質量為0.100之木塊放置於予與水面成37度角之斜面上

Let F be the applied force. Then
frictional force = (1/4) x normal reaction
but normal reaction = 0.1g.cos(37) N = 0.7827 N
where g is the acceleration due to gravity, taken to be 9.8 m/s^2
hence, frictional force = (1/4) x 0.7827 N = 0.1957 N

F + 0.1957 = 0.1g.sin(37)
F = (0.1g.sin(37) - 0.1957) N = 0.394 N