Question

The coefficient of friction between the block m₁ and the inclined plane is µ. If $\frac { m _ { 1 } } { m _ { 2 } }$ = sin q (shown in figure) then the mass m₁ moves downwards with acceleration :

• A. µg cosq
• B. $\frac { \mu \mathrm { m } _ { 1 } } { \mathrm {~m} _ { 1 } + \mathrm { m } _ { 2 } }$ g cosq
• C. $\frac { - \mu \mathrm { m } _ { 2 } } { \mathrm {~m} _ { 1 } + \mathrm { m } _ { 2 } }$ gcosq
• D. $\frac { \mu } { g \cos \theta }$

Answer 1

Answer: (C)

$\frac { - \mu \mathrm { m } _ { 2 } } { \mathrm {~m} _ { 1 } + \mathrm { m } _ { 2 } }$ gcosq

Answer 2

m₁g–T = m₁a ...(1)

T – m₂g sin q – µm₂g cos q = m₂a _...(2)

From (1) + (2)

a = $\frac { m _ { 1 } g - m _ { 2 } g \sin \theta - \mu m _ { 2 } g \cos \theta } { m _ { 1 } + m _ { 2 } }$ = $\frac { \frac { m _ { 1 } } { m _ { 2 } } g - g \sin \theta - \mu g \cos \theta } { 1 + \frac { m _ { 1 } } { m _ { 2 } } }$