"Figure 1 shows two wedge blocks A and B, weighing 1000 N and 50 N, respectively, in contact. The coefficient of friction between all contacting surfaces is 0.1 . What would be the smallest value of P required to push the block A upwards? What would be its value if there were no friction?"
Ans, with friction: 887.3 N (given)
Ans, without friction: 577.4 N (given)
The gray rectangle in the picture is part of a separate problem.
ƩF = 0
F_f = μF_N
The Attempt at a Solution
I have solved problems of this nature before. I have a good understanding of vector addition and statics. I solved the problem sans friction in the following manner:
F_N = W_A /cos30
ƩF_x = 0 ∴ F_N \cdot cos60 = P
P = W_A \cdot cos60 /cos30 = 577.4 N
With friction, however, I run into a problem: The side walls in contact with A exert a frictional force downwards. This of course depends on the normal force exerted by those walls, which partly balance the horizontal components of the normal and frictional forces exerted by B on A. But as far as I can tell, the vertical component of the normal force is equal to the sum of the weight and these frictional forces. So I encounter what appears to be a catch-22. I have tried various methods employing systems of equations but I get nowhere.