Question

Two packages are connected by a massless string that goes over an ideal pulley as shown in the figure. The string acts on package 10kg parallel to the surface of the plane. The coefficient of static friction between package 10kg and the plane is 0.40. F is a force which can be applied on 10kg package

10kg 300 5kg Oke 30°

Question: How would you write the expression for frictional force? (Fs)

Answer 1

For 10 kg block

For equilibrium perpendicular to inclined plane

$R = mg.\cos 30^{0} + F.\sin 30^{0}$

For the equilibrium along parallel to inclined surface

$T +F\cos 30^{0} = mg\sin 30^{0} +F_{s}$

$F_{s} = T +F\cos 30^{0} - mg\sin 30^{0}$

For 5 kg block

$T = 5\times g= 5 kg \times 9.8 m/s^{2} = 49 N$

$F_{s} = 49 N +F\cos 30^{0} - (98 N)\sin 30^{0} = F\cos 30^{0}$