- #1
rando25
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Homework Statement
Given: The friction between the block with mass 6 kg and the wedge with mass 18 kg is 0.27 . The surface between the wedge with mass 18 kg and the horizontal plane is smooth (without friction). The acceleration of gravity is 9.8 m/s^ 2. A block is released on the inclined plane (top side of the wedge). What is the minimum force F which must be exerted on the 18 kg block in order that the 6 kg block does not move down the plane?
Homework Equations
Fnet = ma
and possibly(?): Ff = uFn
The Attempt at a Solution
Having the wedge in the problem is confusing me.
The forces acting on the x-axis of the 18 kg block are F applied and Fg parallel
The forces acting on the x-axis of the 6kg block are F frictional and Fg parallel.
Fnet = ma / Fnet = 0
Fa + Fg(parallel) = 0
I think I am solving for Fa (18kg), I'm not sure.
Fa = Fgsin20
Fa = (18kg)(9.8m/s2)sin20
Fa = 60.33 N
I know the 6kg block plays a role, but i don't know how or where to incorporate it.
I'm really confused, any help is appreciated.