Engineering Statics

E-25


Two 8° wedges of negligible weight are used to move and position the 800 kg block. Knowing that the coefficient of static friction at all surfaces of contact is 0.30, determine the smallest force P that should be applied as shown to one of the wedges.

The FBD of the block and wedge are drawn separately.<BR> 
Examining the end of the force chain, a force we call H<sub>1</sub> 
is required to move the block.
Summing the forces in the y direction, we determine the vertical normal force.
Summing the forces in the x direction using the frictional force, which is 
a function of the vertical normal force, we determin the horizontal force, 
H<sub>1</sub>.
We now draw the free body diagram of the intermediate wedge with its forces.
From the free body diagram, we sum the forces in the x direction...
We simplify the equation...
We simplify further...
We isolate N<sub>1</sub> in terms of H<sub>2</sub>.
We simplify further...
From the free body diagram, we sum the forces in the y direction...
Again, we solve for N<sub>1</sub> in terms of H<sub>2</sub>...
We simplify...
And then we substitute solving for H<sub>2</sub> in newtons.
We create a free body diagram of the last wedge, with the applied force P.