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.

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

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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.

Log in to reply

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

Log in to reply

What have you tried? Where are you stuck in?

Log in to reply