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Consider the situation as shown in the figure. Find the maximum value of m_(1)//m_(2) so that the system remains at rest. What will happen if tan theta lt mu? |
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Answer» Solution :Since we are interested in MAXIMUM value of `m_(1)//m_(2)`, equilibrium is limiting `f = f_(max) = MU N` For `m_(1) : N_(1) = m_(1) g cos THETA ,f_(1) = mu N_(1) = mu m_(1) g cos theta` For `m_(2) : N_(2) = m_(2) g , f_(2) = mu N_(2) = mu m_(2) g` For the system to be at rest `m_(1) g sin theta LE (f_(1) + f_(2))` `m_(1) g sin theta le (mu m_(1) g cos theta + mu m_(2)g)` `m_(1) (sin theta - mu cos theta) le mu m_(2)` `(m_(1))/(m_(2)) le (mu)/((sin theta - mu cos theta))` `((m_(1))/(m_(2)))_(max) = (mu)/( sin theta - mu cos theta)` If `tan theta lt mu` [i.e. `theta lt tan^(-1) mu`] `(sin theta)/( cos theta) lt mu rArr sin theta lt mu cos theta` ` ( sin theta - mu cos theta)` lt 0 i.e. `((m_(1))/(m_(2)))` is negative i.e. the system will not slide for any value of `m_(1)//m_(2) theta lt tan ^(-1) mu`. 
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