Help me with this mechanics problem

I tried drawing FBD and vector diagram but i just can get it to work

#Friction #Mechanics #Pulleys

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Comments

What have you tried? Where are you stuck in?

Calvin Lin Staff - 5 years, 6 months ago

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 - 5 years, 6 months ago

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.

Kuldeep Guha Mazumder - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$