Q. Two blocks of mass 1 kg are tied by a string and are at rest over a rough incline as shown in the figure.The left surface is rough and the right surface is smooth.Find the frictional force acting on the block:-
A) `5 N up`
B) `5(sqrt3 + 1) N up`
C) `5(sqrt3 - 1) N down`
D) `5sqrt3 N down`
See the figure below for a diagram of the forces.
The system is in equilibrium at rest. It means there is no acceleration on either mass.
On the right mass (mass on the smooth surface) there act tension in the string T upwards and parallel to incline component of the weight Gp1 downwards.
`T = m*g*sin(60)`
On the left mass (mass on the rough surface) there act tension on the string T upwards, friction force Ff downwards and parallel to incline component of the weight Gp2 downwards.
`T = miu*m*g*cos(30) +m*g*sin(30)`
where miu is the friction coefficient.
By eliminating the tension T between the above equations one has
`miu*m*g*cos(30) = m*g*(sin(60)-sin(30))`
Or written a bit different
`Ff = m*g*(sin(60)-sin(30)) = 1*10*(sqrt(3)/2 -1/2) =5*(sqrt(3) -1)`
Therefore correct answer is C) `Ff = 5*(sqrt(3)-1)` down