Please help me solve this question!! i cant figure out what2 do!!! HELP!?

Question

Ive got a static friction question and have been working on it for hours and cannot get the correct answer! can you please explain how i do this? ive been doing most questions mathematically. but even 2 do it graphically id need to no the coefficient of static friction, which i don't! anyways, the q is:

A load of 20kg resting on an incline plane begins to slip when the plane is tilted to an angle of 25 degrees to the horizontal. What force parallel to the plane will be necessary to keep the load from slipping down the plane if the angle of the slope is increased to 35 degrees?

Please help me, i need to no this

Answer 1

http://www.youtube.com/watch?v=cIIwwCi2z...

Watch this maybe it will clear your mind and enable you to focus.

Answer 2

you'll begin to understand the problem if you draw a diagram of the situation and the forces in it.

just when the load begins to slip, the force of static friction (f) is equal to the component of the weight (W) along the slope.

W sin25 = f,

f = m*N  (where m = coefficient of static friction)

the normal force is equal to the component of W in its line,

N = W cos25

f = Wm cos25.

W sin25 = Wm cos25

m = tan25

    = 0.4663.

this is the coefficient of static friction.

now that you know the coeffiecient of static friction, you can find the force required to keep the block from slipping when the slope is 35 degrees. let F be the required force:

F + f = W sin35

F = W sin35 - f

  = W sin35 - Wm cos35

  = W (sin35 - mcos35)

  = W (sin35 - tan25*cos35)

  = W ( sin35 - sin25*cos35 / cos25 )

  = W (sin35*cos25 - cos35*sin25) / cos25

  = W sin(35-25) / cos25

  = W sin10/cos25

  = 3.8310 N

Answer 3

This means coeff of friction is tan25.

Now when the angle is 35, the froce down the plank is wsin35 and normal reaction is wcos35. Clearly

F - wsin35 = tan25(wcos35) where w = 20g and F is the reqd force.

Calculate now.