A body of mass m is placed on a rough surface with coefficient of friction `mu` inclined at `theta`. If the mass is in equilibrium , then

Options

A. `theta= tan^(-1)mu`
B. `theta= tan^(-1)((1)/(mu))`
C. `theta= tan^(-1) (m)/(mu)`
D. `theta=tan^(-1)(mu)/(m)`

Answer

A

Explanation

(d) Consider a body of mass m palced on a rough inclined surface and it is just on the point of sliding down, with coefficient of friction `mu` inclined at angle `theta`, as shown in figure.
At the equilibrium point O,
In case of limiting condition, `F=mg sin theta`......(i)
Normal force, `R= mg cos theta`...(ii)
Now, comparing Eqs. (i) and (ii), we get
`(F)/(R)=( mg sin theta)/( mg cos theta)(therefore " force of friction," F=mu R)`
`mu= tan theta implies theta= tan^(-1)(mu)`
This is maximum value of theta for mass m to be a rest. For samller `theta`, body will be at rest, i.e., in equilibrium.
So angle of repose, i.e. ` theta=tan^(-1)mu`