phyiscal question

1. Let F be the required force.
F.cos(25) + Ff = 8g.sin(25)
whee Ff is the frictional force, and g is the acceleration due to gravity

But, Ff = 0.3R, where R is the normal reaction
and R = F.sin(25) + 8g.cos(25)
Thus, F.cos(25) + 0.3[ F.sin(25) + 8g.cos(25)] = 8g.sin(25)
solve for F

2. Let the force be F'
we have, F'.cos(25) = 8g.sin(25) + Ff
i.e. F'.cos(25) = 8g.sin(25) + 0.3[ F'.sin(25) + 8g.cos(25)]
solve for F'

3. The frictional force now becomes kintic friction, use the coefficient of kinetic friction to replace the coefficient of static friction
F''.cos(25) = 8g.sin(25) + 0.2[ F''.sin(25) + 8g.cos(25)]

solve for F'', the new force




2010-01-02 00:32:37 補充：
Thanks to 子路who has reminded me. I have made a mistake in the calculation, as I have mis-read the question and thought that the force is acting horizontally,
In that case, just remove all the cos(25) from the equations.

2010-01-02 00:37:43 補充：
The equations ahould be
Q1: F + Ff = 8g.sin(25), and Ff = 0.3(8g.cos(25))
hence, F + 0.3[ 8g.cos(25) ] = 8g.sin(25)

2010-01-02 00:42:37 補充：
Q2: F' = 8g.sin(25) + Ff, and Ff = 0.3(8g.cos(25))
i.e. F' = 8g.sin(25) + 0.3[ 8g.cos(25)]
Q3: F'' = 8g.sin(25) + 0.2[ 8g.cos(25)]
Note that in Q1, friction is upward along the plane to prevent the block sliding down. In Q2, friction is acting downward to prevent the block from moving upward.

minimum force, parallel to the plane
still need to multiply by cos theta?