C have no relative motion, we consider them to be a single system with mass (10+m)kg where m kg, the mass of C, is to be found out. Now, draw the FBD for this single system, four vectors, viz. (10+m)g downwards, being balanced by reaction from table R upwards, frictional force 0.2R=0.2(10+m)g towards left and tension T towards right (since the rope is massless, tension is T everywhere). Clearly, for the limiting equilibrium of the system, we must have T=5g, and 0.2(10+m)g=T. Thus 0.2(10+m)g=5g, which means m=15. Thus mass of C has to be at least 15 kg. – Kuldeep Guha Mazumder · 1 year, 6 months ago

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Right .. I tried eliminating b by drawing a force vector of 50 N on the right. Then I tried Making equation but when i got to making an equation for C I couldn't find the friction between A and C – T Sidharth · 1 year, 7 months ago

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What have you tried? Where are you stuck in? – Calvin Lin Staff · 1 year, 7 months ago

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